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Mathematics LibreTexts

5.2: Solve Systems of Equations by Substitution

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Learning Objectives

By the end of this section, you will be able to:

  • Solve a system of equations by substitution
  • Solve applications of systems of equations by substitution

Before you get started, take this readiness quiz.

  • Simplify −5(3−x). If you missed this problem, review Example 1.10.43 .
  • Simplify 4−2(n+5). If you missed this problem, review Example 1.10.49 .
  • Solve for y. 8y−8=32−2y If you missed this problem, review Example 2.3.22 .
  • Solve for x. 3x−9y=−3 If you missed this problem, review Example 2.6.22 .

Solving systems of linear equations by graphing is a good way to visualize the types of solutions that may result. However, there are many cases where solving a system by graphing is inconvenient or imprecise. If the graphs extend beyond the small grid with x and y both between −10 and 10, graphing the lines may be cumbersome. And if the solutions to the system are not integers, it can be hard to read their values precisely from a graph.

In this section, we will solve systems of linear equations by the substitution method.

Solve a System of Equations by Substitution

We will use the same system we used first for graphing.

\(\left\{\begin{array}{l}{2 x+y=7} \\ {x-2 y=6}\end{array}\right.\)

We will first solve one of the equations for either x or y . We can choose either equation and solve for either variable—but we’ll try to make a choice that will keep the work easy.

Then we substitute that expression into the other equation. The result is an equation with just one variable—and we know how to solve those!

After we find the value of one variable, we will substitute that value into one of the original equations and solve for the other variable. Finally, we check our solution and make sure it makes both equations true.

We’ll fill in all these steps now in Example \(\PageIndex{1}\).

Example \(\PageIndex{1}\): How to Solve a System of Equations by Substitution

Solve the system by substitution. \(\left\{\begin{array}{l}{2 x+y=7} \\ {x-2 y=6}\end{array}\right.\)

This figure has three columns and six rows. The first row says, “Step 1. Solve one of the equations for either variable.” To the right of this, the middl row reads, “We’ll solve the first equation for y.” The third column shows the two equations: 2x + y = 7 and x – 2y = 6. It shows that 2x + y = 7 becomes y = 7 – 2x.

Try It \(\PageIndex{2}\)

Solve the system by substitution. \(\left\{\begin{array}{l}{-2 x+y=-11} \\ {x+3 y=9}\end{array}\right.\)

Try It \(\PageIndex{3}\)

Solve the system by substitution. \(\left\{\begin{array}{l}{x+3 y=10} \\ {4 x+y=18}\end{array}\right.\)

SOLVE A SYSTEM OF EQUATIONS BY SUBSTITUTION.

  • Solve one of the equations for either variable.
  • Substitute the expression from Step 1 into the other equation.
  • Solve the resulting equation.
  • Substitute the solution in Step 3 into one of the original equations to find the other variable.
  • Write the solution as an ordered pair.
  • Check that the ordered pair is a solution to both original equations.

If one of the equations in the system is given in slope–intercept form, Step 1 is already done! We’ll see this in Example \(\PageIndex{4}\).

Example \(\PageIndex{4}\)

Solve the system by substitution. \(\left\{\begin{array}{l}{x+y=-1} \\ {y=x+5}\end{array}\right.\)

The second equation is already solved for y . We will substitute the expression in place of y in the first equation.

Try It \(\PageIndex{5}\)

Solve the system by substitution. \(\left\{\begin{array}{l}{x+y=6} \\ {y=3 x-2}\end{array}\right.\)

Try It \(\PageIndex{6}\)

Solve the system by substitution. \(\left\{\begin{array}{l}{2 x-y=1} \\ {y=-3 x-6}\end{array}\right.\)

(−1,−3)

If the equations are given in standard form, we’ll need to start by solving for one of the variables. In this next example, we’ll solve the first equation for y .

Example \(\PageIndex{7}\)

Solve the system by substitution. \(\left\{\begin{array}{l}{3 x+y=5} \\ {2 x+4 y=-10}\end{array}\right.\)

We need to solve one equation for one variable. Then we will substitute that expression into the other equation.

Try It \(\PageIndex{8}\)

Solve the system by substitution. \(\left\{\begin{array}{l}{4 x+y=2} \\ {3 x+2 y=-1}\end{array}\right.\)

(1,−2)

Try It \(\PageIndex{9}\)

Solve the system by substitution. \(\left\{\begin{array}{l}{-x+y=4} \\ {4 x-y=2}\end{array}\right.\)

In Example \(\PageIndex{7}\) it was easiest to solve for y in the first equation because it had a coefficient of 1. In Example \(\PageIndex{10}\) it will be easier to solve for x .

Example \(\PageIndex{10}\)

Solve the system by substitution. \(\left\{\begin{array}{l}{x-2 y=-2} \\ {3 x+2 y=34}\end{array}\right.\)

We will solve the first equation for \(x\) and then substitute the expression into the second equation.

