algebra 1 solving systems by substitution

Substitution Method

One of the methods to solve a system of linear equations in two variables algebraically is the "substitution method". In this method, we find the value of any one of the variables by isolating it on one side and taking every other term on the other side of the equation. Then we substitute that value in the second equation.

The substitution method is preferable when one of the variables in one of the equations has a coefficient of 1. It involves simple steps to find the values of variables of a system of linear equations by substitution method. Let's learn about it in detail in this article.

What is Substitution Method?

The substitution method is a simple way to solve a system of  linear equations algebraically and find the solutions of the variables. As the name suggests, it involves finding the value of the x-variable in terms of the y-variable from the first equation and then substituting or replacing the value of the x-variable in the second equation. In this way, we can solve and find the value of the y-variable. And at last, we can put the value of y in any of the given equations to find x This process can be interchanged as well where we first solve for x and then solve for y.

In simple words, the substitution method involves substituting the value of any one of the variables from one equation into the other equation. Let us take an example of solving two equations x-2y=8 and x+y=5 using the substitution method.

☛ Note:  The other three algebraic methods of solving linear equations . To learn each of these methods, click on the respective links given below.

• Elimination method
• Cross multiplication method
• Graphical method

Solving Systems of Equations by Substitution Method

The steps to apply or use the substitution method to solve a system of equations are given below:

• Step 1:  Simplify the given equation by expanding the parenthesis if needed.
• Step 2: Solve any one of the equations for any one of the variables. You can use any variable based on the ease of calculation.
• Step 3: Substitute the obtained value of x or y in the other equation.
• Step 4: Now, simplify the new equation obtained using arithmetic operations  and solve the equation for one variable.
• Step 5: Now, substitute the value of the variable from  Step 4  in any of the given equations to solve for the other variable.

Here is an example of solving system of equations by using substitution method: 2x+3(y+5)=0 and x+4y+2=0.

Step 1:  Simplify the first equation to get 2x + 3y + 15 = 0. Now we have two equations as,

2x + 3y + 15 = 0 _____ (1)

x + 4y + 2 = 0 ______ (2)

Step 2: We are solving equation (2) for x. So, we get x = -4y - 2.

Step 3: Substitute the obtained value of x in the equation (1). i.e., we are substituting x = -4y-2 in the equation 2x + 3y + 15 = 0, we get, 2(-4y-2) + 3y + 15 = 0.

Step 4: Now, simplify the new equation. We get, -8y-4+3y+15=0

-5y + 11 = 0

Step 5: Now, substitute the value of y in any of the given equations. Let us substitute the value of y in equation (2).

x + 4y + 2 = 0

x + 4 × (11/5) + 2 = 0

x + 44/5 + 2 = 0

x + 54/5 = 0

Therefore, after solving the given system of equations by substitution method, we get x = -54/5 and y= 11/5.

Difference Between Elimination and Substitution Method

Both elimination and substitution methods are ways to solve linear equations algebraically. When the substitution method becomes a little difficult to apply in equations involving large numbers or fractions, we can use the elimination method to ease our calculations. Let us understand the difference between these two methods through the table given below:

Important Notes on Substitution Method:

• To start with the substitution method, first, select the equation that has coefficient 1 for at least one of the variables and solve for the same variable (with coefficient 1). This makes the process easier.
• Before starting with the substitution method, combine all like terms (if any).
• After solving for one variable, we can select any of the given equations or any equation in the whole process to find the other variable.
• If we get any true statement like 3 = 3, 0 = 0, etc while solving using the substitution method, then it means that the system has infinitely many solutions.
• If we get any false statement like 3 = 2, 0 = 1, etc then the system has no solution.

☛  Related Topics:

• Substitution Method Calculator
• Substitution Method Class 10
• System of Equations Solver

Substitution Method Examples

Example 1:  Solve the system of linear equations by substitution method: 5m−2n=17 and 3m+n=8.

