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## Solving Systems of Equations Real World Problems

Wow! You have learned many different strategies for solving systems of equations! First we started with Graphing Systems of Equations . Then we moved onto solving systems using the Substitution Method . In our last lesson we used the Linear Combinations or Addition Method to solve systems of equations.

Now we are ready to apply these strategies to solve real world problems! Are you ready? First let's look at some guidelines for solving real world problems and then we'll look at a few examples.

## Steps For Solving Real World Problems

- Highlight the important information in the problem that will help write two equations.
- Define your variables
- Write two equations
- Use one of the methods for solving systems of equations to solve.
- Check your answers by substituting your ordered pair into the original equations.
- Answer the questions in the real world problems. Always write your answer in complete sentences!

Ok... let's look at a few examples. Follow along with me. (Having a calculator will make it easier for you to follow along.)

## Example 1: Systems Word Problems

You are running a concession stand at a basketball game. You are selling hot dogs and sodas. Each hot dog costs $1.50 and each soda costs $0.50. At the end of the night you made a total of $78.50. You sold a total of 87 hot dogs and sodas combined. You must report the number of hot dogs sold and the number of sodas sold. How many hot dogs were sold and how many sodas were sold?

1. Let's start by identifying the important information:

- hot dogs cost $1.50
- Sodas cost $0.50
- Made a total of $78.50
- Sold 87 hot dogs and sodas combined

2. Define your variables.

- Ask yourself, "What am I trying to solve for? What don't I know?

In this problem, I don't know how many hot dogs or sodas were sold. So this is what each variable will stand for. (Usually the question at the end will give you this information).

Let x = the number of hot dogs sold

Let y = the number of sodas sold

3. Write two equations.

One equation will be related to the price and one equation will be related to the quantity (or number) of hot dogs and sodas sold.

1.50x + 0.50y = 78.50 (Equation related to cost)

x + y = 87 (Equation related to the number sold)

4. Solve!

We can choose any method that we like to solve the system of equations. I am going to choose the substitution method since I can easily solve the 2nd equation for y.

5. Think about what this solution means.

x is the number of hot dogs and x = 35. That means that 35 hot dogs were sold.

y is the number of sodas and y = 52. That means that 52 sodas were sold.

6. Write your answer in a complete sentence.

35 hot dogs were sold and 52 sodas were sold.

7. Check your work by substituting.

1.50x + 0.50y = 78.50

1.50(35) + 0.50(52) = 78.50

52.50 + 26 = 78.50

35 + 52 = 87

Since both equations check properly, we know that our answers are correct!

That wasn't too bad, was it? The hardest part is writing the equations. From there you already know the strategies for solving. Think carefully about what's happening in the problem when trying to write the two equations.

## Example 2: Another Word Problem

You and a friend go to Tacos Galore for lunch. You order three soft tacos and three burritos and your total bill is $11.25. Your friend's bill is $10.00 for four soft tacos and two burritos. How much do soft tacos cost? How much do burritos cost?

- 3 soft tacos + 3 burritos cost $11.25
- 4 soft tacos + 2 burritos cost $10.00

In this problem, I don't know the price of the soft tacos or the price of the burritos.

Let x = the price of 1 soft taco

Let y = the price of 1 burrito

One equation will be related your lunch and one equation will be related to your friend's lunch.

3x + 3y = 11.25 (Equation representing your lunch)

4x + 2y = 10 (Equation representing your friend's lunch)

We can choose any method that we like to solve the system of equations. I am going to choose the combinations method.

5. Think about what the solution means in context of the problem.

x = the price of 1 soft taco and x = 1.25.

That means that 1 soft tacos costs $1.25.

y = the price of 1 burrito and y = 2.5.

That means that 1 burrito costs $2.50.

Yes, I know that word problems can be intimidating, but this is the whole reason why we are learning these skills. You must be able to apply your knowledge!

If you have difficulty with real world problems, you can find more examples and practice problems in the Algebra Class E-course.

## Take a look at the questions that other students have submitted:

Problem about the WNBA

Systems problem about ages

Problem about milk consumption in the U.S.

Vans and Buses? How many rode in each?

Telephone Plans problem

Systems problem about hats and scarves

Apples and guavas please!

How much did Alice spend on shoes?

All about stamps

Going to the movies

Small pitchers and large pitchers - how much will they hold?

Chickens and dogs in the farm yard

- System of Equations
- Systems Word Problems

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## SAT Mathematics : Solving Systems of Equations in Word Problems

Study concepts, example questions & explanations for sat mathematics, all sat mathematics resources, example questions, example question #21 : solving word problems.

Blin and Alex raised a total of $240,000 for a charity. If Blin raised $60,000 more than Alex did, how much money did Blin raise?

This problem provides you with a classic SAT Systems of Equations setup. You're given the sum of two entities - here that's Blin's Total + Alex's Total = $240,000 - and a relationship between those two entities. Here that's Blin raised $60,000 more than Alex did.

The sum equation is generally straightforward; you can express that as B + A = 240,000. For the relationship equation, you can say that Blin's total is $60,000 more than Alex's, which translates to B = 60,000 + A.

Now you have two equations:

B + A = 240,000

B = 60,000 + A

This structure sets up well for Substitution; you already have B in terms of A, so you can plug in the second equation for the first:

(60,000 + A) + A = 240,000

60,000 + 2A = 240,000

2A = 180,000

Of course, at this point you need to check whether you solved for the proper variable. The question asks you about Blin's total, not Alex's, so you need to add $60,000 to get to Blin's total of $150,000. Then check your work: does Blin's total of $150,000 plus Alex's total of $90,000 arrive at the sum we know to be $240,000? It does, so you can confidently choose $150,000.

## Example Question #1 : Solving Systems Of Equations In Word Problems

Judy and Nancy participated in a walk-a-thon to raise money for charity. They combined for 42 total miles between them, with Judy walking twice as many miles as Nancy. How many miles did Nancy walk?

This problem provides you with a classic SAT Systems of Equations setup: you get a sum equation about a combined total of two entities, and then an equation that gives a relationship between the two. Here you know that Judy and Nancy combined for 42 miles, which you can turn into an equation as:

Then you're told that Judy walked twice as many miles as Nancy did. That equation is J = 2N (be careful about putting the two in the wrong spot - it's best to think about "what do I have to do to Nancy's total to make it equal Judy's?? You'd have to double it to catch up, so twice Nancy's total is what it takes to equal Judy's --> J = 2N).

Now you can substitute the second equation for the first. If you plug in 2N where the J appears in the first, you have:

2N + N = 42

And since this problem asks you for Nancy's total, you have it: Nancy walked 14 miles.

Siblings Tamika and David tracked every hour they spent studying during a particular school year. Combined, they studied for a total of 932 hours, with Tamika studying for 28 more hours than her brother David did. How many hours did Tamika study?

This problem provides you with two equations, setting up nicely for a Systems of Equations approach. The first equation is about their combined total, and you can set it up as:

T + D = 932

Then the second equation is:

T = D + 28. (you can think of this as "to equal Tamika's total, David would have had to add 28 hours")

One helpful tip on this problem is that while it sets up nicely for a Substitution Method approach (substitute D + 28 for T in the first equation), you want to solve for T, not for D. So you can with one quick algebra step set it up to use the Elimination Method. Just subtract D from both sides in the second equation, and it becomes T - D = 28

Now you can stack your equations and sum them:

These sum to:

Divide both sides by 2 and you get T = 480.