Try It \(\PageIndex{11}\)

Solve the system by substitution. \(\left\{\begin{array}{l}{x-5 y=13} \\ {4 x-3 y=1}\end{array}\right.\)

(−2,−3)

Try It \(\PageIndex{12}\)

Solve the system by substitution. \(\left\{\begin{array}{l}{x-6 y=-6} \\ {2 x-4 y=4}\end{array}\right.\)

When both equations are already solved for the same variable, it is easy to substitute!

Example \(\PageIndex{13}\)

Solve the system by substitution. \(\left\{\begin{array}{l}{y=-2 x+5} \\ {y=\frac{1}{2} x}\end{array}\right.\)

Since both equations are solved for y , we can substitute one into the other.

Try It \(\PageIndex{14}\)

Solve the system by substitution. \(\left\{\begin{array}{l}{y=3 x-16} \\ {y=\frac{1}{3} x}\end{array}\right.\)

Try It \(\PageIndex{15}\)

Solve the system by substitution. \(\left\{\begin{array}{l}{y=-x+10} \\ {y=\frac{1}{4} x}\end{array}\right.\)

Be very careful with the signs in the next example.

Example \(\PageIndex{16}\)

Solve the system by substitution. \(\left\{\begin{array}{l}{4 x+2 y=4} \\ {6 x-y=8}\end{array}\right.\)

We need to solve one equation for one variable. We will solve the first equation for y .

Try It \(\PageIndex{17}\)

Solve the system by substitution. \(\left\{\begin{array}{l}{x-4 y=-4} \\ {-3 x+4 y=0}\end{array}\right.\)

\((2,\frac{3}{2})\)

Try It \(\PageIndex{18}\)

Solve the system by substitution. \(\left\{\begin{array}{l}{4 x-y=0} \\ {2 x-3 y=5}\end{array}\right.\)

\((−\frac{1}{2},−2)\)

In Example , it will take a little more work to solve one equation for x or y .

Example \(\PageIndex{19}\)

Solve the system by substitution. \(\left\{\begin{array}{l}{4 x-3 y=6} \\ {15 y-20 x=-30}\end{array}\right.\)

We need to solve one equation for one variable. We will solve the first equation for x .

Try It \(\PageIndex{20}\)

Solve the system by substitution. \(\left\{\begin{array}{l}{2 x-3 y=12} \\ {-12 y+8 x=48}\end{array}\right.\)

infinitely many solutions

Try It \(\PageIndex{21}\)

Solve the system by substitution. \(\left\{\begin{array}{l}{5 x+2 y=12} \\ {-4 y-10 x=-24}\end{array}\right.\)

Look back at the equations in Example \(\PageIndex{22}\). Is there any way to recognize that they are the same line?

Let’s see what happens in the next example.

Example \(\PageIndex{22}\)

Solve the system by substitution. \(\left\{\begin{array}{l}{5 x-2 y=-10} \\ {y=\frac{5}{2} x}\end{array}\right.\)

The second equation is already solved for y , so we can substitute for y in the first equation.

Try It \(\PageIndex{23}\)

Solve the system by substitution. \(\left\{\begin{array}{l}{3 x+2 y=9} \\ {y=-\frac{3}{2} x+1}\end{array}\right.\)

no solution

Try It \(\PageIndex{24}\)

Solve the system by substitution. \(\left\{\begin{array}{l}{5 x-3 y=2} \\ {y=\frac{5}{3} x-4}\end{array}\right.\)

Solve Applications of Systems of Equations by Substitution

We’ll copy here the problem solving strategy we used in the Solving Systems of Equations by Graphing section for solving systems of equations. Now that we know how to solve systems by substitution, that’s what we’ll do in Step 5.

HOW TO USE A PROBLEM SOLVING STRATEGY FOR SYSTEMS OF LINEAR EQUATIONS.

  • Read the problem. Make sure all the words and ideas are understood.
  • Identify what we are looking for.
  • Name what we are looking for. Choose variables to represent those quantities.
  • Translate into a system of equations.
  • Solve the system of equations using good algebra techniques.
  • Check the answer in the problem and make sure it makes sense.
  • Answer the question with a complete sentence.

Some people find setting up word problems with two variables easier than setting them up with just one variable. Choosing the variable names is easier when all you need to do is write down two letters. Think about this in the next example—how would you have done it with just one variable?

Example \(\PageIndex{25}\)

The sum of two numbers is zero. One number is nine less than the other. Find the numbers.

Try It \(\PageIndex{26}\)

The sum of two numbers is 10. One number is 4 less than the other. Find the numbers.

The numbers are 3 and 7.

Try It \(\PageIndex{27}\)

The sum of two number is −6. One number is 10 less than the other. Find the numbers.

The numbers are 2 and −8.

In the Example \(\PageIndex{28}\), we’ll use the formula for the perimeter of a rectangle, P = 2 L + 2 W .

Example \(\PageIndex{28}\)

The perimeter of a rectangle is 88. The length is five more than twice the width. Find the length and width of the rectangle.

Try It \(\PageIndex{29}\)

The perimeter of a rectangle is 40. The length is 4 more than the width. Find the length and width of the rectangle.