The given two equations are:

5m−2n=17 ____ (1)

3m+n=8 _____ (2)

The solution of the given two equations can be found by the following steps:

• From equation 2 we can find the value of n in terms of m, where n = 8 - 3m
• Substitute the value of n in equation 1. We get, 5m - 2(8-3m)=17

5m - 2(8-3m)=17

5m - 16 + 6m =17

11m = 17 + 16

• Substitute the value of m in equation 2, we get, 3×3+n=8

Answer:  ∴ The solution is m=3 and n=-1.

Example 2: Jacky has two numbers such that the sum of two numbers is 20 and the difference between them is 10. Find the numbers.

Let the two numbers be x and y such that x>y. It is given that,

x+y=20 ___ (1)

and x−y=10 ___ (2).

We will now solve by substitution.

From equation 1, we get x = 20-y. Substitute this value in equation 2 to find the value of y.

Now, substitute the value of y in equation 1, we get x+5=20, which gives us x=15.

Answer:  Therefore, the two numbers are 15 and 5.

Example 3: Solve the given system of linear equations by substitution method:

- 2x - 5 + 3x + y = 0 ___ (1)

3x + y = 11 ___ (2)

As we can see that the first equation can be further simplified by combining like terms . After simplifying it, we get x+y-5=0. From this equation, let us find the value of x in terms of y, which is x = 5-y. Now substitute this value in equation 2, we get 3(5-y)+y=11.

Now, let us substitute the value of y in equation 1. We get x+2-5=0, which can be simplified to x = 3.

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FAQs on Substitution Method

What is the substitution method in algebra.

In algebra, the substitution method is one of the ways to solve linear equations in two variables . In this method, we substitute the value of a variable found by one equation in the second equation. It is very easy to use when we have smaller numbers, but in the case of large numbers or fractional coefficients, it becomes tedious to apply the substitution method.

What are the Steps for the Substitution Method?

The three simple steps for the substitution method are given below:

• Find the value of any one variable from any of the equations in terms of the other variable.
• Substitute it in the other equation and solve.
• Substitute the value of the second variable again in any of the equations.

When would you Use the Substitution Method?

The substitution method can be applied to any pair of linear equations with two variables . It is advisable to use the substitution method when we have smaller coefficients in terms or when the equations are given in form x = ay+c and/or y=bx+p.

What do we Substitute in the Substitution Method?

In the substitution method, we substitute the value of one variable found by simplifying an equation in the other equation. For example, if there are two variables in the equations, say m and n, then we can first find the value of m in terms of n from any one of the equations, and then we substitute that value in the second equation to get an answer of n. Then, we again substitute the value of n in any of the given equations to find m.

What do the Substitution Method and the Elimination Method have in Common?

Both methods involve the process of substitution. In both methods, we find the value of one variable first and then substitute it in any of the given equations. 

What is the First Step in the Substitution Method?

The first step in the substitution method is to find the value of any one of the variables from one equation in terms of the other variable. For example, if there are two equations x+y=7 and x-y=8, then from the first equation we can find that x=7-y. The further steps involve substituting this in the other equation and then solving.

What is the Process of Solving Systems by Substitution?

With a system of equations with variables x and y, we first find the value of x in terms of y from any one of the equations given. Then, we substitute that value in the other equation to find the value of y. At last, we again substitute the value of y in any given equation to find x.

Is the Substitution Method Only for Linear Equations?

No, substitution method can be applied for any type of equations . For example, the equations y = x 2  and y = 3x + 4 can be solved by using the substitution method.

Systems of Linear Equations: Two Variables

Solving systems of equations by substitution.

Solving a linear system in two variables by graphing works well when the solution consists of integer values, but if our solution contains decimals or fractions, it is not the most precise method. We will consider two more methods of solving a system of linear equations that are more precise than graphing. One such method is solving a system of equations by the substitution method , in which we solve one of the equations for one variable and then substitute the result into the second equation to solve for the second variable. Recall that we can solve for only one variable at a time, which is the reason the substitution method is both valuable and practical.

How To: Given a system of two equations in two variables, solve using the substitution method.

• Solve one of the two equations for one of the variables in terms of the other.
• Substitute the expression for this variable into the second equation, then solve for the remaining variable.
• Substitute that solution into either of the original equations to find the value of the first variable. If possible, write the solution as an ordered pair.
• Check the solution in both equations.