## Example Question #24 : Solving Word Problems

Joanna has two dogs, Rover and Spot, who weigh a combined total of 87 pounds. If Rover weighs 9 pounds more than Spot, how many pounds does Spot weigh?

This problem provides you with two equations, allowing you to use Systems of Equations logic to solve once you've constructed the equations. Often on the SAT, these Systems problems will start by giving you a combined total as this one does. If the two dogs weigh a total of 87 pounds, you can express that as:

R + S = 87 (in which R = Rover's weight and S = Spot's weight)

Then if Rover weighs 9 pounds more than Spot, you have a few choices for how to express that. One is to say that "Rover's weight is 9 pounds more than Spot's weight" and then use the fact that "is" means "equals" so that the words translate directly to the equation: R = 9 + S. Another way is to see that Rover weighs more, and the difference is 9, meaning that R - S = 9. The equation with subtraction allows you to go straight into the Elimination Method as you have a + S in the first equation and a -S in the second, so that may be your preferred choice. If you structure it that way, you can stack the equations and sum them:

Which sums to:

That means that R = 48, but remember in Systems of Equations problems to always check that your algebra has answered the specific question, as a problem with two variables allows you to mistakenly solve for the wrong one! The question asks you about Spot's weight, so you can subtract 9 from 48 to see that Spot weighs 39 pounds.

## Example Question #25 : Solving Word Problems

The snack bar at a certain lake sells slices of pizza in two varieties: vegetarian pizza costs $1.25 per slice, and pepperoni pizza costs $1.50 per slice. If the snack bar sold a total of 20 slices of pizza for a total of $28.25 on a particular day, how many more slices of pepperoni were sold than slices of vegetarian?

This problem provides you with the opportunity to set up two equations: the total number of slices of pizza, and the total amount of money. If you use V to represent the number of vegetarian slices and P to represent the number of pepperoni slices, you can express the total number of slices as:

Then for the money, the total revenue will be a combination of how much was made from selling vegetarian slices ($1.25 times the number of vegetarian) and how much was made from selling pepperoni ($1.50 times the number of pepperoni). This equation is then:

1.25V + 1.5P = 28.25

This problem sets up nicely for using the Substitution Method for systems of equations. Since V is likely easier to work with in the second equation (the substituted term is likely easier to multiply by 1.5 than by 1.25), you can leverage the first equation to get P in terms of V:

When you plug this in to the second equation, you get:

1.25V + 1.5(20 - V) = 28.25

This expands to:

1.25V + 30 - 1.5V = 28.25

Combine like terms to get:

-0.25V = -1.75

So V = 7, meaning that P = 13.

Now as with any systems of equations problem, it's imperative that you go back to double check what the question wants. It's asking for the difference between the two types of pizza (how many more pepperoni slices were sold than vegetarian), so your last step is to subtract 13 - 7 to get the correct answer, 6.

## Example Question #26 : Solving Word Problems

An outdoor concert venue sells two types of tickets: balcony tickets cost $30 and main floor tickets cost $55. When the venue sells all 500 available tickets for a given show it earns a total of $20,000. How many main floor tickets does the venue offer for each show?

This problem provides you with the opportunity to set up two equations: one, the number of total tickets and two, the revenue from selling all of those tickets. If you use B to represent the number of balcony tickets and F to represent the number of floor tickets, your equations can look like:

Number of tickets:

B + F = 500

Revenue (the number of each type of ticket multiplied by the price of that type of ticket):

30B + 55F = 20000

If you multiply the entire first equation by -30, it sets you up nicely to be able to use the Elimination Method. That gives you:

-30B - 30F = -15000

You can then solve for F, which is 200. Of course, on any systems of equations problem you should always double check the question being asked, since with two variables there are multiple different questions they could ask you. Here they ask about the number of main floor tickets so you can safely answer 200.

## Example Question #27 : Solving Word Problems

Jim spent $60 purchasing lattes and coffees for his coworkers. If lattes cost $6 and coffees cost $3, and Jim purchased two more coffees than lattes, how many total drinks did he purchase?

If you then multiply both sides by 2, you have:

Which sets up for the elimination method when you stack and add the two equations:

A certain bakery sells only cupcakes and brownies, and charges $5 for each cupcake and $4 for each brownie. If Jeremy spent $34 purchasing 8 items at the bakery, how many brownies did he purchase?

With those two equations, you should see that they stack up nicely so that you can use the Elimination Method to solve. You have:

And then if you multiply the entire bottom equation by -4, you'll have:

## Example Question #29 : Solving Word Problems

A movie theater sold 120 tickets to the matinee showing of a popular children's movie, and 150 tickets to the evening showing. The theater sold the same number of adult tickets for each show, but for the evening show was just 20 children's tickets short of selling twice as many children's tickets as it did for the matinee. How many children's tickets were sold to the evening show?

In this word problem, it's first helpful to assign meaningful variables. Since the matinee equation is the "cleaner" of the two, you might start with:

So you have two equations now:

So use the Elimination Method and subtract the simpler, matinee equation from the more-complex, evening equation and you'll be left with:

## Example Question #30 : Solving Word Problems

A sporting good store sells one type of baseball bat and one type of baseball. The cost for 2 bats and 4 balls is $160. The cost for 1 bat and 6 balls is $160, as well. If someone were to buy an equal number of bats and balls, at most how many bats can he purchase if he has a budget of $240 for the purchase?

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## Systems of Linear Equations and Word Problems

Note that we saw how to solve linear inequalities here in the Coordinate System and Graphing Lines section . Note also that we solve Algebra Word Problems without Systems here , and we solve systems using matrices in the Matrices and Solving Systems with Matrices section here.

## Introduction to Systems

“Systems of equations” just means that we are dealing with more than one equation and variable. So far, we’ve basically just played around with the equation for a line, which is $ y=mx+b$. Let’s say we have the following situation:

Now, you can always do “guess and check” to see what would work, but you might as well use algebra! It’s much better to learn the algebra way, because even though this problem is fairly simple to solve, the algebra way will let you solve any algebra problem – even the really complicated ones.

The first trick in problems like this is to figure out what we want to know. This will help us decide what variables (unknowns) to use. What we want to know is how many pairs of jeans we want to buy (let’s say “$ j$”) and how many dresses we want to buy (let’s say “$ d$”). Always write down what your variables will be:

Let $ j=$ the number of jeans you will buy Let $ d=$ the number of dresses you’ll buy

Like we did before, let’s translate word-for-word from math to English:

Now we have the 2 equations as shown below. Notice that the $ j$ variable is just like the $ x$ variable and the $ d$ variable is just like the $ y$. It’s easier to put in $ j$ and $ d$ so we can remember what they stand for when we get the answers.

This is what we call a system, since we have to solve for more than one variable – we have to solve for 2 here. The cool thing is to solve for 2 variables , you typically need 2 equations , to solve for 3 variables , you need 3 equations , and so on. That’s easy to remember, right?

We need to get an answer that works in both equations ; this is what we’re doing when we’re solving; this is called solving simultaneous systems , or solving system simultaneously . There are several ways to solve systems; we’ll talk about graphing first.

## Solving Systems by Graphing

Remember that when you graph a line, you see all the different coordinates (or $ x/y$ combinations) that make the equation work. In systems, you have to make both equations work, so the intersection of the two lines shows the point that fits both equations (assuming the lines do in fact intersect; we’ll talk about that later). The points of intersections satisfy both equations simultaneously.