The length is 12 and the width is 8.

Try It \(\PageIndex{30}\)

The perimeter of a rectangle is 58. The length is 5 more than three times the width. Find the length and width of the rectangle.

The length is 23 and the width is 6.

For Example \(\PageIndex{31}\) we need to remember that the sum of the measures of the angles of a triangle is 180 degrees and that a right triangle has one 90 degree angle.

Example \(\PageIndex{31}\)

The measure of one of the small angles of a right triangle is ten more than three times the measure of the other small angle. Find the measures of both angles.

We will draw and label a figure.

Try It \(\PageIndex{32}\)

The measure of one of the small angles of a right triangle is 2 more than 3 times the measure of the other small angle. Find the measure of both angles.

The measure of the angles are 22 degrees and 68 degrees.

Try It \(\PageIndex{33}\)

The measure of one of the small angles of a right triangle is 18 less than twice the measure of the other small angle. Find the measure of both angles.

The measure of the angles are 36 degrees and 54 degrees.

Example \(\PageIndex{34}\)

Heather has been offered two options for her salary as a trainer at the gym. Option A would pay her $25,000 plus $15 for each training session. Option B would pay her $10,000 + $40 for each training session. How many training sessions would make the salary options equal?

Try It \(\PageIndex{35}\)

Geraldine has been offered positions by two insurance companies. The first company pays a salary of $12,000 plus a commission of $100 for each policy sold. The second pays a salary of $20,000 plus a commission of $50 for each policy sold. How many policies would need to be sold to make the total pay the same?

There would need to be 160 policies sold to make the total pay the same.

Try It \(\PageIndex{36}\)

Kenneth currently sells suits for company A at a salary of $22,000 plus a $10 commission for each suit sold. Company B offers him a position with a salary of $28,000 plus a $4 commission for each suit sold. How many suits would Kenneth need to sell for the options to be equal?

Kenneth would need to sell 1,000 suits.

Access these online resources for additional instruction and practice with solving systems of equations by substitution.

  • Instructional Video-Solve Linear Systems by Substitution
  • Instructional Video-Solve by Substitution

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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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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! 

5.2 Solving Systems of Equations by Substitution

Learning objectives.

By the end of this section, you will be able to:

  • Solve a system of equations by substitution
  • Solve applications of systems of equations by substitution

Be Prepared 5.4

Before you get started, take this readiness quiz.

Simplify −5 ( 3 − x ) −5 ( 3 − x ) . If you missed this problem, review Example 1.136 .

Be Prepared 5.5

Simplify 4 − 2 ( n + 5 ) 4 − 2 ( n + 5 ) . If you missed this problem, review Example 1.123 .

Be Prepared 5.6

Solve for y y : 8 y − 8 = 32 − 2 y 8 y − 8 = 32 − 2 y If you missed this problem, review Example 2.34 .

Be Prepared 5.7

Solve for x x : 3 x − 9 y = −3 3 x − 9 y = −3 If you missed this problem, review Example 2.65 .

Solving systems of linear equations by graphing is a good way to visualize the types of solutions that may result. However, there are many cases where solving a system by graphing is inconvenient or imprecise. If the graphs extend beyond the small grid with x and y both between −10 and 10, graphing the lines may be cumbersome. And if the solutions to the system are not integers, it can be hard to read their values precisely from a graph.

In this section, we will solve systems of linear equations by the substitution method.

Solve a System of Equations by Substitution

We will use the same system we used first for graphing.

We will first solve one of the equations for either x or y . We can choose either equation and solve for either variable—but we’ll try to make a choice that will keep the work easy.

Then we substitute that expression into the other equation. The result is an equation with just one variable—and we know how to solve those!

After we find the value of one variable, we will substitute that value into one of the original equations and solve for the other variable. Finally, we check our solution and make sure it makes both equations true.

We’ll fill in all these steps now in Example 5.13 .

Example 5.13

How to solve a system of equations by substitution.

Solve the system by substitution. { 2 x + y = 7 x − 2 y = 6 { 2 x + y = 7 x − 2 y = 6

Try It 5.25

Solve the system by substitution. { −2 x + y = −11 x + 3 y = 9 { −2 x + y = −11 x + 3 y = 9

Try It 5.26

Solve the system by substitution. { x + 3 y = 10 4 x + y = 18 { x + 3 y = 10 4 x + y = 18

Solve a system of equations by substitution.

  • Step 1. Solve one of the equations for either variable.
  • Step 2. Substitute the expression from Step 1 into the other equation.
  • Step 3. Solve the resulting equation.
  • Step 4. Substitute the solution in Step 3 into one of the original equations to find the other variable.
  • Step 5. Write the solution as an ordered pair.
  • Step 6. Check that the ordered pair is a solution to both original equations.

If one of the equations in the system is given in slope–intercept form, Step 1 is already done! We’ll see this in Example 5.14 .

Example 5.14

Solve the system by substitution.

{ x + y = −1 y = x + 5 { x + y = −1 y = x + 5

The second equation is already solved for y . We will substitute the expression in place of y in the first equation.