Example 3: Solving a System of Equations in Two Variables by Substitution

Solve the following system of equations by substitution.

First, we will solve the first equation for [latex]y[/latex].

Now we can substitute the expression [latex]x - 5[/latex] for [latex]y[/latex] in the second equation.

Now, we substitute [latex]x=8[/latex] into the first equation and solve for [latex]y[/latex].

Our solution is [latex]\left(8,3\right)[/latex].

Check the solution by substituting [latex]\left(8,3\right)[/latex] into both equations.

Can the substitution method be used to solve any linear system in two variables?

Yes, but the method works best if one of the equations contains a coefficient of 1 or –1 so that we do not have to deal with fractions.

• Precalculus. Authored by : OpenStax College. Provided by : OpenStax. Located at : http://cnx.org/contents/[email protected]:1/Preface . License : CC BY: Attribution

Solving Systems of Equations by Substitution

These algebra lessons, with videos, examples and step-by-step solutions, introduce the technique of solving systems of equations by substitution.

Related Pages Solve Systems of Equations by Addition Systems of Equations (Graphical Method) Worksheets to practice solving systems of equations More Algebra Lessons

In some word problems, we may need to translate the sentences into more than one equation . If we have two unknown variables then we would need at least two equations to solve the variable. In general, if we have n unknown variables then we would need at least n equations to solve the variable.

The following example show the steps to solve a system of equations using the substitution method. Scroll down the page for more examples and solutions.

In the Substitution Method, we isolate one of the variables in one of the equations and substitute the results in the other equation. We usually try to choose the equation where the coefficient of a variable is 1 and isolate that variable. This is to avoid dealing with fractions whenever possible. If none of the variables has a coefficient of 1 then you may want to consider the Addition Method or Elimination Method .

Steps to solving Systems of Equations by Substitution:

• Isolate a variable in one of the equations. (Either y = or x =).
• Substitute the isolated variable in the other equation.
• This will result in an equation with one variable. Solve the equation.
• Substitute the solution from step 3 into another equation to solve for the other variable.
• Recommended: Check the solution.

Example: 3x + 2y = 2      (equation 1) y + 8 = 3x        (equation 2)

Solution: Step 1: Try to choose the equation where the coefficient of a variable is 1.

Choose equation 2 and isolate variable y y = 3x - 8         (equation 3)

Step 2: From equation 3, we know that y is the same as 3x - 8

We can then substitute the variable y in equation 1 with 3x - 8 3x + 2(3x - 8) = 2

Step 3: Remove brackets using distributive property

3x + 6x - 16 = 2

Step 4: Combine like terms

9x - 16 = 2

Step 5: Isolate variable x

Step 6: Substitute x = 2 into

equation 3 to get the value for y

y = 3(2) - 8 y = 6 - 8 = -2

Step 7: Check your answer with equation 1

3(2) + 2(-2) = 6 - 4 = 2

Answer: x = 2 and y = -2

Solving systems of equations using Substitution Method through a series of mathematical steps to teach students algebra

Example: 2x + 5y = 6 9y + 2x = 22

How to solve systems of equations by substitution?

Example: y = 2x + 5 3x - 2y = -9

Steps to solve a linear system of equations using the substitution method

Example: x + 3y = 12 2x + y = 6

Example of a system of equations that is solved using the substitution method.

Example: 2x + 3y = 13 -2x + y = -9

Solving Linear Systems of Equations Using Substitution Include an explanation of the graphs - one solution, no solution, infinite solutions.

Examples: 2x + 4y = 4 y = x - 2 x + 3y = 6 2x + 6y = -12 2x - 3y = 6 4x - 6y = 12

Example of how to solve a system of linear equation using the substitution method. x + 2y = -20 y = 2x

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4.2: Solving Systems by Substitution

• Last updated
• Save as PDF
• Page ID 19871

• David Arnold
• College of the Redwoods

In this section we introduce an algebraic technique for solving systems of two equations in two unknowns called the substitution method. The substitution method is fairly straightforward to use. First, you solve either equation for either variable, then substitute the result into the other equation. The result is an equation in a single variable. Solve that equation, then substitute the result into any of the other equations to find the remaining unknown variable.