Put these equations into the $ y=mx+b$ ($ d=mj+b$) format, by solving for the $ d$ (which is like the $ y$):

$ \displaystyle j+d=6;\text{ }\,\text{ }\text{solve for }d:\text{ }d=-j+6\text{ }$

$ \displaystyle 25j+50d=200;\text{ }\,\,\text{solve for }d:\text{ }d=\frac{{200-25j}}{{50}}=-\frac{1}{2}j+4$

Now graph both lines:

Note that with non-linear equations, there will most likely be more than one intersection; an example of how to get more than one solution via the Graphing Calculator can be found in the Exponents and Radicals in Algebra section. Also, t here are some examples of systems of inequality here in the Coordinate System and Graphing Lines section .

## Solving Systems with Substitution

Substitution is the favorite way to solve for many students! It involves exactly what it says: substituting one variable in another equation so that you only have one variable in that equation. Here is the same problem:

You’re going to the mall with your friends and you have $200 to spend from your recent birthday money. You discover a store that has all jeans for $25 and all dresses for $50 . You really, really want to take home 6 items of clothing because you “need” that many new things. How many pairs of jeans and how many dresses you can buy so you use the whole $200 (tax not included)?

Below are our two equations, and let’s solve for “$ d$” in terms of “$ j$” in the first equation. Then, let’s substitute what we got for “$ d$” into the next equation. Even though it doesn’t matter which equation you start with, remember to always pick the “easiest” equation first (one that we can easily solve for a variable) to get a variable by itself.

We could buy 4 pairs of jeans and 2 dresses . Note that we could have also solved for “$ j$” first; it really doesn’t matter. You’ll want to pick the variable that’s most easily solved for. Let’s try another substitution problem that’s a little bit different:

## Solving Systems with Linear Combination or Elimination

Probably the most useful way to solve systems is using linear combination, or linear elimination. The reason it’s most useful is that usually in real life we don’t have one variable in terms of another (in other words, a “$ y=$” situation).

The main purpose of the linear combination method is to add or subtract the equations so that one variable is eliminated. We can add, subtract, or multiply both sides of equations by the same numbers – let’s use real numbers as shown below. We are using the Additive Property of Equality , Subtraction Property of Equality , Multiplicative Property of Equality , and/or Division Property of Equality that we saw here in the Types of Numbers and Algebraic Properties section :

If we have a set of 2 equations with 2 unknowns, for example, we can manipulate them by adding, multiplying or subtracting (we usually prefer adding) so that we get one equation with one variable. Let’s use our previous problem:

We could buy 4 pairs of jeans and 2 dresses .

Here’s another example:

## Types of equations

In the example above, we found one unique solution to the set of equations. Sometimes, however, for a set of equations, there are no solutions (when lines are parallel) or an infinite number of solutions or infinitely many solutions (when the two lines are actually the same line, and one is just a “multiple” of the other).

When there is at least one solution , the equations are consistent equations , since they have a solution. When there is only one solution, the system is called independent , since they cross at only one point. When equations have infinite solutions, they are the same equation, are consistent , and are called dependent or coincident (think of one just sitting on top of the other).

When equations have no solutions , they are called inconsistent equations , since we can never get a solution .

Here are graphs of inconsistent and dependent equations that were created on a graphing calculator:

## Systems with Three Equations

Let’s get a little more complicated with systems; in real life, we rarely just have two unknowns to solve for.

Let’s say at the same store, they also had pairs of shoes for $20 and we managed to get $60 more to spend! Now we have a new problem. To spend the even $260 , how many pairs of jeans, dresses, and pairs of shoes should we get if want, for example, exactly 10 total items (Remember that jeans cost $25 each and dresses cost $50 each).

Let’s let $ j=$ the number of pair of jeans, $ d=$ the number of dresses, and $ s=$ the number of pairs of shoes we should buy. So far, we’ll have the following equations:

$ \displaystyle \begin{array}{c}j+d+s=10\text{ }\\25j+\text{ }50d+\,20s=260\end{array}$

We’ll need another equation, since for three variables, we need three equations (otherwise, we theoretically might have infinite ways to solve the problem). In this type of problem, you would also need something like this: We want twice as many pairs of jeans as pairs of shoes . Now, since we have the same number of equations as variables , we can potentially get one solution for the system of equations. Here are the three equations:

We’ll learn later how to put these in our calculator to easily solve using matrices (see the Matrices and Solving Systems with Matrices section). For now, we can use two equations at a time to eliminate a variable (using substitution and/or elimination), and keep doing this until we’ve solved for all variables. These can get really difficult to solve, but remember that in “real life”, there are computers to do all this work!

Remember again, that if we ever get to a point where we end up with something like this, it means there are an infinite number of solutions : $ 4=4$ (variables are gone and a number equals another number and they are the same). And if we up with something like this, it means there are no solutions : $ 5=2$ (variables are gone and two numbers are left and they don’t equal each other).

Let’s solve our system: $ \displaystyle \begin{array}{c}j+d+s=10\text{ }\\25j+\text{ }50d+20s=260\\j=2s\end{array}$ :

We could buy 6 pairs of jeans, 1 dress, and 3 pairs of shoes .

Here’s one more example of a three-variable system of equations, where we’ll only use linear elimination:

$ \displaystyle \begin{align}5x-6y-\,7z\,&=\,7\\6x-4y+10z&=\,-34\\2x+4y-\,3z\,&=\,29\end{align}$

I know – this is really difficult stuff! But if you do it step-by-step and keep using the equations you need with the right variables, you can do it. Think of it like a puzzle – you may not know exactly where you’re going, but do what you can in baby steps, and you’ll get there (sort of like life sometimes, right?!). And we’ll learn much easier ways to do these types of problems.

## Algebra Word Problems with Systems

Let’s do more word problems; you’ll notice that many of these are the same type that we did earlier in the Algebra Word Problems section , but now we can use more than one variable. This will actually make the problems easier! Again, when doing these word problems:

- If you’re wondering what the variables (or unknowns) should be when working on a word problem, look at what the problem is asking. These are usually (but not always) what your variables are!
- If you’re not sure how to set up the equations, use regular numbers (simple ones!) and see what you’re doing. Then put the variables back in!

Here are some problems:

## Investment Word Problem

We also could have set up this problem with a table:

## Mixture Word Problems

Here’s a mixture word problem . With mixture problems, remember if the problem calls for a pure solution or concentrate , use 100% (if the percentage is that solution) or 0% (if the percentage is another solution).

Let’s do the math (use substitution )!

$ \displaystyle \begin{array}{c}x\,\,+\,\,y=10\\.01x+.035y=10(.02)\end{array}$ $ \displaystyle \begin{array}{c}\,y=10-x\\.01x+.035(10-x)=.2\\.01x\,+\,.35\,\,-\,.035x=.2\\\,-.025x=-.15;\,\,\,\,\,x=6\\\,y=10-6=4\end{array}$

We would need 6 liters of the 1% milk, and 4 liters of the 3.5% milk.

Here’s another mixture problem:

$ \displaystyle \begin{array}{c}x+y=50\\8x+4y=50\left( {6.4} \right)\end{array}$ $ \displaystyle \begin{array}{c}y=50-x\\8x+4\left( {50-x} \right)=320\\8x+200-4x=320\\4x=120\,;\,\,\,\,x=30\\y=50-30=20\\8x+4y=50(6.4)\end{array}$

We would need 30 pounds of the $8 coffee bean, and 20 pounds of the $4 coffee bean. See how similar this problem is to the one where we use percentages?