Try It 5.27

Solve the system by substitution. { x + y = 6 y = 3 x − 2 { x + y = 6 y = 3 x − 2

Try It 5.28

Solve the system by substitution. { 2 x − y = 1 y = −3 x − 6 { 2 x − y = 1 y = −3 x − 6

If the equations are given in standard form, we’ll need to start by solving for one of the variables. In this next example, we’ll solve the first equation for y .

Example 5.15

Solve the system by substitution. { 3 x + y = 5 2 x + 4 y = −10 { 3 x + y = 5 2 x + 4 y = −10

We need to solve one equation for one variable. Then we will substitute that expression into the other equation.

Try It 5.29

Solve the system by substitution. { 4 x + y = 2 3 x + 2 y = −1 { 4 x + y = 2 3 x + 2 y = −1

Try It 5.30

Solve the system by substitution. { − x + y = 4 4 x − y = 2 { − x + y = 4 4 x − y = 2

In Example 5.15 it was easiest to solve for y in the first equation because it had a coefficient of 1. In Example 5.16 it will be easier to solve for x .

Example 5.16

Solve the system by substitution. { x − 2 y = −2 3 x + 2 y = 34 { x − 2 y = −2 3 x + 2 y = 34

We will solve the first equation for x x and then substitute the expression into the second equation.

Try It 5.31

Solve the system by substitution. { x − 5 y = 13 4 x − 3 y = 1 { x − 5 y = 13 4 x − 3 y = 1

Try It 5.32

Solve the system by substitution. { x − 6 y = −6 2 x − 4 y = 4 { x − 6 y = −6 2 x − 4 y = 4

When both equations are already solved for the same variable, it is easy to substitute!

Example 5.17

Solve the system by substitution. { y = −2 x + 5 y = 1 2 x { y = −2 x + 5 y = 1 2 x

Since both equations are solved for y , we can substitute one into the other.

Try It 5.33

Solve the system by substitution. { y = 3 x − 16 y = 1 3 x { y = 3 x − 16 y = 1 3 x

Try It 5.34

Solve the system by substitution. { y = − x + 10 y = 1 4 x { y = − x + 10 y = 1 4 x

Be very careful with the signs in the next example.

Example 5.18

Solve the system by substitution. { 4 x + 2 y = 4 6 x − y = 8 { 4 x + 2 y = 4 6 x − y = 8

We need to solve one equation for one variable. We will solve the first equation for y .

Try It 5.35

Solve the system by substitution. { x − 4 y = −4 −3 x + 4 y = 0 { x − 4 y = −4 −3 x + 4 y = 0

Try It 5.36

Solve the system by substitution. { 4 x − y = 0 2 x − 3 y = 5 { 4 x − y = 0 2 x − 3 y = 5

In Example 5.19 , it will take a little more work to solve one equation for x or y .

Example 5.19

Solve the system by substitution. { 4 x − 3 y = 6 15 y − 20 x = −30 { 4 x − 3 y = 6 15 y − 20 x = −30

We need to solve one equation for one variable. We will solve the first equation for x .

Since 0 = 0 is a true statement, the system is consistent. The equations are dependent. The graphs of these two equations would give the same line. The system has infinitely many solutions.

Try It 5.37

Solve the system by substitution. { 2 x − 3 y = 12 −12 y + 8 x = 48 { 2 x − 3 y = 12 −12 y + 8 x = 48

Try It 5.38

Solve the system by substitution. { 5 x + 2 y = 12 −4 y − 10 x = −24 { 5 x + 2 y = 12 −4 y − 10 x = −24

Look back at the equations in Example 5.19 . Is there any way to recognize that they are the same line?

Let’s see what happens in the next example.

Example 5.20

Solve the system by substitution. { 5 x − 2 y = −10 y = 5 2 x { 5 x − 2 y = −10 y = 5 2 x

The second equation is already solved for y , so we can substitute for y in the first equation.

Since 0 = −10 is a false statement the equations are inconsistent. The graphs of the two equation would be parallel lines. The system has no solutions.

Try It 5.39

Solve the system by substitution. { 3 x + 2 y = 9 y = − 3 2 x + 1 { 3 x + 2 y = 9 y = − 3 2 x + 1

Try It 5.40

Solve the system by substitution. { 5 x − 3 y = 2 y = 5 3 x − 4 { 5 x − 3 y = 2 y = 5 3 x − 4

Solve Applications of Systems of Equations by Substitution

We’ll copy here the problem solving strategy we used in the Solving Systems of Equations by Graphing section for solving systems of equations. Now that we know how to solve systems by substitution, that’s what we’ll do in Step 5.

How to use a problem solving strategy for systems of linear equations.

  • Step 1. Read the problem. Make sure all the words and ideas are understood.
  • Step 2. Identify what we are looking for.
  • Step 3. Name what we are looking for. Choose variables to represent those quantities.
  • Step 4. Translate into a system of equations.
  • Step 5. Solve the system of equations using good algebra techniques.
  • Step 6. Check the answer in the problem and make sure it makes sense.
  • Step 7. Answer the question with a complete sentence.