Example \(\PageIndex{1}\)

Solve the following system of equations:

\[2x-5 y=-8 \label{Eq4.2.1} \]

\[y=3x-1 \label{Eq4.2.2} \]

Equation \ref{Eq4.2.2} is already solved for \(y\). Substitute Equation \ref{Eq4.2.2} into Equation \ref{Eq4.2.1}. This means we will substitute \(3x−1\) for \(y\) in Equation \ref{Eq4.2.1}.

\[\begin{aligned} 2x-5y &= -8 \quad {\color {Red} \text { Equation }} \ref{Eq4.2.1} \\ 2x-5({\color {Red}3x-1}) &= -8 \quad {\color {Red} \text { Substitute } 3x-1 \text { for } y \text { in }} \ref{Eq4.2.1} \end{aligned} \nonumber \]

Now solve for \(x\).

\[\begin{aligned} 2x-15x+5 &= -8 \quad \color {Red} \text { Distribute }-5 \\ -13x+5 &= -8 \quad \color {Red} \text { Simplify. } \\ -13x &= -13 \quad \color {Red} \text { Subtract } 5 \text { from both sides, } \\ x &= 1 \quad \color {Red} \text { Divide both sides by }-13 \end{aligned} \nonumber \]

As we saw in Solving Systems by Graphing , the solution to the system is the point of intersection of the two lines represented by the equations in the system. This means that we can substitute the answer \(x = 1\) into either equation to find the corresponding value of \(y\). We choose to substitute \(1\) for \(x\) in Equation \ref{Eq4.2.2}, then solve for \(y\), but you will get exactly the same result if you substitute \(1\) for \(x\) in Equation \ref{Eq4.2.1}.

\[\begin{aligned} y &= 3x-1 \quad {\color {Red} \text { Equation }}\ref{Eq4.2.2} \\ y &= 3(1)-1 \quad \color {Red} \text { Substitute } 1 \text { for } x \\ y &= 2 \quad \color {Red} \text { Simplify. } \end{aligned} \nonumber \]

Hence, \((x,y) = (1 ,2)\) is the solution of the system.

Check: To show that the solution \((x,y) = (1 ,2)\) is a solution of the system, we need to show that \((x,y) = (1 ,2)\) satisfies both equations \ref{Eq4.2.1} and \ref{Eq4.2.2}.

Substitute \((x,y) = (1 ,2)\) in Equation \ref{Eq4.2.1}:

\[\begin{aligned} 2 x-5 y &=-8 \\ 2(1)-5(2) &=-8 \\ 2-10 &=-8 \\-8 &=-8 \end{aligned} \nonumber \]

Thus, (1,2) satisfies Equation \ref{Eq4.2.1}.

Substitute \((x,y) = (1 ,2)\) in Equation \ref{Eq4.2.2}:

\[\begin{array}{l}{y=3 x-1} \\ {2=3(1)-1} \\ {2=3-1} \\ {2=2}\end{array} \nonumber \]

Thus, (1,2) satisfies Equation \ref{Eq4.2.2}.

Because \((x,y) = (1 ,2)\) satisfies both equations, it is a solution of the system.

Exercise \(\PageIndex{1}\)

\[\begin{aligned} 9 x+2 y &=-19 \\ y &=13+3 x \end{aligned} \nonumber \]

Substitution Method

The substitution method involves these steps:

• Solve either equation for either variable.
• Substitute the result from step one into the other equation. Solve the resulting equation.
• Substitute the result from step two into either of the original system equations or the resulting equation from step one (whichever seems easiest), then solve to find the remaining unknown variable.