## Distance Word Problem:

Here’s a distance word problem using systems ; distance problems have to do with an object’s speed, time, and distance. Note that, as well as the distance word problem here in the Algebra Word Problems section , there’s an example of a Parametric Distance Problem here in the Parametric Equations section .

## Which Plumber Problem

Many word problems you’ll have to solve have to do with an initial charge or setup charge, and a charge or rate per time period. In these cases, the initial charge will be the $ \boldsymbol {y}$ -intercept , and the rate will be the slope . Here is an example:

## Geometry Word Problem:

Many times, we’ll have a geometry problem as an algebra word problem; these might involve perimeter, area, or sometimes angle measurements (so don’t forget these things!). Let’s do one involving angle measurements.

See – these are getting easier! Here’s one that’s a little tricky though:

## Work Problem :

Let’s do a “ work problem ” that is typically seen when studying Rational Equations (fraction with variables in them) and can be found here in the Rational Functions, E quations and Inequalities section .

Note that there’s also a simpler version of this problem here in the Direct, Inverse, Joint and Combined Variation section .

## Three Variable Word Problem:

Let’s do one more with three equations and three unknowns:

## The “Candy” Problem

Sometimes we get lucky and can solve a system of equations where we have more unknowns (variables) then equations. (Actually, I think it’s not so much luck, but having good problem writers!) Here’s one like that:

There are more Systems Word Problems in the Matrices and Solving Systems with Matrices section , Linear Programming section , and Right Triangle Trigonometry section .

Understand these problems, and practice, practice, practice!

For Practice : Use the Mathway widget below to try a Systems of Equations problem. Click on Submit (the blue arrow to the right of the problem) and click on Solve by Substitution or Solve by Addition/Elimination to see the answer .

You can also type in your own problem, or click on the three dots in the upper right hand corner and click on “Examples” to drill down by topic.

If you click on Tap to view steps , or Click Here , you can register at Mathway for a free trial , and then upgrade to a paid subscription at any time (to get any type of math problem solved!).

On to Algebraic Functions, including Domain and Range – you’re ready!

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## System-of-Equations Word Problems

Exercises More Exercises

Many problems lend themselves to being solved with systems of linear equations. In "real life", these problems can be incredibly complex. This is one reason why linear algebra (the study of linear systems and related concepts) is its own branch of mathematics.

In your studies, however, you will generally be faced with much simpler problems. What follows are some typical examples.

## The admission fee at a small fair is $1.50 for children and $4.00 for adults. On a certain day, 2200 people enter the fair and $5050 is collected. How many children and how many adults attended?

Content Continues Below

## MathHelp.com

System of Equations Word Problems

In the past , I would have set this up by picking a variable for one of the groups (say, " c " for "children") and then use "(total) less (what I've already accounted for)" (in this case, " 2200 – c ") for the other group. Using a system of equations, however, allows me to use two different variables for the two different unknowns.

number of adults: a

number of children: c

With these variables, I can create equations for the totals they've given me:

total number: a + c = 2200

total income: 4 a + 1.5 c = 5050

Now I can solve the system for the number of adults and the number of children. I will solve the first equation for one of the variables, and then substitute the result into the other equation:

a = 2200 – c

4(2200 – c ) + 1.5 c = 5050

8800 – 4 c + 1.5 c = 5050

8800 – 2.5 c = 5050

–2.5 c = –3750

Now I can back-solve for the value of the other variable:

a = 2200 – (1500) = 700

I have values for my two variables. I can look back at my definitions for the variables to interpret these values. To answer the original question, there were:

1500 children and 700 adults.

You will probably start out with problems which, like the one above, seem very familiar. But you will then move on to more complicated problems.

## The sum of the digits of a two-digit number is 7 . When the digits are reversed, the number is increased by 27 . Find the number.

The trick here is to work with the digits explicitly. I'll use " t " for the "tens" digit of the original number and " u " for the "units" (or "ones") digit. I then have:

The ten's digit stands for "ten times of this digit's value". Just as "26" is "10 times 2, plus 6 times 1", so also the two-digit number they've given me will be ten times the "tens" digit, plus one times the "units" digit. In other words:

original number: 10 t + 1 u

The new number has the values of the digits (represented by the variables) in reverse order. This gives me:

new number: 10 u + 1 t

And this new number is twenty-seven more than the original number. The keyword "is" means "equals", so I get:

(new number) is (old number) increased by (twenty-seven)

10 u + 1 t = (10 t + 1 u ) + 27

Now I have a system of equations that I can solve:

10 u + t = 10 t + u + 27

First I'll simplify the second equation:

9 u – 9 t = 27

u – t = 3

After reordering the variables in the first equation, I now have:

Adding down , I get:

Then t = 2 . Back-solving, this means that the original number was 25 and the new number (gotten by switching the digits) is 52 . Since 52 – 25 = 27 , this solution checks out.

The number is 25 .

## Find the equation of the parabola that passes through the points (–1, 9) , (1, 5) , and (2, 12) .

Recalling that a parabola has a quadratic as its equation, I know that I am looking for an equation of the form ax 2 + bx + c = y . Also, I know that points are of the form ( x , y ) . Practically speaking, this mean that, in each of these points, they have given me values for x and y that make the quadratic equation true. Plugging the three points in the general equation for a quadratic, I get a system of three equations, where the variables stand for the unknown coefficients of that quadratic:

a (–1) 2 + b (–1) + c = 9

a (1) 2 + b (1) + c = 5

a (2) 2 + b (2) + c = 12

Simplifying the three equations, I get:

1 a – b + c = 9

1 a + b + c = 5

4 a + 2 b + c = 12

I won't display the solving of this problem, but the result is that a = 3, b = –2, and c = 4 , so the equation they're wanting is:

y = 3 x 2 – 2 x + 4

You may also see similar exercises referring to circles, using:

x 2 + y 2 + bx + cy + d = 0

...or other conics, though parabolas are the most common. Keep in mind that projectile problems (like shooting an arrow up in the air or dropping a penny from the roof of a tall building) are also parabola problems, using:

–( 1 / 2 ) gt 2 + v 0 t + h 0 = s

...where h 0 is the original height, v 0 is the initial velocity, s is the height at time t , usually measured in seconds, and g refers to gravity, being 9.8 if you're working in meters and 32 if you're working in feet).

All of these different permutations of the above example work the same way: Take the general equation for the curve, plug in the given points, and solve the resulting system of equations for the values of the coefficients. Warning: If you see an exercise of this sort in the homework, be advised that you may be expected to know the forms of the general equations (such as " ax 2 + bx + c = y " for parabolas) on the next text.

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Chapter 6: Polynomials

## 6.8 Mixture and Solution Word Problems

Solving mixture problems generally involves solving systems of equations. Mixture problems are ones in which two different solutions are mixed together, resulting in a new, final solution. Using a table will help to set up and solve these problems. The basic structure of this table is shown below:

The first column in the table (Name) is used to identify the fluids or objects being mixed in the problem. The second column (Amount) identifies the amounts of each of the fluids or objects. The third column (Value) is used for the value of each object or the percentage of concentration of each fluid. The last column (Equation) contains the product of the Amount times the Value or Concentration.

Example 6.8.1

Jasnah has 70 mL of a 50% methane solution. How much of a 80% solution must she add so the final solution is 60% methane? Find the equation.