Some people find setting up word problems with two variables easier than setting them up with just one variable. Choosing the variable names is easier when all you need to do is write down two letters. Think about this in the next example—how would you have done it with just one variable?

Example 5.21

The sum of two numbers is zero. One number is nine less than the other. Find the numbers.

Try It 5.41

The sum of two numbers is 10. One number is 4 less than the other. Find the numbers.

Try It 5.42

The sum of two number is −6. One number is 10 less than the other. Find the numbers.

In the Example 5.22 , we’ll use the formula for the perimeter of a rectangle, P = 2 L + 2 W .

Example 5.22

The perimeter of a rectangle is 88. The length is five more than twice the width. Find the length and the width.

Try It 5.43

The perimeter of a rectangle is 40. The length is 4 more than the width. Find the length and width of the rectangle.

Try It 5.44

The perimeter of a rectangle is 58. The length is 5 more than three times the width. Find the length and width of the rectangle.

For Example 5.23 we need to remember that the sum of the measures of the angles of a triangle is 180 degrees and that a right triangle has one 90 degree angle.

Example 5.23

The measure of one of the small angles of a right triangle is ten more than three times the measure of the other small angle. Find the measures of both angles.

We will draw and label a figure.

Try It 5.45

The measure of one of the small angles of a right triangle is 2 more than 3 times the measure of the other small angle. Find the measure of both angles.

Try It 5.46

The measure of one of the small angles of a right triangle is 18 less than twice the measure of the other small angle. Find the measure of both angles.

Example 5.24

Heather has been offered two options for her salary as a trainer at the gym. Option A would pay her $25,000 plus $15 for each training session. Option B would pay her $10,000 + $40 for each training session. How many training sessions would make the salary options equal?

Try It 5.47

Geraldine has been offered positions by two insurance companies. The first company pays a salary of $12,000 plus a commission of $100 for each policy sold. The second pays a salary of $20,000 plus a commission of $50 for each policy sold. How many policies would need to be sold to make the total pay the same?

Try It 5.48

Kenneth currently sells suits for company A at a salary of $22,000 plus a $10 commission for each suit sold. Company B offers him a position with a salary of $28,000 plus a $4 commission for each suit sold. How many suits would Kenneth need to sell for the options to be equal?

Access these online resources for additional instruction and practice with solving systems of equations by substitution.

  • Instructional Video-Solve Linear Systems by Substitution
  • Instructional Video-Solve by Substitution

Section 5.2 Exercises

Practice makes perfect.

In the following exercises, solve the systems of equations by substitution.

