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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?

Going to the movies

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

Chickens and dogs in the farm yard

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## RWM102: Algebra

Solving linear word problems with substitution.

Watch this video for examples of these types of problems. Recall the five step method we discussed in section 3.3. When you solve these types of problems, the first key step is to identify the variable.

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Word Problems Worksheet 1 – This 6 problem algebra worksheet will help you practice solving real-life systems of equations problems using the “ substitution ” method.  All of the coefficients and answers are positive integers. Word Problems Worksheet 1 RTF Word Problems Worksheet 1 PDF View Answers

Word Problems Worksheet 3 – This 6 problem algebra worksheet will help you practice solving real life systems of equations problems using the “ substitution ” method. Most of the problems involve money and require the use of the distributive property . Word Problems Worksheet 3 RTF Word Problems Worksheet 3 PDF View Answers

Word Problems Worksheet 4 – This 6 problem algebra worksheet will help you practice solving real life systems of equations problems using the “ substitution ” method. Most of the problems involve money. A few twists are incorporated, so read carefully! Word Problems Worksheet 4 RTF Word Problems Worksheet 4 PDF View Answers

Word Problems Worksheet 5 – This 8 problem algebra worksheet features more abstract word problems like “ The sum of x and y is 42.  y is 12 less than x.  Find x and y .”  There are quite a few negative integers, so be careful! Word Problems Worksheet 5 RTF Word Problems Worksheet 5 PDF View Answers

Word Problems Worksheet 6 – This 8 problem algebra worksheet features more abstract word problems like “ The sum of  three times a number, x, and half of another number, y, is 96.  y is 60 less than x.  Find x and y. ” One of the problems even has an infinite number of solutions! Word Problems Worksheet 6 RTF Word Problems Worksheet 6 PDF View Answers

These free  systems of equations   worksheets  will help you practice solving real-life systems of equations using the “ substitution ” method.  You will  need to create and solve a system of equations to represent each situation.  The exercises can also be solved using other algebraic methods if you choose.

This is a progressive series that starts simple with problems involving buying pizza and figuring out ages.  Eventually the distributive property and “ the opposite of x ” come into play.  Watch out for those perimeter problems!

Each worksheet will help students master Common Core skills in the Algebra strand.  They are great for ambitious students in pre-algebra or algebra classes.

These free  substitution  worksheets are printable and available in a variety of formats.  Each sheet includes an example to help you get started.  Of course, answer keys are provided as well.

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## 1.20: Word Problems for Linear Equations

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Word problems are important applications of linear equations. We start with examples of translating an English sentence or phrase into an algebraic expression.

Example 18.1

Translate the phrase into an algebraic expression:

a) Twice a variable is added to 4

Solution: We call the variable $$x .$$ Twice the variable is $$2 x .$$ Adding $$2 x$$ to 4 gives:

$4 + 2x\nonumber$

b) Three times a number is subtracted from 7.

Solution: Three times a number is $$3 x .$$ We need to subtract $$3 x$$ from 7. This means:\

$7-3 x\nonumber$

c) 8 less than a number.

Solution: The number is denoted by $$x .8$$ less than $$x$$ mean, that we need to subtract 8 from it. We get:

$x-8\nonumber$

For example, 8 less than 10 is $$10-8=2$$.

d) Subtract $$5 p^{2}-7 p+2$$ from $$3 p^{2}+4 p$$ and simplify.

Solution: We need to calculate $$3 p^{2}+4 p$$ minus $$5 p^{2}-7 p+2:$$

$\left(3 p^{2}+4 p\right)-\left(5 p^{2}-7 p+2\right)\nonumber$

Simplifying this expression gives:

$\left(3 p^{2}+4 p\right)-\left(5 p^{2}-7 p+2\right)=3 p^{2}+4 p-5 p^{2}+7 p-2 =-2 p^{2}+11 p-2\nonumber$

e) The amount of money given by $$x$$ dimes and $$y$$ quarters.

Solution: Each dime is worth 10 cents, so that this gives a total of $$10 x$$ cents. Each quarter is worth 25 cents, so that this gives a total of $$25 y$$ cents. Adding the two amounts gives a total of

$10 x+25 y \text{ cents or } .10x + .25y \text{ dollars}\nonumber$

Now we deal with word problems that directly describe an equation involving one variable, which we can then solve.