- The solution names are 50% (S 50 ), 60% (S 60 ), and 80% (S 80 ).
- The amounts are S 50 = 70 mL, S 80 , and S 60 = 70 mL + S 80 .
- The concentrations are S 50 = 0.50, S 60 = 0.60, and S 80 = 0.80.

The equation derived from this data is 0.50 (70 mL) + 0.80 (S 80 ) = 0.60 (70 mL + S 80 ).

Example 6.8.2

Sally and Terry blended a coffee mix that sells for [latex]\$2.50[/latex] by mixing two types of coffee. If they used 40 mL of a coffee that costs [latex]\$3.00,[/latex] how much of another coffee costing [latex]\$1.50[/latex] did they mix with the first?

The equation derived from this data is:

[latex]\begin{array}{ccccccc} 1.50(C_{1.50})&+&3.00(40)&=&2.50(40&+&C_{1.50}) \\ 1.50(C_{1.50})&+&120&=&100&+&2.50(C_{1.50}) \\ -2.50(C_{1.50})&-&120&=&-120&-&2.50(C_{1.50}) \\ \hline &&-1.00(C_{1.50})&=&-20&& \\ \\ &&(C_{1.50})&=&\dfrac{-20}{-1}&& \\ \\ &&C_{1.50}&=&20&& \end{array}[/latex]

This means 20 mL of the coffee selling for [latex]\$1.50[/latex] is needed for the mix.

Example 6.8.3

Nick and Chloe have two grades of milk from their small dairy herd: one that is 24% butterfat and another that is 18% butterfat. How much of each should they use to end up with 42 litres of 20% butterfat?

The equation derived from this data is:

[latex]\begin{array}{rrrrrrr} 0.24(B_{24})&+&0.18(42&- &B_{24})&=&0.20(42) \\ 0.24(B_{24})&+&7.56&-&0.18(B_{24})&=&8.4 \\ &-&7.56&&&&-7.56 \\ \hline &&&&0.06(B_{24})&=&0.84 \\ \\ &&&&B_{24}&=&\dfrac{0.84}{0.06} \\ \\ &&&&B_{24}&=&14 \end{array}[/latex]

This means 14 litres of the 24% buttermilk, and 28 litres of the 18% buttermilk is needed.

Example 6.8.4

In Natasha’s candy shop, chocolate, which sells for [latex]\$4[/latex] a kilogram, is mixed with nuts, which are sold for [latex]\$2.50[/latex] a kilogram. Chocolate and nuts are combined to form a chocolate-nut candy, which sells for [latex]\$3.50[/latex] a kilogram. How much of each are used to make 30 kilograms of the mixture?

[latex]\begin{array}{rrrrrrl} 4.00(C)&+&2.50(30&-&C)&=&3.50(30) \\ 4.00(C)&+&75&-&2.50(C)&=&105 \\ &-&75&&&&-75 \\ \hline &&&&1.50(C)&=&30 \\ \\ &&&&C&=&\dfrac{30}{1.50} \\ \\ &&&&C&=&20 \end{array}[/latex]

Therefore, 20 kg of chocolate is needed for the mixture.

With mixture problems, there is often mixing with a pure solution or using water, which contains none of the chemical of interest. For pure solutions, the concentration is 100%. For water, the concentration is 0%. This is shown in the following example.

Example 6.8.5

Joey is making a a 65% antifreeze solution using pure antifreeze mixed with water. How much of each should be used to make 70 litres?

[latex]\begin{array}{rrrrl} 1.00(A)&+&0.00(70-A)&=&0.65(0.70) \\ &&1.00A&=&45.5 \\ &&A&=&45.5 \\ \end{array}[/latex]

This means the amount of water added is 70 L − 45.5 L = 24.5 L.

For questions 1 to 9, write the equations that define the relationship.

- A tank contains 8000 litres of a solution that is 40% acid. How much water should be added to make a solution that is 30% acid?
- How much pure antifreeze should be added to 5 litres of a 30% mixture of antifreeze to make a solution that is 50% antifreeze?
- You have 12 kilograms of 10% saline solution and another solution of 3% strength. How many kilograms of the second should be added to the first in order to get a 5% solution?
- How much pure alcohol must be added to 24 litres of a 14% solution of alcohol in order to produce a 20% solution?
- How many litres of a blue dye that costs [latex]\$1.60[/latex] per litre must be mixed with 18 litres of magenta dye that costs [latex]\$2.50[/latex] per litre to make a mixture that costs [latex]\$1.90[/latex] per litre?
- How many grams of pure acid must be added to 40 grams of a 20% acid solution to make a solution which is 36% acid?
- A 100-kg bag of animal feed is 40% oats. How many kilograms of pure oats must be added to this feed to produce a blend of 50% oats?
- A 20-gram alloy of platinum that costs [latex]\$220[/latex] per gram is mixed with an alloy that costs [latex]\$400[/latex] per gram. How many grams of the [latex]\$400[/latex] alloy should be used to make an alloy that costs [latex]\$300[/latex] per gram?
- How many kilograms of tea that cost [latex]\$4.20[/latex] per kilogram must be mixed with 12 kilograms of tea that cost [latex]\$2.25[/latex] per kilogram to make a mixture that costs [latex]\$3.40[/latex] per kilogram?

Solve questions 10 to 21.

- How many litres of a solvent that costs [latex]\$80[/latex] per litre must be mixed with 6 litres of a solvent that costs [latex]\$25[/latex] per litre to make a solvent that costs [latex]\$36[/latex] per litre?
- How many kilograms of hard candy that cost [latex]\$7.50[/latex] per kg must be mixed with 24 kg of jelly beans that cost [latex]\$3.25[/latex] per kg to make a mixture that sells for [latex]\$4.50[/latex] per kg?
- How many kilograms of soil supplement that costs [latex]\$7.00[/latex] per kg must be mixed with 20 kg of aluminum nitrate that costs [latex]\$3.50[/latex] per kg to make a fertilizer that costs [latex]\$4.50[/latex] per kg?
- A candy mix sells for [latex]\$2.20[/latex] per kg. It contains chocolates worth [latex]\$1.80[/latex] per kg and other candy worth [latex]\$3.00[/latex] per kg. How much of each are in 15 kg of the mixture?
- A certain grade of milk contains 10% butterfat and a certain grade of cream 60% butterfat. How many litres of each must be taken so as to obtain a mixture of 100 litres that will be 45% butterfat?
- Solution A is 50% acid and solution B is 80% acid. How much of each should be used to make 100 cc of a solution that is 68% acid?
- A paint that contains 21% green dye is mixed with a paint that contains 15% green dye. How many litres of each must be used to make 600 litres of paint that is 19% green dye?
- How many kilograms of coffee that is 40% java beans must be mixed with coffee that is 30% java beans to make an 80-kg coffee blend that is 32% java beans?
- A caterer needs to make a slightly alcoholic fruit punch that has a strength of 6% alcohol. How many litres of fruit juice must be added to 3.75 litres of 40% alcohol?
- A mechanic needs to dilute a 70% antifreeze solution to make 20 litres of 18% strength. How many litres of water must be added?
- How many millilitres of water must be added to 50 millilitres of 100% acid to make a 40% solution?
- How many litres of water need to be evaporated from 50 litres of a 12% salt solution to produce a 15% salt solution?

Answer Key 6.8

Intermediate Algebra by Terrance Berg is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License , except where otherwise noted.

## Share This Book

## Systems of Equations (Word Problems)

These lessons, with videos, examples and step-by-step solutions help Grade 8 students learn how to analyze and solve pairs of simultaneous linear equations.