{ 2 x + y = −4 3 x − 2 y = −6 { 2 x + y = −4 3 x − 2 y = −6

{ 2 x + y = −2 3 x − y = 7 { 2 x + y = −2 3 x − y = 7

{ x − 2 y = −5 2 x − 3 y = −4 { x − 2 y = −5 2 x − 3 y = −4

{ x − 3 y = −9 2 x + 5 y = 4 { x − 3 y = −9 2 x + 5 y = 4

{ 5 x − 2 y = −6 y = 3 x + 3 { 5 x − 2 y = −6 y = 3 x + 3

{ −2 x + 2 y = 6 y = −3 x + 1 { −2 x + 2 y = 6 y = −3 x + 1

{ 2 x + 3 y = 3 y = − x + 3 { 2 x + 3 y = 3 y = − x + 3

{ 2 x + 5 y = −14 y = −2 x + 2 { 2 x + 5 y = −14 y = −2 x + 2

{ 2 x + 5 y = 1 y = 1 3 x − 2 { 2 x + 5 y = 1 y = 1 3 x − 2

{ 3 x + 4 y = 1 y = − 2 5 x + 2 { 3 x + 4 y = 1 y = − 2 5 x + 2

{ 3 x − 2 y = 6 y = 2 3 x + 2 { 3 x − 2 y = 6 y = 2 3 x + 2

{ −3 x − 5 y = 3 y = 1 2 x − 5 { −3 x − 5 y = 3 y = 1 2 x − 5

{ 2 x + y = 10 − x + y = −5 { 2 x + y = 10 − x + y = −5

{ −2 x + y = 10 − x + 2 y = 16 { −2 x + y = 10 − x + 2 y = 16

{ 3 x + y = 1 −4 x + y = 15 { 3 x + y = 1 −4 x + y = 15

{ x + y = 0 2 x + 3 y = −4 { x + y = 0 2 x + 3 y = −4

{ x + 3 y = 1 3 x + 5 y = −5 { x + 3 y = 1 3 x + 5 y = −5

{ x + 2 y = −1 2 x + 3 y = 1 { x + 2 y = −1 2 x + 3 y = 1

{ 2 x + y = 5 x − 2 y = −15 { 2 x + y = 5 x − 2 y = −15

{ 4 x + y = 10 x − 2 y = −20 { 4 x + y = 10 x − 2 y = −20

{ y = −2 x − 1 y = − 1 3 x + 4 { y = −2 x − 1 y = − 1 3 x + 4

{ y = x − 6 y = − 3 2 x + 4 { y = x − 6 y = − 3 2 x + 4

{ y = 2 x − 8 y = 3 5 x + 6 { y = 2 x − 8 y = 3 5 x + 6

{ y = − x − 1 y = x + 7 { y = − x − 1 y = x + 7

{ 4 x + 2 y = 8 8 x − y = 1 { 4 x + 2 y = 8 8 x − y = 1

{ − x − 12 y = −1 2 x − 8 y = −6 { − x − 12 y = −1 2 x − 8 y = −6

{ 15 x + 2 y = 6 −5 x + 2 y = −4 { 15 x + 2 y = 6 −5 x + 2 y = −4

{ 2 x − 15 y = 7 12 x + 2 y = −4 { 2 x − 15 y = 7 12 x + 2 y = −4

{ y = 3 x 6 x − 2 y = 0 { y = 3 x 6 x − 2 y = 0

{ x = 2 y 4 x − 8 y = 0 { x = 2 y 4 x − 8 y = 0

{ 2 x + 16 y = 8 − x − 8 y = −4 { 2 x + 16 y = 8 − x − 8 y = −4

{ 15 x + 4 y = 6 −30 x − 8 y = −12 { 15 x + 4 y = 6 −30 x − 8 y = −12

{ y = −4 x 4 x + y = 1 { y = −4 x 4 x + y = 1

{ y = − 1 4 x x + 4 y = 8 { y = − 1 4 x x + 4 y = 8

{ y = 7 8 x + 4 −7 x + 8 y = 6 { y = 7 8 x + 4 −7 x + 8 y = 6

{ y = − 2 3 x + 5 2 x + 3 y = 11 { y = − 2 3 x + 5 2 x + 3 y = 11

In the following exercises, translate to a system of equations and solve.

The sum of two numbers is 15. One number is 3 less than the other. Find the numbers.

The sum of two numbers is 30. One number is 4 less than the other. Find the numbers.

The sum of two numbers is −26. One number is 12 less than the other. Find the numbers.

The perimeter of a rectangle is 50. The length is 5 more than the width. Find the length and width.

The perimeter of a rectangle is 60. The length is 10 more than the width. Find the length and width.

The perimeter of a rectangle is 58. The length is 5 more than three times the width. Find the length and width.

The perimeter of a rectangle is 84. The length is 10 more than three times the width. Find the length and width.

The measure of one of the small angles of a right triangle is 14 more than 3 times the measure of the other small angle. Find the measure of both angles.

The measure of one of the small angles of a right triangle is 26 more than 3 times the measure of the other small angle. Find the measure of both angles.

The measure of one of the small angles of a right triangle is 15 less than twice the measure of the other small angle. Find the measure of both angles.

The measure of one of the small angles of a right triangle is 45 less than twice the measure of the other small angle. Find the measure of both angles.

Maxim has been offered positions by two car dealers. The first company pays a salary of $10,000 plus a commission of $1,000 for each car sold. The second pays a salary of $20,000 plus a commission of $500 for each car sold. How many cars would need to be sold to make the total pay the same?

Jackie has been offered positions by two cable companies. The first company pays a salary of $ 14,000 plus a commission of $100 for each cable package sold. The second pays a salary of $20,000 plus a commission of $25 for each cable package sold. How many cable packages would need to be sold to make the total pay the same?

Amara currently sells televisions for company A at a salary of $17,000 plus a $100 commission for each television she sells. Company B offers her a position with a salary of $29,000 plus a $20 commission for each television she sells. How many televisions would Amara need to sell for the options to be equal?

Mitchell currently sells stoves for company A at a salary of $12,000 plus a $150 commission for each stove he sells. Company B offers him a position with a salary of $24,000 plus a $50 commission for each stove he sells. How many stoves would Mitchell need to sell for the options to be equal?

Everyday Math

When Gloria spent 15 minutes on the elliptical trainer and then did circuit training for 30 minutes, her fitness app says she burned 435 calories. When she spent 30 minutes on the elliptical trainer and 40 minutes circuit training she burned 690 calories. Solve the system { 15 e + 30 c = 435 30 e + 40 c = 690 { 15 e + 30 c = 435 30 e + 40 c = 690 for e e , the number of calories she burns for each minute on the elliptical trainer, and c c , the number of calories she burns for each minute of circuit training.

Stephanie left Riverside, California, driving her motorhome north on Interstate 15 towards Salt Lake City at a speed of 56 miles per hour. Half an hour later, Tina left Riverside in her car on the same route as Stephanie, driving 70 miles per hour. Solve the system { 56 s = 70 t s = t + 1 2 { 56 s = 70 t s = t + 1 2 .

  • ⓐ for t t to find out how long it will take Tina to catch up to Stephanie.
  • ⓑ what is the value of s s , the number of hours Stephanie will have driven before Tina catches up to her?

Writing Exercises

Solve the system of equations { x + y = 10 x − y = 6 { x + y = 10 x − y = 6

ⓐ by graphing. ⓑ by substitution. ⓒ Which method do you prefer? Why?