Example 18.2

Solve the following word problems:

a) Five times an unknown number is equal to 60. Find the number.

Solution: We translate the problem to algebra:

$5x = 60\nonumber$

We solve this for $$x$$ :

$x=\frac{60}{5}=12\nonumber$

b) If 5 is subtracted from twice an unknown number, the difference is $$13 .$$ Find the number.

Solution: Translating the problem into an algebraic equation gives:

$2x − 5 = 13\nonumber$

We solve this for $$x$$. First, add 5 to both sides.

$2x = 13 + 5, \text{ so that } 2x = 18\nonumber$

Dividing by 2 gives $$x=\frac{18}{2}=9$$.

c) A number subtracted from 9 is equal to 2 times the number. Find the number.

Solution: We translate the problem to algebra.

$9 − x = 2x\nonumber$

We solve this as follows. First, add $$x$$ :

$9 = 2x + x \text{ so that } 9 = 3x\nonumber$

Then the answer is $$x=\frac{9}{3}=3$$

d) Multiply an unknown number by five is equal to adding twelve to the unknown number. Find the number.

Solution: We have the equation:

$5x = x + 12.\nonumber$

Subtracting $$x$$ gives

$4x = 12.\nonumber$

Dividing both sides by 4 gives the answer: $$x=3$$.

e) Adding nine to a number gives the same result as subtracting seven from three times the number. Find the number.

Solution: Adding 9 to a number is written as $$x+9,$$ while subtracting 7 from three times the number is written as $$3 x-7$$. We therefore get the equation:

$x + 9 = 3x − 7.\nonumber$

We solve for $$x$$ by adding 7 on both sides of the equation:

$x + 16 = 3x.\nonumber$

Then we subtract $$x:$$

$16 = 2x.\nonumber$

After dividing by $$2,$$ we obtain the answer $$x=8$$

The following word problems consider real world applications. They require to model a given situation in the form of an equation.

Example 18.3

a) Due to inflation, the price of a loaf of bread has increased by $$5 \%$$. How much does the loaf of bread cost now, when its price was $$\ 2.40$$ last year?

Solution: We calculate the price increase as $$5 \% \cdot \ 2.40 .$$ We have

$5 \% \cdot 2.40=0.05 \cdot 2.40=0.1200=0.12\nonumber$

We must add the price increase to the old price.

$2.40+0.12=2.52\nonumber$

The new price is therefore $$\ 2.52$$.

b) To complete a job, three workers get paid at a rate of $$\ 12$$ per hour. If the total pay for the job was $$\ 180,$$ then how many hours did the three workers spend on the job?

Solution: We denote the number of hours by $$x$$. Then the total price is calculated as the price per hour $$(\ 12)$$ times the number of workers times the number of hours $$(3) .$$ We obtain the equation

$12 \cdot 3 \cdot x=180\nonumber$

Simplifying this yields

$36 x=180\nonumber$

Dividing by 36 gives

$x=\frac{180}{36}=5\nonumber$

Therefore, the three workers needed 5 hours for the job.

c) A farmer cuts a 300 foot fence into two pieces of different sizes. The longer piece should be four times as long as the shorter piece. How long are the two pieces?

$x+4 x=300\nonumber$

Combining the like terms on the left, we get

$5 x=300\nonumber$

Dividing by 5, we obtain that

$x=\frac{300}{5}=60\nonumber$

Therefore, the shorter piece has a length of 60 feet, while the longer piece has four times this length, that is $$4 \times 60$$ feet $$=240$$ feet.

d) If 4 blocks weigh 28 ounces, how many blocks weigh 70 ounces?

Solution: We denote the weight of a block by $$x .$$ If 4 blocks weigh $$28,$$ then a block weighs $$x=\frac{28}{4}=7$$

How many blocks weigh $$70 ?$$ Well, we only need to find $$\frac{70}{7}=10 .$$ So, the answer is $$10 .$$

Note You can solve this problem by setting up and solving the fractional equation $$\frac{28}{4}=\frac{70}{x}$$. Solving such equations is addressed in chapter 24.

e) If a rectangle has a length that is three more than twice the width and the perimeter is 20 in, what are the dimensions of the rectangle?