Related Pages Systems of Equations - Graphical Method Solving Equations Common Core for Grade 8 Common Core for Mathematics More Math Lessons for Grade 8

A. Understand that solutions to a system of two linear equations in two variables correspond to points of intersection of their graphs, because points of intersection satisfy both equations simultaneously.

B. Solve systems of two linear equations in two variables algebraically, and estimate solutions by graphing the equations. Solve simple cases by inspection. For example, 3x + 2y = 5 and 3x + 2y = 6 have no solution because 3x + 2y cannot simultaneously be 5 and 6 .

C. Solve real-world and mathematical problems leading to two linear equations in two variables. For example, given coordinates for two pairs of points, determine whether the line through the first pair of points intersects the line through the second pair.

Common Core: 8.EE.8c

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- \mathrm{Lauren's\:age\:is\:half\:of\:Joe's\:age.\:Emma\:is\:four\:years\:older\:than\:Joe.\:The\:sum\:of\:Lauren,\:Emma,\:and\:Joe's\:age\:is\:54.\:How\:old\:is\:Joe?}
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- To solve word problems start by reading the problem carefully and understanding what it's asking. Try underlining or highlighting key information, such as numbers and key words that indicate what operation is needed to perform. Translate the problem into mathematical expressions or equations, and use the information and equations generated to solve for the answer.
- How do you identify word problems in math?
- Word problems in math can be identified by the use of language that describes a situation or scenario. Word problems often use words and phrases which indicate that performing calculations is needed to find a solution. Additionally, word problems will often include specific information such as numbers, measurements, and units that needed to be used to solve the problem.
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- Symbolab is the best calculator for solving a wide range of word problems, including age problems, distance problems, cost problems, investments problems, number problems, and percent problems.
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- An age problem is a type of word problem in math that involves calculating the age of one or more people at a specific point in time. These problems often use phrases such as 'x years ago,' 'in y years,' or 'y years later,' which indicate that the problem is related to time and age.

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## SAT Math Word Problems

Most students tremble with fear when they think about preparing for the Math portion of the SAT . After all, there are 20+ topics tested, and the test requires mastery of topics ranging from arithmetic to trigonometry. Some topics cover basic math questions from algebra class, such as solving linear equations or recognizing graphs of parabolas. Solving questions such as these is tough enough, but the SAT adds a layer of difficulty by presenting word problems from a wide range of disciplines that require you to use your algebra skills to solve them.

To tackle these word problems, you must first understand the context of the word problem and then create the equations that you’ll need to solve, based on the information provided in the question.

In this article, I’ll first cover two critical algebra skills you need to master. Then I’ll provide a number of strategies for efficiently solving hard SAT word problems, as well as examples, so we can see these strategies in action and avoid common mistakes.

## Here are the topics we’ll cover:

Example 1: linear equation with one variable, sales tax: example 1, substitution method practice 1, substitution method practice 2, when dealing with word problems, translation is key, balancing equations: example 1, balancing equations: example 2, balancing equations: example 3, “number of items”: example 1, “number of items”: example 2, age problem: example 1, money problem: example 1, in conclusion: sat math word problems, what’s next.

We’ll start with solving basic linear equations, then we’ll review the substitution method of solving SAT algebra problems. Finally, we’ll use these two techniques to solve math word problem examples.

## Basic Algebra – Solving Single-Variable Linear Equations

The simplest algebra questions on the SAT are single-variable linear equations. Your goal is to solve for the value of the variable:

- 2v + 5 = 3v – 7

Note that solving single-variable equations generally requires that we combine all terms containing the variable on one side of the equation and all constant terms on the other side.

Let’s practice simplifying and solving an equation with one variable.

If 8q – 2 = 10q + 14, then what is the value of q?

First, we combine our constant terms by adding 2 to both sides:

8q – 2 + 2 = 10q + 14 + 2

8q = 10q + 16

Next, we combine the variable terms by subtracting 10q from both sides of the equation:

8q – 10q = 10q + 16 – 10q

Finally, we divide both sides of the equation by -2 to isolate the variable:

-2q / -2 = 16 / -2

When solving for a single variable, we combine like terms and isolate the variable we are solving for.

Now, let’s practice with an example.

If -2x – 11 = 5x + 3, then x equals which of the following?

-2x – 11 = 5x + 3

-2x = 5x + 14

Now let’s turn our attention to applying algebra to solve a straightforward word problem involving sales tax.

## Sales Tax Questions

A common single-variable word problem that you might be asked on the SAT involves the calculation of sales tax. When you buy an item, you not only pay for the item itself but you also generally pay a surcharge called sales tax, typically stated as a percentage of the cost of the item. We add the sales tax to the cost of the item to determine the actual amount we will pay at the register.

The equation for purchasing an item in an area where sales tax is charged is:

Cost of Item + Sales Tax = Total

Let’s look at a basic example of the calculation of sales tax.

Laura buys a blender for $40.00 in a town where the sales tax rate is 4.5%. How much will she have to pay, including sales tax?

We see that the cost of the item is $40.00, and the sales tax rate is 4.5%. Thus, the sales tax amount is 4.5% of $40.00, or 40 x 0.045. Let’s put these into the equation:

40 + (40)(0.045) = Total

40 + 1.8 = Total

41.8 = Total

Thus, Laura will pay a total of $41.80: the item cost of $40.00 and the sales tax of $1.80.

Know the sales tax formula: Cost of Item + Sales Tax = Total

Let’s look at a sample SAT question about sales tax.

Alex buys a pair of shoes. The sales tax rate in his town is 6%. The register total for his purchase, including sales tax, is $115.54. How much sales tax did Alex pay?

We set up the linear equation by noting that the total purchase price is equal to the cost of the item plus the sales tax:

Total = Cost of Item + Sales Tax

We are given the total and the sales tax rate. However, we don’t know the actual cost of the item, so we will let x = the cost of the item:

115.54 = x + (0.06)(x)

We combine like terms on the right side of the equation and solve for x:

115.54 = 1.06x

115.54 / 1.06 = x

We see that the cost of the item is $109.00. Now we calculate the sales tax:

109 x 0.06 = 6.54

The sales tax is $6.54.

Next, let’s discuss linear equations with two variables.

## Linear Equations with Two Variables/Systems of Linear Equations

Two-variable equations have not just one but two different variables:

- v – u = 12
- 2x + y = 10
- 5z – 3y = 42

These are called two-variable equations because they contain two different variables . If we have a second equation that contains one or both of the variables, the two equations are called a system of linear equations. One common way to solve for the values of the variables is by using the substitution method . Let’s discuss the substitution method now.

## Using the Substitution Method to Solve a System of Linear Equations

When we’re working with SAT systems of equations word problems, the essence of the substitution method is that we first isolate one variable in one of the equations. Then we substitute whatever that variable is equal to into the other equation. Let’s practice with an example.

In the system of linear equations below, determine the value of a.

b = 3a (equation 1)

a + b = 12 (equation 2)

Looking at our two equations, we see that b is already isolated in equation 1. So, we can substitute 3a (from equation one) for b (in equation two). This gives us:

a + 3a = 12

For the system of linear equations below, what is the value of x?

3x + y = 11 (equation 1)

2x + 5y = 3 (equation 2)

First, let’s isolate y in equation 1 by subtracting 3x from both sides of the equation:

y = 11 – 3x

Now, we can substitute 11 – 3x for y in equation 2. This gives us:

2x + 5(11 – 3x) = 3

2x + 55 – 15x = 3

We can use the substitution method to solve a system of linear equations.

