Skip to main content
\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)
Mathematics LibreTexts

5.2: Solve Systems of Equations by Substitution

  • Page ID
    18956
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)

    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

    Note

    Before you get started, take this readiness quiz.

    1. Simplify −5(3−x).
      If you missed this problem, review Exercise 1.10.43.
    2. Simplify 4−2(n+5).
      If you missed this problem, review Exercise 1.10.41.
    3. Solve for y. 8y−8=32−2y
      If you missed this problem, review Exercise 2.3.22.
    4. 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. \(\left\{\begin{array}{l}{2 x+y=7} \\ {x-2 y=6}\end{array}\right.\)

    Answer

    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.The second row reads, “Step 2. Substitute the expression from Step 1 into the other equation.” Then, “We replace y in the second equation with the expression 7 – 2x.” It then shows the x – 2y = 6 becomes x – 2(7 – 2x) = 6.The third row says, “Step 3: Solve the resulting equation.” Then “Now we have an equation with just 1 variable. WE know how to solve this!” It then shows that x – 2(7 – 2x) = 6 becomes x – 14 + 4x = 6 which becomes 5x = 20. Thus x = 4.The fourth row says, “Step 4. Substitute the solution in Step 3 into one of the original quaitons to find the other variable.” Then, “We’ll use the first equation and replace x with 4.” Then it shows that 2x + y = 7 becomes 2(4) + y = 7. This becomes 8 + y = 7, and thus y = −1.The fifth row reads, “Step 5. Write the solution as an ordered pair.” Then “The ordered air is (x, y).” Then (4, −1).The sixth row reads, “Step 6. Check that the order pair is a solution to both original equations.” Then, “Substitute (4, −1) into both equations and make sure they are both true.” It then shows that 2x + y = 7 becomxe 2(4) + −1 = 7, and thus 7 = 7. It also shows that x – 2y = 6 becomes 4 – 2(−1) = 6, and thus 6−6. It also states, “Both equations are ture. (4, −1) is the solution to the system.”

    Exercise \(\PageIndex{2}\)

    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.

    1. Solve one of the equations for either variable.
    2. Substitute the expression from Step 1 into the other equation.
    3. Solve the resulting equation.
    4. Substitute the solution in Step 3 into one of the original equations to find the other variable.
    5. Write the solution as an ordered pair.
    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 Exercise \(\PageIndex{4}\).

    Exercise \(\PageIndex{4}\)

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

    Answer

    The 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. \(\left\{\begin{array}{l}{3 x+y=5} \\ {2 x+4 y=-10}\end{array}\right.\)

    Answer

    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 −3x + 5. .
    Solve the resulting equation for x. .
    .
    .
    Substitute x = 3 into 3x + y = 5 to find y. Example5.15.jpg
    .
    .
    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. \(\left\{\begin{array}{l}{x-2 y=-2} \\ {3 x+2 y=34}\end{array}\right.\)

    Answer

    We 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. \(\left\{\begin{array}{l}{y=-2 x+5} \\ {y=\frac{1}{2} x}\end{array}\right.\)

    Answer

    Since 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 \(\PageIndex{16}\)

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

    Answer

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

    .
    Solve the first equation for y. .
    Substitute −2x + 2 for y in the second equation. .
    Replace the y with −2x + 2. .
    Solve the equation for x. .
    .
    .


    Substitute \(x = \frac{5}{4}\) into 4x + 2y = 4 to find y.
    .
    .
    .
    .
    .
    The ordered pair is \((\frac{5}{4},−\frac{1}{2})\).
    Check the ordered pair in both equations.

    \(\begin{array} {rllrll} 4x+2y &=&4& 6x-y&=&8\\4(\frac{5}{4}) +2(-\frac{1}{2})&\stackrel{?}{=}&4 &6(\frac{5}{4}) - (-\frac{1}{2})& \stackrel{?}{=} & 8\\5-1&\stackrel{?}{=}&4 &\frac{15}{4} - (-\frac{1}{2}) &\stackrel{?}{=} & 8\\4 &=&4\checkmark &\frac{16}{2} &\stackrel{?}{=}&8\\ &&&8&=&8\checkmark \end{array}\)
    The solution is (54,−12).

    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 \(\PageIndex{19}\)

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

    Answer

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

    .
    Solve the first equation for x. .
    Substitute \(\frac{3}{4} y+\frac{3}{2}\) for x in the second equation. .
    Replace the x with \(\frac{3}{4} y+\frac{3}{2}\) .
    Solve for y. .
    .
    .
    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.