Solution: We denote the width by $$x$$. Then the length is $$2 x+3$$. The perimeter is 20 in on one hand and $$2($$length$$)+2($$width$$)$$ on the other. So we have

$20=2 x+2(2 x+3)\nonumber$

Distributing and collecting like terms give

$20=6 x+6\nonumber$

Subtracting 6 from both sides of the equation and then dividing both sides of the resulting equation by 6 gives:

$20-6=6 x \Longrightarrow 14=6 x \Longrightarrow x=\frac{14}{6} \text { in }=\frac{7}{3} \text { in }=2 \frac{1}{3} \text { in. }\nonumber$

f) If a circle has circumference 4in, what is its radius?

Solution: We know that $$C=2 \pi r$$ where $$C$$ is the circumference and $$r$$ is the radius. So in this case

$4=2 \pi r\nonumber$

Dividing both sides by $$2 \pi$$ gives

$r=\frac{4}{2 \pi}=\frac{2}{\pi} \text { in } \approx 0.63 \mathrm{in}\nonumber$

g) The perimeter of an equilateral triangle is 60 meters. How long is each side?

Solution: Let $$x$$ equal the side of the triangle. Then the perimeter is, on the one hand, $$60,$$ and on other hand $$3 x .$$ So $$3 x=60$$ and dividing both sides of the equation by 3 gives $$x=20$$ meters.

h) If a gardener has $$\ 600$$ to spend on a fence which costs $$\ 10$$ per linear foot and the area to be fenced in is rectangular and should be twice as long as it is wide, what are the dimensions of the largest fenced in area?

Solution: The perimeter of a rectangle is $$P=2 L+2 W$$. Let $$x$$ be the width of the rectangle. Then the length is $$2 x .$$ The perimeter is $$P=2(2 x)+2 x=6 x$$. The largest perimeter is $$\ 600 /(\ 10 / f t)=60$$ ft. So $$60=6 x$$ and dividing both sides by 6 gives $$x=60 / 6=10$$. So the dimensions are 10 feet by 20 feet.

i) A trapezoid has an area of 20.2 square inches with one base measuring 3.2 in and the height of 4 in. Find the length of the other base.

Solution: Let $$b$$ be the length of the unknown base. The area of the trapezoid is on the one hand 20.2 square inches. On the other hand it is $$\frac{1}{2}(3.2+b) \cdot 4=$$ $$6.4+2 b .$$ So

$20.2=6.4+2 b\nonumber$

Multiplying both sides by 10 gives

$202=64+20 b\nonumber$

Subtracting 64 from both sides gives

$b=\frac{138}{20}=\frac{69}{10}=6.9 \text { in }\nonumber$

and dividing by 20 gives

Exit Problem

Write an equation and solve: A car uses 12 gallons of gas to travel 100 miles. How many gallons would be needed to travel 450 miles?

## SOLVING WORD PROBLEMS USING SUBSTITUTION METHOD

In this section, you will learn how to solve linear linear equations in two variables using the concept substitution.

We use the following steps to solve a system of linear equations.

Solve one of the equations for one of its variables.

Substitute the expression from step 1 into the other equation and solve for the other variable.

Substitute the value from step 2 into either original equations and solve for the variable in step 1.

Problem 1 :

Solve 2x  + 3y  =  11 and 2x - 4y  =  -24 and hence find the value of "m" for which y  =  mx + 3.

2x + 3y  =  11 ----------- (1)

2x - 4y  =  -24 ----------- (2)

Find the value of one variable in terms of other variable

3y  =  11 - 2 x

y  =  (11 - 2 x)/3

Let us apply the value of y in (2),

2x - 4(11 - 2x)/3  =  -24

14x - 44  =  -24 (3)

14x - 44  =  -72

14x  =  -72 + 44

14x  =  - 28

Divide 14 on both sides, we get

x  =  -2

Substitute x  =  -2  in the equation y  =  (11 - 2 x)/3

y  =  [11 - 2(-2)]/3

y  =  15/3

Now we have to apply these values in the equation

y   =  m x + 3

5  =  m (-2) + 3

5  =  -2 m + 3

-2m  =  2

-2 m  =  2

m  =  2/(-2)

m  =  -1

Problem 2 :

Form the pair of linear equations of the following problems and find their solution by substitution method.