Solve the system of equations { 3 x + y = 12 x = y − 8 { 3 x + y = 12 x = y − 8 by substitution and explain all your steps in words.

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ After reviewing this checklist, what will you do to become confident for all objectives?

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  • Authors: Lynn Marecek, MaryAnne Anthony-Smith, Andrea Honeycutt Mathis
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  • Book title: Elementary Algebra 2e
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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.

Solving a systems using substitution

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.

Solving Systems Using Combinations

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:

solving systems of linear equations by substitution word problems

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

solving systems of linear equations by substitution word problems

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Systems of Linear Equations Word Problems - Practice - Expii

Systems of linear equations word problems - practice, explanations (3).

solving systems of linear equations by substitution word problems

(Videos) Set up Word Problems Using a System

by mathman1024

solving systems of linear equations by substitution word problems

This video by mathman1024 works through word problems with systems of equations.

The general guideline to follow when doing these word problems is:

  • Define your variables . Since this is a system, there will be two or more variables.
  • Write your equations. Again, since this is a system, there will be two or more equations. This is also the tricky part which will be focused on in the videos.
  • Solve the system. This can be done using the elimination method , substitution method , or graphing .
  • State your answer.

The first problem that he goes over in the video is, " The sum of two numbers is 79, and their difference is 23. What are the two numbers? "

We want to write algebraic expressions for this system.

Step one is defining the variables. The question asks for two numbers so we can see that these are the variables. Let x=one numbery=the other number The first sentence is, " The sum of two numbers is 79. " Since sum means addition , we write, x+y=79 The next part states, " The difference of those two numbers is 23. " Since difference means subtraction, we write, x−y=23 And now we have our system of equations: {x+y=79x−y=23. This can be solved with substitution but you might notice it would be easier with elimination. x+y=79+x−y=232x+0y=1022x=1022x2=1022x=51 Finally, plug this back in to either equation to find the y value. (51)+y=7951−51+y=79−51y=28 The solution to this word problem's system of equations is (51,28).

The next problem is a little trickier. It isn't as obvious how to set up the equations. It says, " A minor league ballpark attracts 88 fans and draws in $553 in revenue from ticket sales. A child's ticket costs $4 and an adult's ticket is $7. How many of each type of ticket were sold? "

First, we define our variables. The problem asks for how many of each type of ticket. Since there are two types of tickets we can write, Let C=# of childrenA=# of adults The first snippet is, " A minor league ballpark attracts 88 fans ". We know there's a total of 88 fans which are made up of children and adults. So we can write, C+A=88 Next we have, " and draws in $553 in revenue from ticket sales. A child's ticket costs $4 and an adult's ticket is $7. " So we have to factor in the money. We see that we get $4 from each child and we get $7 from each adult. Total, we have $553. So we can write, 4C+7A=553 Now we have the system: {C+A=884C+7A=553. This can be solved by either elimination or substitution. For this one, I personally would choose substitution, but either works.

Let's solve the first equation for C. C+A=88C+A−A=88−AC=88−A Plug this back into the other equation to solve for A. 4(88−A)+7A=553352−4A+7A=553352+3A=553352−352+3A=553−3523A=2013A3=2013A=67 Finally, plug this back into either equation to solve for C. C=88−(67)C=21 Our solution is (21,67).

Remember, the best way to double check your answer to either of these example problems is to plug in your solved coordinates back into the original equations.

Related Lessons

As with most word problems, the most effective approach to take is translation. The goal is to take written information and translate into a new language: math equations.

Let's work through an example to see how this translation works.

solving systems of linear equations by substitution word problems

Image by Clker-Free-Vector-Images via Pixabay ( CC0 )

The key to translating a word problem is to identify the given information. Here's what we know from reading the problem:

  • Cupcakes cost $2
  • Pies cost $7.50
  • William bought 20 items
  • Total cost (pies + cupcakes) = $73

We can get two equations from this information.

First, we know that, all together, William bought 20 items. If we say that the number of cupcakes is C and the number of pies is P, we can write the equation:

solving systems of linear equations by substitution word problems

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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?}
  • \mathrm{Kira\:went\:for\:a\:drive\:in\:her\:new\:car.\:She\:drove\:for\:142.5\:miles\:at\:a\:speed\:of\:57\:mph.\:For\:how\:many\:hours\:did\:she\:drive?}
  • \mathrm{The\:sum\:of\:two\:numbers\:is\:249\:.\:Twice\:the\:larger\:number\:plus\:three\:times\:the\:smaller\:number\:is\:591\:.\:Find\:the\:numbers.}
  • \mathrm{If\:2\:tacos\:and\:3\:drinks\:cost\:12\:and\:3\:tacos\:and\:2\:drinks\:cost\:13\:how\:much\:does\:a\:taco\:cost?}
  • \mathrm{You\:deposit\:3000\:in\:an\:account\:earning\:2\%\:interest\:compounded\:monthly.\:How\:much\:will\:you\:have\:in\:the\:account\:in\:15\:years?}
  • 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. 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.
  • Is there a calculator that can solve word problems?
  • 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.
  • What is an age problem?
  • 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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  • 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. Solving Systems of Linear Equations Using Substitution Word Problems