    Exercise \(\PageIndex{20}\)

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

    Answer

    infinitely many solutions

    Exercise \(\PageIndex{21}\)

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

    Answer

    infinitely 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. \(\left\{\begin{array}{l}{5 x-2 y=-10} \\ {y=\frac{5}{2} x}\end{array}\right.\)

    Answer

    The 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. .
    .
    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.

    Exercise \(\PageIndex{23}\)

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

    Answer

    no solution

    Exercise \(\PageIndex{24}\)

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

    Answer

    no 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.

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

    Exercise \(\PageIndex{25}\)

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

    Answer
    Step 1. Read the problem.
    Step 2. Identify what we are looking for. We are looking for two numbers.
    Step 3. Name what we are looking for. Let n= the first number
    Let m= the second number
    Step 4. Translate into a system of equations. The sum of two numbers is zero.
    .
    One number is nine less than the other.
    .
    The system is: .
    Step 5. Solve the system of
    equations. We will use substitution
    since the second equation is solved
    for n.
    Substitute m − 9 for n in the first equation. .
    Solve for m. .
    .
    .
    Substitute \(m=\frac{9}{2}\) into the second equation
    and then solve for n.
    .
    .
    .
    .
    Step 6. Check the answer in the problem. Do these numbers make sense in
    the problem? We will leave this to you!
    Step 7. Answer the question. The numbers are \(\frac{9}{2}\) and \(-\frac{9}{2}\).

    Exercise \(\PageIndex{26}\)

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

    Answer

    The 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.

    Answer

    The 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 \(\PageIndex{28}\)

    Add exercises text here.

    Answer
    Step 1. Read the problem. .
    Step 2. Identify what you are looking for. We are looking for the length and width.
    Step 3. Name what we are looking for. Let L= the length
      W= the width
    Step 4. Translate into a system of equations. The perimeter of a rectangle is 88.
       2L + 2W = P
    .
    The length is five more than twice the width.
    .
    The system is: .
    Step 5. Solve the system of equations.
    We will use substitution since the second
    equation is solved for L.

    Substitute 2W + 5 for L in the first equation.
    .
    Solve for W. .
    .
    .
    .
    Substitute W = 13 into the second
    equation and then solve for L.
    .
    .
    .
    Step 6. Check the answer in the problem. Does a rectangle with length 31 and width
    13 have perimeter 88? Yes.
    Step 7. Answer the equation. The length is 31 and the width is 13.

    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.

    Answer

    The 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.

    Answer

    The 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 \(\PageIndex{31}\)

    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.

    Answer

    We will draw and label a figure.

    Step 1. Read the problem. .
    Step 2. Identify what you are looking for. We are looking for the measures of the angles.
    Step 3. Name what we are looking for. Let a= the measure of the 1st angle
    b= the measure of the 2nd angle
    Step 4. Translate into a system of equations. 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.
    .
    The sum of the measures of the angles of
    a triangle is 180.
    .
    The system is: .
    Step 5. Solve the system of equations.
    We will use substitution since the first
    equation is solved for a.
    .
    Substitute 3b + 10 for a in the
    second equation.
    .
    Solve for b. .
    .
    .
    Substitute b = 20 into the first
    equation and then solve for a.
    .
    .
    Step 6. Check the answer in the problem. We will leave this to you!
    Step 7. Answer the question. The measures of the small angles are
    20 and 70.

    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.

    Answer

    The 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.

    Answer

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

    Exercise \(\PageIndex{34}\)

    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?

    Answer
    Step 1. Read the problem.
    Step 2. Identify what you are looking for. We are looking for the number of training sessions
    that would make the pay equal.
    Step 3. Name what we are looking for. Let s= Heather’s salary.
    n= the number of training sessions
    Step 4. Translate into a system of equations. 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
    .
    The system is: .
    Step 5. Solve the system of equations.
    We will use substitution.
    .
    Substitute 25,000 + 15n for s in the second equation. .
    Solve for n. .
    .
    .
    Step 6. Check the answer. Are 600 training sessions a year reasonable?
    Are the two options equal when n = 600?
    Step 7. Answer the question. The salary options would be equal for 600 training sessions.

    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?

    Answer

    There 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?

    Answer

    Kenneth would need to sell 1,000 suits.

    Note

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

    Key Concepts

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