(i) The difference between two numbers is 26 and one number is three times the other. Find them

Let the two numbers are "x" and "y"

Difference between two number is 26

x - y  =  26 -------- (1)

One number is three times the other

x  =  3 y -------- (2)

Let us apply (2) in (1)

3 y - y  =  26

2 y  =  26

Divide by 2 on both sides, we get

y  =  13

By applying the value of y in (2), we get

x  =  3 (13)

x  =  39

So, required two numbers are 39 and 13.

Problem 3 :

The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them

Let the two supplementary angles are "x" and "y"

Sum of these two angles is 180

x + y  =  180 -----(1)

the larger angle exceeds the smaller by 18

x  =  y + 18 -----(2)

Now,we are going to apply the value of x in the first equation

y + 18 + y  =  180

2y  =  180 - 18

2y  =  162

y  =  162/2

y  =  81

x  =  81 + 18

x  =  99

So, two supplementary angles are 99 and 81.

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## Most Used Actions

Number line.

• substitution\:x+y+z=25,\:5x+3y+2z=0,\:y-z=6
• substitution\:x+2y=2x-5,\:x-y=3
• substitution\:5x+3y=7,\:3x-5y=-23
• substitution\:x+z=1,\:x+2z=4

substitution-system-of-equations-calculator

• High School Math Solutions – Systems of Equations Calculator, Elimination A system of equations is a collection of two or more equations with the same set of variables. In this blog post,...

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1. IXL

Solve a system of equations using substitution: word problems. IXL's SmartScore is a dynamic measure of progress towards mastery, rather than a percentage grade. It tracks your skill level as you tackle progressively more difficult questions. Consistently answer questions correctly to reach excellence (90), or conquer the Challenge Zone to ...

2. 5.2: Solve Systems of Equations by Substitution

Example 5.2.7. Solve the system by substitution. {3x + y = 5 2x + 4y = − 10. Solution. We need to solve one equation for one variable. Then we will substitute that expression into the other equation. Solve for y. Substitute into the other equation. Replace the y with −3 x + 5. Solve the resulting equation for x.

3. Systems of equations with substitution

You might need: Calculator. Solve the system of equations. 5 x − 7 y = 58 y = − x + 2. x =. y =. Learn for free about math, art, computer programming, economics, physics, chemistry, biology, medicine, finance, history, and more. Khan Academy is a nonprofit with the mission of providing a free, world-class education for anyone, anywhere.

4. Systems of Linear Equations Word Problems: Substitution

This two-page algebra worksheet will give students valuable practice writing equations to model real-world problems and solving systems of equations using substitution. For more practice solving systems of linear equations word problems, students can also complete the Systems of Linear Equations Word Problems: Graphing and Systems of Linear ...

5. Systems of equations word problems

Systems of equations word problems. Google Classroom. Microsoft Teams. You might need: Calculator. Malcolm and Ravi raced each other. The average of their maximum speeds was 260 km/h . If doubled, Malcolm's maximum speed would be 80 km/h more than Ravi's maximum speed. What were Malcolm's and Ravi's maximum speeds?

6. Substitution Method Practice Problems With Answers

Do you want to learn how to solve systems of equations using the substitution method? Check out this webpage for ten (10) practice problems with detailed answers and explanations. You will also find links to other related topics in intermediate algebra, such as rational inequalities, distance formula, graphing a line, and literal equations.

7. Solving Systems of Equations Word Problems

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!

8. Substitution method review (systems of equations)

In order to use the substitution method, we'll need to solve for either x or y in one of the equations. Let's solve for y in the second equation: − 2 x + y = 9 y = 2 x + 9. Now we can substitute the expression 2 x + 9 in for y in the first equation of our system: 7 x + 10 y = 36 7 x + 10 ( 2 x + 9) = 36 7 x + 20 x + 90 = 36 27 x + 90 = 36 3 x ...

9. Using Systems of Equations to Solve Word Problems

https://www.patreon.com/ProfessorLeonardUsing the Substitution Method to solve some word problems and explore how systems of linear equations give us a diffe...

