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A way to solve a linear system algebraically is to use the substitution method. The substitution method functions by substituting the one y-value with the other. We're going to explain this by using an example. $$y=2x+4$$ $$3x+y=9$$ We can substitute y in the second equation with the first equation since y = y. $$3x+y=9$$ $$3x+\left (
{\color{green} {2x+4}} \right )=9$$ $$5x+4=9$$ $$5x=5$$ $$x=1$$ This value of x can then be used to find y by substituting 1 with x e.g. in the first equation $$y=2x+4$$ $$y=2\cdot {\color{green} 1}+4$$ $$y=6$$ The solution of the linear system is (1, 6). You can use the substitution method even if both equations of the linear system are in standard form. Just begin by solving one of the equations for one of its variables. Solve the linear system using the substitution method $$2y - 4x = 2$$ $$y = -x + 4$$Video
lesson
Learning Objectives By the end of this section, you will be able to:
Note
Before you get started, take this readiness quiz.
- Simplify −5(3−x).
If you missed this problem, review Exercise 1.10.43. - Simplify 4−2(n+5).
If you missed this problem, review Exercise 1.10.41. - Solve for y. 8y−8=32−2y
If you missed this problem, review Exercise 2.3.22. - Solve for x. 3x−9y=−3
If you missed this problem, review Exercise 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 Exercise \(\PageIndex{1}\).
Exercise \(\PageIndex{1}\): How to Solve a System of Equations by Substitution
Solve the system by substitution. \(\left\{\begin{array}{l}{2 x+y=7} \\ {x-2 y=6}\end{array}\right.\)
AnswerExercise \(\PageIndex{2}\)
Solve the system by substitution. \(\left\{\begin{array}{l}{-2 x+y=-11} \\ {x+3 y=9}\end{array}\right.\)
Answer(6,1)
Exercise \(\PageIndex{3}\)
Solve the system by substitution. \(\left\{\begin{array}{l}{x+3 y=10} \\ {4 x+y=18}\end{array}\right.\)
Answer(4,2)
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 Exercise \(\PageIndex{4}\).
Exercise \(\PageIndex{4}\)
Solve the system by substitution. \(\left\{\begin{array}{l}{x+y=-1} \\ {y=x+5}\end{array}\right.\)
AnswerThe second equation is already solved for y. We will substitute the expression in place of y in the first equation.
| The second equation is already solved for y. We will substitute into the first equation. | |
| Replace the y with x + 5. | |
| Solve the resulting equation for x. | |
| Substitute x = −3 into y = x + 5 to find y. | |
| The ordered pair is (−3, 2). | |
| Check the ordered pair in both equations: \(\begin{array} {rllrll} x+y &=&-1 & y&=&x+5\\-3+2 &\stackrel{?}{=}&-1 &2& \stackrel{?}{=} & -3 + 5\\-1 &=&-1\checkmark &2 &=&2\checkmark \end{array}\) | |
| The solution is (−3, 2). |
Exercise \(\PageIndex{5}\)
Solve the system by substitution. \(\left\{\begin{array}{l}{x+y=6} \\ {y=3 x-2}\end{array}\right.\)
Answer(2,4)
Exercise \(\PageIndex{6}\)
Solve the system by substitution. \(\left\{\begin{array}{l}{2 x-y=1} \\ {y=-3 x-6}\end{array}\right.\)
Answer(−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.
Exercise \(\PageIndex{7}\)
Solve the system by substitution. \(\left\{\begin{array}{l}{3 x+y=5} \\ {2 x+4 y=-10}\end{array}\right.\)
AnswerWe 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 −3x + 5. | |
| Solve the resulting equation for x. | |
| Substitute x = 3 into 3x + y = 5 to find y. | |
| The ordered pair is (3, −4). | |
Check the ordered pair in both equations: \(\begin{array} {rllrll} 3x+y &=&5 & 2x+4y&=&-10\\3\cdot3+(-4) &\stackrel{?}{=}&5 &2\cdot3 + 4(-4)& \stackrel{?}{=} & -10\\9-4&\stackrel{?}{=}&5 &6-16& \stackrel{?}{=} & -10\\5 &=&5\checkmark &-10&=&-10\checkmark \end{array}\) | |
| The solution is (3, −4). |
Exercise \(\PageIndex{8}\)
Solve the system by substitution. \(\left\{\begin{array}{l}{4 x+y=2} \\ {3 x+2 y=-1}\end{array}\right.\)
Answer(1,−2)
Exercise \(\PageIndex{9}\)
Solve the system by substitution. \(\left\{\begin{array}{l}{-x+y=4} \\ {4 x-y=2}\end{array}\right.\)
Answer(2,6)
In Exercise \(\PageIndex{7}\) it was easiest to solve for y in the first equation because it had a coefficient of 1. In Exercise \(\PageIndex{10}\) it will be easier to solve for x.