Next, let’s discuss the many topics on the SAT in which we use algebra to determine the solution.

Here are a couple of sample SAT math word problems that you might encounter:

Luca buys apples, which cost $2 each, and bananas, which cost $3 each, at the market. If he buys twice as many bananas as apples and spends $48 at the market, how many bananas does he buy?

Tyrone and Greg have a total of 40 marbles. If Tyrone has 4 times the number of marbles that Greg has, how many marbles does Greg have?

The bottom line is that a word problem presents a scenario requiring us to translate the given information into an algebraic equation that we then solve.

Word problems are not just about solving equations; they are also about translating words into equations! Let’s look at some common translations:

- “Is” translates to equals ( = )

Kendra is the same age as Carla

Kendra’s age = Carla’s age

- “More” translates to addition ( + )

Arita has 6 more marbles than Pablo

Arita = Pablo + 6

- “Less/fewer” translates to subtraction (-)

Sammy has 3 fewer coins than Rati

Sammy = Rati – 3

- “Times as many” translates to multiplication (✖)

Harold has 5 times as many newspapers as Phoebe

Harold = Phoebe ✖ 5

Know the common translations of words to algebraic equations.

Before jumping into word problem practice questions, let’s discuss one point of confusion students have when translating words into equations: properly balancing the equations.

## You Must Make Sure Your Equations are Balanced

A basic principle for translating words into equations is that the equations must be balanced correctly. If they are not, you might obtain reversed values for your variables’ values, or you might get a negative answer. Let’s look at a few correct and incorrect ways to balance equations, and we’ll explain why the equations are balanced correctly.

Sherry has 30 more dollars than Melanie.

When we are setting up this equation, we must understand that Sherry has more money than Melanie. For example, if Melanie has 20 dollars, then Sherry has 50 dollars.

So, to properly balance the equation, we must add 30 dollars to Melanie’s amount and set it equal to Sherry’s amount.

If we let S = Sherry’s money and M = Melanie’s money, we have:

Nala has 4 fewer toys than Frank.

When considering this equation, we must understand that Nala has fewer toys than Frank.

In other words, Frank has more toys than Nala.

So, to set up a balanced equation, we must subtract 4 from Frank’s amount and set that equal to Nala’s.

If we let N = the number of Nala’s toys and F = the number of Frank’s toys, we have:

N – 4 = F

There are 5 times as many baseballs as tennis balls.

When considering this equation, we must understand that there are more baseballs than tennis balls.

So, to set up a correct equation, we multiply the number of tennis balls by 5 and set that amount equal to the number of baseballs.

If we let B = the number of baseballs and T = the number of tennis balls, we have:

Now that we are familiar with how to translate, balance, and solve equations, let’s jump into the various types of general word problems you will encounter on the SAT, and how to solve them.

## “Number of Items” Questions

We have already encountered some simple examples of “number of items” questions in this article. Now, let’s attack a full-fledged problem like those you might encounter on test day.

Julia and Tory have a total of 30 candy bars. If Julia has 5 times as many candy bars as Tory, how many candy bars does Tory have?

First, we define our variables.

Let’s let J = the number of candy bars Julia has and T = the number of candy bars Tory has.

Next, let’s create equations based on the information given in the question stem.

“Julia and Tory have a total of 30 candy bars” is translated as:

J + T = 30 (equation 1)

“Julia has 5 times as many candy bars as Tory” is translated as:

J = 5T (equation 2)

Next, we use the substitution method by substituting 5T for J in equation 1:

5T + T = 30

Let’s practice one more. This time, we will add a twist.

Leti has 4 times as many cookies as Henrik. If Leti gives Henrik 7 cookies, Henrik will have 4 fewer cookies than Leti. How many cookies did Leti originally have?

We can let L = the number of cookies Leti has and H = the number of cookies Henrik has. Next, we use the information from the question stem to create equations.

Because Leti has 4 times as many cookies as Henrik, we have:

L = 4H (equation 1)

After Leti gives Henrik 7 cookies, Leti will have L – 7 cookies. After Henrik gets the 7 cookies from Leti, he will have H + 7 cookies. He will still have 4 fewer cookies than Leti.

Thus, our second equation is:

L – 7 = (H + 7) +4

L = H + 18 (equation 2)

Next, we can use the substitution method, substituting 4H for L in equation 2:

4H = H + 18

Henrik originally had 6 cookies. Thus, Leti originally had 4 x 6 = 24 cookies.

Next, let’s discuss another common type of SAT word problem: age problems.

## Age Problems

Age problems are a common type of SAT problem with a unique spin. Instead of having a straightforward translation, as we practiced above, an age problem will usually make us translate words into equations based on an age in the past, an age in the future, or even both.

For example, we might need to consider Ann’s age 10 years from now. To do this, we let Ann’s age today = A, so her age in 10 years will be A + 10.

Similarly, Ann’s age 12 years ago would be expressed as A – 12.

Age problems generally require us to compare ages in the past or future.

Chantal is 6 years younger than Yolanda. If, in 5 years, Yolanda will be twice as old as Chantal, how old is Chantal today?

Let’s let C = Chantal’s age today and Y = Yolanda’s age today. Next, let’s create some equations.

Because Chantal is 6 years younger than Yolanda, we have:

C = Y – 6 (equation 1)

Now we express what each age will be in 5 years:Chantal will be C + 5 and Yolanda will be Y + 6. The relationship between their ages in 5 years is that Yolanda will be twice Chantal’s age:

Y + 5 = 2(C + 5)

Y + 5 = 2C + 10 (equation 2)

Next, we use the substitution method and substitute Y – 6 for C in equation 2, and we have:

Y + 5 = 2(Y – 6) + 10

Y + 5 = 2Y – 12 + 10

Y + 5 = 2Y – 2

Yolanda is 7 years old today. Thus, Chantal is 7 – 6 = 1 year old today.

Next, let’s discuss money problems.

## Money Problems

In general, money problems on the SAT involve two different commodities, such as adult tickets and child tickets, and the revenue earned from selling them. We can create two equations, one relating the number of items and the second relating the revenue earned from the sale of the items.

We will see that the substitution method is extremely useful for solving the two equations that are created from the given information.

Be prepared to use the substitution method when dealing with money problems.

At an ice cream stand, customers can buy either a small cone for $3.00 or a large cone for $4.50. Yesterday, the owner reported he had sold 720 cones, and his revenue for the day was $2,340. How many small cones did he sell?

First, let’s define our variables. We can let the number of small cones sold = x and the number of large cones sold = y.

Next, let’s create our equations. Because the owner sold 720 ice cream cones, we have:

x + y = 720 (equation 1)

His total revenue was $2,340 from selling x small cones for $3.00 each and y large cones for $4.50 each:

3x + 4.5y = 2,340 (equation 2)

Let’s isolate x in equation 1:

x = 720 – y

Now, let’s substitute 720 – y for x into equation 2 and solve:

3(720 – y) + 4.5y = 2,340

2,160 – 3y + 4.5y = 2,340

Since the number of large cones sold is 120, the number of small cones sold is 720 – 120 = 600.

Of the 20+ major math topics on the SAT, one of the most challenging is word problems. In order to be successful with this type of problem, you need to be skilled both at translating words into equations and at solving algebra equations. The ability to solve SAT word problems is a must for scoring well on the exam.