10. RWM102: Solving Linear Word Problems with Substitution

Solving Linear Word Problems with Substitution. Watch this video for examples of these types of problems. Recall the five step method we discussed in section 3.3. When you solve these types of problems, the first key step is to identify the variable.

11. PDF Systems of Equations Word Problems

Systems of Equations Word Problems Date_____ Period____ 1) Find the value of two numbers if their sum is 12 and their difference is 4. 4 and 8 2) The difference of two numbers is 3. Their sum is 13. Find the numbers. 5 and 8 3) Flying to Kampala with a tailwind a plane averaged 158 km/h. On the return trip the plane only

12. Substitution Worksheets

Systems of Equations Substitution Worksheets. Substitution Worksheets; Word Problems Worksheets; These free systems of equations worksheets will help you practice solving real-life systems of equations using the "substitution" method. Each set of free algebra worksheets is a progressive series that starts with simple problems featuring positive integers.

13. Solving a System of Linear Equations Using Substitution in a Word

Step 1: Define your variables. Step 2: Write equations with the variables that represent the scenario in the word problem. Step 3: Pick one of the equations and solve it for one of the variables ...

14. Word Problems

Word Problems. Word Problems Worksheet 1 - This 6 problem algebra worksheet will help you practice solving real-life systems of equations problems using the " substitution " method. All of the coefficients and answers are positive integers. Word Problems Worksheet 1 RTF. Word Problems Worksheet 1 PDF. View Answers.

15. Systems of equations with substitution: coins

You then have an equation with a single variable to find. If you use substitution method, you solve one of the equations for a single variable. For example, change K+L=450 into K=450-L. You can then use the value of "k" to substitute into the other equation. The substitution forces "k" out of the equation leaving you with a single variable to find.

16. Systems of Linear Equations and Word Problems

Systems with Three Equations. The "Candy" Problem. Algebra Word Problems with Systems: Right Triangle Trigonometry Systems Problem. Investment Word Problem. Inequality Word Problem (in Linear Programming section) Mixture Word Problems. More Practice. Note that we saw how to solve linear inequalities here in the Coordinate System and ...

17. 1.20: Word Problems for Linear Equations

Solution: Translating the problem into an algebraic equation gives: 2x − 5 = 13 2 x − 5 = 13. We solve this for x x. First, add 5 to both sides. 2x = 13 + 5, so that 2x = 18 2 x = 13 + 5, so that 2 x = 18. Dividing by 2 gives x = 182 = 9 x = 18 2 = 9. c) A number subtracted from 9 is equal to 2 times the number.

18. Solving Word Problems Using Substitution Method

Problem 2 : Form the pair of linear equations of the following problems and find their solution by substitution method. (i) The difference between two numbers is 26 and one number is three times the other. Find them. Solution : Let the two numbers are "x" and "y". Difference between two number is 26. x - y = 26 -------- (1)

19. Mr Barton Maths

Simultaneous Equations - Word Problems : 1: 2: 3: Simultaneous Equations - Graphical ... Simultaneous Equations - Linear and Non-Linear : 1: 2: 3: Corbett Maths keyboard_arrow_up. ... Simultaneous equations (substitution, both linear) Questions: Simultaneous equations (linear and non-linear) Questions: Solutions . If you find ...

20. Linear equations word problems

Linear equations word problems. Google Classroom. Microsoft Teams. Andrei wants to fill a glass tank with spherical marbles and then fill the remaining space with water. The variable w models the amount of water (in liters) Andrei uses if he uses n marbles. w = 32 − 0.05 n. What is the volume of each marble?

21. System of Equations Substitution Calculator

Free system of equations substitution calculator - solve system of equations using substitution method step-by-step ... (Product) Notation Induction Logical Sets Word Problems. Pre Calculus. ... In a previous post, we learned about how to solve a system of linear equations. In this post, we will learn how... Enter a problem. Cooking Calculators.

22. Linear simultaneous equations

Systems of equations with substitution: -3x-4y=-2 & y=2x-5 (Opens a modal) ... Word problems: Writing equations reducible to linear form Get 3 of 4 questions to level up! Quiz 3. Level up on the above skills and collect up to 320 Mastery points Start quiz. Linear equations word problems. Learn. Age word problem: Ben & William (Opens a modal ...