Exercise \(\PageIndex{10}\)
Solve the system by substitution. \(\left\{\begin{array}{l}{x-2 y=-2} \\ {3 x+2 y=34}\end{array}\right.\)
AnswerWe will solve the first equation for xx and then substitute the expression into the second equation.
| Solve for x. Substitute into the other equation. | |
| Replace the x with 2y − 2. | |
| Solve the resulting equation for y. | |
Substitute y = 5 into x − 2y = −2 to find x. | |
| The ordered pair is (8, 5). | |
| Check the ordered pair in both equations: \(\begin{array} {rllrll} x-2y &=&-2 & 3x+2y&=&34\\8-2\cdot 5 &\stackrel{?}{=}&-2 &3\cdot8 + 2\cdot5& \stackrel{?}{=} & 34\\8-10&\stackrel{?}{=}&-2 &24+10& \stackrel{?}{=} & 34\\-2 &=&-2\checkmark &34&=&34\checkmark \end{array}\) | |
| The solution is (8, 5). |
Exercise \(\PageIndex{11}\)
Solve the system by substitution. \(\left\{\begin{array}{l}{x-5 y=13} \\ {4 x-3 y=1}\end{array}\right.\)
Answer(−2,−3)
Exercise \(\PageIndex{12}\)
Solve the system by substitution. \(\left\{\begin{array}{l}{x-6 y=-6} \\ {2 x-4 y=4}\end{array}\right.\)
Answer(6,2)
When both equations are already solved for the same variable, it is easy to substitute!
Exercise \(\PageIndex{13}\)
Solve the system by substitution. \(\left\{\begin{array}{l}{y=-2 x+5} \\ {y=\frac{1}{2} x}\end{array}\right.\)
AnswerSince both equations are solved for y, we can substitute one into the other.
| Substitute \(\frac{1}{2}x\) for y in the first equation. | |
| Replace the y with \(\frac{1}{2}x\) | |
| Solve the resulting equation. Start by clearing the fraction. | |
| Solve for x. | |
| Substitute x = 2 into \(y = \frac{1}{2}x\) to find y. | |
| The ordered pair is (2,1). | |
| Check the ordered pair in both equations: \(\begin{array} {rllrll} y &=&\frac{1}{2}x & y&=&-2x+5\\1 &\stackrel{?}{=}&\frac{1}{2}\cdot2 &1& \stackrel{?}{=} & -2\cdot2+5\\1 &=&1\checkmark &1 &=&-4+5\\ &&&1&=&1\checkmark \end{array}\) | |
| The solution is (2,1). |
Exercise \(\PageIndex{14}\)
Solve the system by substitution. \(\left\{\begin{array}{l}{y=3 x-16} \\ {y=\frac{1}{3} x}\end{array}\right.\)
Answer(6,2)
Exercise \(\PageIndex{15}\)
Solve the system by substitution. \(\left\{\begin{array}{l}{y=-x+10} \\ {y=\frac{1}{4} x}\end{array}\right.\)
Answer(8,2)
Be very careful with the signs in the next example.
Exercise \(\PageIndex{17}\)
Solve the system by substitution. \(\left\{\begin{array}{l}{x-4 y=-4} \\ {-3 x+4 y=0}\end{array}\right.\)
Answer\((2,\frac{3}{2})\)
Exercise \(\PageIndex{18}\)
Solve the system by substitution. \(\left\{\begin{array}{l}{4 x-y=0} \\ {2 x-3 y=5}\end{array}\right.\)
Answer\((−\frac{1}{2},−2)\)
In Example, it will take a little more work to solve one equation for x or y.
Exercise \(\PageIndex{20}\)
Solve the system by substitution. \(\left\{\begin{array}{l}{2 x-3 y=12} \\ {-12 y+8 x=48}\end{array}\right.\)
Answerinfinitely many solutions
Exercise \(\PageIndex{21}\)
Solve the system by substitution. \(\left\{\begin{array}{l}{5 x+2 y=12} \\ {-4 y-10 x=-24}\end{array}\right.\)
Answerinfinitely many solutions
Look back at the equations in Exercise \(\PageIndex{22}\). Is there any way to recognize that they are the same line?
Let’s see what happens in the next example.
Exercise \(\PageIndex{22}\)
Solve the system by substitution. \(\left\{\begin{array}{l}{5 x-2 y=-10} \\ {y=\frac{5}{2} x}\end{array}\right.\)
AnswerThe second equation is already solved for y, so we can substitute for y in the first equation.
| Substitute x for y in the first equation. | |
| Replace the y with \(\frac{5}{2}x\). | |
| Solve for x. | |
Exercise \(\PageIndex{23}\)
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
Exercise \(\PageIndex{24}\)
Solve the system by substitution. \(\left\{\begin{array}{l}{5 x-3 y=2} \\ {y=\frac{5}{3} x-4}\end{array}\right.\)
Answerno solution
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?
Exercise \(\PageIndex{26}\)
The sum of two numbers is 10. One number is 4 less than the other. Find the numbers.
AnswerThe numbers are 3 and 7.
Exercise \(\PageIndex{27}\)
The sum of two number is −6. One number is 10 less than the other. Find the numbers.
AnswerThe numbers are 2 and −8.
In the Exercise \(\PageIndex{28}\), we’ll use the formula for the perimeter of a rectangle, P = 2L + 2W.
Exercise \(\PageIndex{29}\)
The perimeter of a rectangle is 40. The length is 4 more than the width. Find the length and width of the rectangle.
AnswerThe length is 12 and the width is 8.
Exercise \(\PageIndex{30}\)
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.
AnswerThe length is 23 and the width is 6.
For Exercise \(\PageIndex{31}\) 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.
Exercise \(\PageIndex{32}\)
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.
AnswerThe measure of the angles are 22 degrees and 68 degrees.
Exercise \(\PageIndex{33}\)
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.
AnswerThe measure of the angles are 36 degrees and 54 degrees.
Exercise \(\PageIndex{35}\)
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?
AnswerThere would need to be 160 policies sold to make the total pay the same.
Exercise \(\PageIndex{36}\)
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?
AnswerKenneth would need to sell 1,000 suits.
Key Concepts
- 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.