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  2. Solving Systems of Equations by Substitution Worksheets

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  3. How To Solve Linear Equations Using Substitution

    solving systems of linear equations by substitution word problems

  4. Solving Systems of Equations by Substitution Worksheets

    solving systems of linear equations by substitution word problems

  5. Solving Linear Systems with Substitution

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  6. Solving Systems of Equations by Substitution Worksheets

    solving systems of linear equations by substitution word problems

VIDEO

  1. January 25, 2024

  2. Algebra Basics: solving systems of Linear Equations by Substitution

  3. January 26, 2024

  4. Solving Systems of Linear Equations by Substitution ALL (4.2)

  5. Algebra 1 Lesson #24 How to Solve Word Problems with Linear Systems of Equations Two Variables

  6. Solving a Linear System in Three Variables

COMMENTS

  1. 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 ...

  2. 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.

  3. 5.2: Solve Systems of Equations by Substitution

    Solve a system of equations by substitution. Solve one of the equations for either variable. Substitute the expression from Step 1 into the other equation. Solve the resulting equation. Substitute the solution in Step 3 into one of the original equations to find the other variable. Write the solution as an ordered pair.

  4. 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

  5. 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.

  6. Systems of Linear Equations and Word Problems

    Then use substitution to solve the system for Megan's time: after dividing both sides by 5, multiply both sides by 6 to get rid of the fractions. ... Systems of Linear Equations and Word Problems; Algebraic Functions, including Domain and Range; Scatter Plots, Correlation, and Regression;

  7. IXL

    Social studies. Spanish. Recommendations. Skill plans. IXL plans. Washington state standards. Textbooks. Test prep. Improve your math knowledge with free questions in "Solve a system of equations using substitution: word problems" and thousands of other math skills.

  8. Systems of equations word problems

    Learn for free about math, art, computer programming, economics, physics, chemistry, biology, medicine, finance, history, and more. ... Systems of equations word problems (with zero and infinite solutions) ... TV & DVD. Systems of equations with elimination: apples and oranges. Systems of equations with substitution: coins. Systems of equations ...

  9. Systems of equations with substitution

    You might need: Calculator. Solve the system of equations. 5 x − 7 y = 58 y = − x + 2. x =. y =. 4:21. Report a problem. 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 ...

  10. 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 ...

  11. 5.2 Solving Systems of Equations by Substitution

    How to use a problem solving strategy for systems of linear equations. Step 1. Read the problem. Make sure all the words and ideas are understood. Step 2. Identify what we are looking for. Step 3. Name what we are looking for. Choose variables to represent those quantities. Step 4. Translate into a system of equations. Step 5.

  12. 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...

  13. 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!

  14. 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 ...

  15. PDF Solving systems of equations word problems worksheet For all 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. 1. A large pizza at Palanzio's Pizzeria costs $6.80 plus $0.90 for each topping.

  16. Systems of Linear Equations Word Problems

    This can be solved with substitution but you might notice it would be easier with elimination. x+y=79+x−y=232x+0y=1022x=1022x2=1022x=51 Finally, plug this back in to either equation to find the y value. (51)+y=7951−51+y=79−51y=28 The solution to this word problem's system of equations is (51,28).

  17. Systems of Equations Word Problems • Activity Builder by ...

    Learn how to solve systems of equations word problems with Desmos. Interactive and engaging activities for students and teachers.

  18. Systems of equations with substitution: coins

    System of equations word problem: infinite solutions. Systems of equations word problems (with zero and infinite solutions) Systems of equations with elimination: TV & DVD. ... That is going to be equal to $2.00. We're solving this system by substitution. Now let's see if I can simplify this. So we have 0.05n plus-- let's distribute the 0.25 ...

  19. Solving systems of equation three ways: substitution, elimination, and

    There are three ways to solve systems of linear equations: substitution, elimination, and graphing. Substitution will have you substitute one equation into the other; elimination will have you add or subtract the equations to eliminate a variable; graphing will have you sketch both curves to visually find the points of intersection.

  20. Grade 10 Math Unit 1

    Free lessons, worksheets, and video tutorials for students and teachers. Topics in this unit include: solving linear systems by graphing, substitution, elimination, and solving application questions. This follows chapter 1 of the principles of math grade 10 McGraw Hill textbook.

  21. JMAP A.REI.C.6: Solving Linear Systems, Graphing Linear Systems

    STANDARD A.REI.C.6. AI. Solve systems of linear equations in two variables both algebraically and graphically. Algebraic methods include both elimination and substitution. AII. Solve systems of linear equations exactly and approximately (e.g., with graphs), focusing on pairs of linear equations in two variables. WORKSHEETS.

  22. System of Equations Substitution Calculator

    Free system of equations substitution calculator - solve system of equations using substitution method step-by-step

  23. Word Problems Calculator

    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.