The two most common skills for solving math word problems are knowing how to solve a linear equation and using the substitution method for solving systems of linear equations. You must translate the question into an equation or equations first, then use the appropriate algebraic method to arrive at the answer.

Some of the common types of word problems that we have covered here are sales tax questions, “number of items” questions, age questions, and money questions.

In this article, we have covered how to solve word problems on the SAT. You have a good start in doing well on this particular topic. But don’t lose your perspective: word problems are only one of many topics you must master in order to get a great SAT score. Read our article on how to improve your SAT score for more strategies, tips, and tricks.

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## What are systems of equations?

A system of equations is a set of one or more equations involving a number of variables..

The solutions to systems of equations are the variable mappings such that all component equations are satisfied—in other words, the locations at which all of these equations intersect. To solve a system is to find all such common solutions or points of intersection.

Systems of linear equations are a common and applicable subset of systems of equations. In the case of two variables, these systems can be thought of as lines drawn in two-dimensional space. If all lines converge to a common point, the system is said to be consistent and has a solution at this point of intersection. The system is said to be inconsistent otherwise, having no solutions. Systems of linear equations involving more than two variables work similarly, having either one solution, no solutions or infinite solutions (the latter in the case that all component equations are equivalent).

More general systems involving nonlinear functions are possible as well. These possess more complicated solution sets involving one, zero, infinite or any number of solutions, but work similarly to linear systems in that their solutions are the points satisfying all equations involved. Going further, more general systems of constraints are possible, such as ones that involve inequalities or have requirements that certain variables be integers.

Solving systems of equations is a very general and important idea, and one that is fundamental in many areas of mathematics, engineering and science.

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## COMMENTS

Wow! You have learned many different strategies for solving systems of equations! First we started with Graphing Systems of Equations . Then we moved onto solving systems using the Substitution Method. In our last lesson we used the Linear Combinations or Addition Method to solve systems of equations.

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Systems of Equations Word Problems 2) The difference 3) Flying to Kampala with a tailwind a plane averaged 158 km/h. On the return trip the plane only averaged 112 km/h while flying back into the same wind. Find the speed of the wind and the speed of the plane in still air.

Unit 1 Algebra foundations Unit 2 Solving equations & inequalities Unit 3 Working with units Unit 4 Linear equations & graphs Unit 5 Forms of linear equations Unit 6 Systems of equations Unit 7 Inequalities (systems & graphs) Unit 8 Functions Unit 9 Sequences Unit 10 Absolute value & piecewise functions Unit 11 Exponents & radicals

Systems of Equations Word Problems - Example 1: Tickets to a movie cost \ ($8\) for adults and \ ($5\) for students. A group of friends purchased \ (20\) tickets for \ ($115.00\). How many adult tickets did they buy? Answer: Let \ (x\) be the number of adult tickets and \ (y\) be the number of student tickets. There are \ (20\) tickets.

Very commonly, system-of-equations word problems involves mixtures or combinations of some sort. For instance: A landscaping company placed two orders with a nursery. The first order was for 13 bushes and 4 trees, and totalled $487. The second order was for 6 bushes and 2 trees, and totalled $232. The bills do not list the per-item price.

Possible Answers: This problem provides you with a classic SAT Systems of Equations setup. You're given the sum of two entities - here that's Blin's Total + Alex's Total = $240,000 - and a relationship between those two entities. Here that's Blin raised $60,000 more than Alex did.

Systems of Linear Equations and Word Problems $ \displaystyle \begin {array} {c}j+d+s=10\text { }\\25j+50d+\,20s=260\\j=2s\end {array}$ Note that when we say "we have twice as many pairs of jeans as pair of shoes", it doesn't translate that well into math.

System-of-Equations Word Problems | Purplemath Find Local Tutors Standardized Test Prep . Practically speaking, this mean that, in each of these points, they have given me values for that make the quadratic equation true.

Systems of Equations Word Problems Solve each word problem. Tickets to a movie cost $5 for adults and $3 for students. A group of friends purchased 18 tickets for $82.00. How many adults ticket did they buy? ____ At a store, Eva bought two shirts and five hats for $154.00. Nicole bought three same shirts and four same hats for $168.00.

Systems of equations word problems (with zero and infinite solutions) Liam's bookstore sold 40 notebooks and 20 newspapers for a total of $ 130 . A day later, the bookstore sold 8 notebooks and 4 newspapers at the same prices, for a total of $ 28 .

Systems of equations word problems (with zero and infinite solutions) Systems of equations with elimination: TV & DVD Systems of equations with elimination: apples and oranges Systems of equations with substitution: coins Systems of equations with elimination: coffee and croissants Systems of equations: FAQ Math > Algebra 1 > Systems of equations >

6.8 Mixture and Solution Word Problems. Solving mixture problems generally involves solving systems of equations. Mixture problems are ones in which two different solutions are mixed together, resulting in a new, final solution. Using a table will help to set up and solve these problems. The basic structure of this table is shown below: Example ...

Here are a set of practice problems for the Systems of Equations chapter of the Algebra notes. If you'd like a pdf document containing the solutions the download tab above contains links to pdf's containing the solutions for the full book, chapter and section. At this time, I do not offer pdf's for solutions to individual problems.

Solving systems of equations word problems worksheet For all problems, define variables, write the system of equations and solve for all variables. The directions are from TAKS so do all three (variables, equations and solve) no matter what is asked in the problem. A large pizza at Palanzio's Pizzeria costs $6.80 plus $0.90 for each topping.

B. Solve systems of two linear equations in two variables algebraically, and estimate solutions by graphing the equations. Solve simple cases by inspection. For example, 3x + 2y = 5 and 3x + 2y = 6 have no solution because 3x + 2y cannot simultaneously be 5 and 6. C. Solve real-world and mathematical problems leading to two linear equations in ...

Systems of Equations Word Problems. 2) Flying with the wind a plane went 183 km/h. Flying into the same wind the plane only went 141 km/h. Find the speed of the plane in still air and the speed of the wind. 3) Castel and Gabriella are selling pies for a school fundraiser. Customers can buy apple pies and lemon meringue pies.

Math Algebra (all content) Unit 5: System of equations About this unit This topic covers: - Solutions of linear systems - Graphing linear systems - Solving linear systems algebraically - Analyzing the number of solutions to systems - Linear systems word problems Systems of equations overview Learn Systems of equations: trolls, tolls (1 of 2)

How do you solve word problems? To solve word problems start by reading the problem carefully and understanding what it's asking. Try underlining or highlighting key information, such as numbers and key words that indicate what operation is needed to perform.

When we're working with SAT systems of equations word problems, the essence of the substitution method is that we first isolate one variable in one of the equations. ... Solution: Looking at our two equations, we see that b is already isolated in equation 1. So, we can substitute 3a (from equation one) for b (in equation two). This gives us ...

8 years ago I get confused by this type of solution because of the units. It's funny how you can just have a system of equations despite all their different units. Like he sums one equation that deals with distance (km) and another that deals with time (hours). How is that not conflicting? This cause my brain to shut down a little... • ( 70 votes)

The solutions to systems of equations are the variable mappings such that all component equations are satisfied—in other words, the locations at which all of these equations intersect. To solve a system is to find all such common solutions or points of intersection. Systems of linear equations are a common and applicable subset of systems of ...

Carlos Hernandez 8 years ago Ok. I understand that the scenario is imposible. According to previos vídeos there only could be 3 possible solutions graphing the system of equations. The only case with no solutions is represented by two parallel lines.