Example \(\PageIndex{2}\)

\[5x-2y=12 \label{Eq4.2.3} \]

\[4x+y=6 \label{Eq4.2.4} \]

The first step is to solve either equation for either variable. This means that we can solve the first equation for \(x\) or \(y\), but it also means that we could first solve the second equation for \(x\) or \(y\). Of these four possible choices, solving the second Equation \ref{Eq4.2.4} for \(y\) seems the easiest way to start.

\[\begin{aligned} 4x+y &= 6 \quad {\color {Red} \text { Equation }}\ref{Eq4.2.4} \\ y &= 6-4x \quad \color {Red} \text { Subtract } 4x \text { from both sides. } \end{aligned} \nonumber \]

Next, substitute \(6−4x\) for \(y\) in Equation \ref{Eq4.2.3}.

\[\begin{aligned} 5x-2y &= 12 \quad {\color {Red} \text { Equation }}\ref{Eq4.2.3} \\ 5x-2(6-4x) &= 12\quad {\color {Red} \text { Substitute } 6-4 x \text { for } y \text { in }}\ref{Eq4.2.3} \\ 5x-12+8x &= 12 \quad \color {Red} \text { Distribute }-2 \\ 13x-12 &= 12 \quad \color {Red} \text { Simplify. }\\ 13x &= 24 \quad \color {Red} \text { Add } 12 \text { to both sides. } \\ x &= \dfrac{24}{13} \quad \color {Red} \text { Divide both sides by } 13 \end{aligned} \nonumber \]

Finally, to find the \(y\)-value, substitute \(24/13\) for \(x\) in the equation \(y =6−4x\) (you can also substitute \(24/13\) for \(x\) in equations \ref{Eq4.2.3} or \ref{Eq4.2.4}).

\[\begin{aligned} y &= 6-4x \\ y &= 6-4\left(\dfrac{24}{13}\right) \quad \color {Red} \text { Substitute } 24 / 13 \text { for } x \text { in } y=6-4x \\ y &= \dfrac{78}{13}-\dfrac{96}{13}\quad \color {Red} \text { Multiply, then make equivalent fractions. } \\ y &= -\dfrac{18}{13} \quad \color {Red} \text { Simplify. } \end{aligned} \nonumber \]

Hence, \((x,y) = (24 /13,−18/13)\) is the solution of the system.

Check: Let’s use the graphing calculator to check the solution. First, we store \(24/13\) in \(X\) with the following keystrokes (see the result in Figure \(\PageIndex{3}\)).

Now, clear the calculator screen by pressing the CLEAR button, then enter the left-hand side of Equation \ref{Eq4.2.3} with the following keystrokes (see the result in Figure \(\PageIndex{4}\)).

Exercise \(\PageIndex{2}\)

\[\begin{aligned} x-2 y &=13 \\ 4 x-3 y &=26 \end{aligned} \nonumber \]

\((13 / 5,-26 / 5)\)

Example \(\PageIndex{3}\)

\[3x-2y=6 \label{Eq4.2.5} \]

\[4x+5y=20 \label{Eq4.2.6} \]

Dividing by \(−2\) gives easier fractions to deal with than dividing by \(3\), \(4\), or \(5\), so let’s start by solving equation (\ref{Eq4.2.5}) for \(y\).

\[\begin{aligned} 3x-2y &= 6\quad {\color {Red} \text { Equation }}\ref{Eq4.2.5} \\ -2y &= 6-3x \quad \color {Red} \text { Subtract } 3 x \text { from both sides. } \\ y &= \dfrac{6-3 x}{-2} \quad \color {Red} \text { Divide both sides by }-2 \\ y &= -3+\dfrac{3}{2} x \quad \color {Red} \text { Divide both } 6 \text { and }-3 x \text { by }-2 \text { using distributive property. } \end{aligned} \nonumber \]

Substitute \(-3+\dfrac{3}{2} x\) for \(y\) in Equation \ref{Eq4.2.6}

\[\begin{aligned} 4x+5y &= 20 \quad {\color {Red} \text { Equation }}\ref{Eq4.2.6} \\ 4x+5\left(-3+\dfrac{3}{2} x\right) &= 20 \quad \color {Red} \text { Substitute }-3+\dfrac{3}{2} x \text { for } y \\ 4x-15+\dfrac{15}{2} x &= 20 \quad \color {Red} \text { Distribute the } 5\\ 8x-30+15x &= 40 \quad \color {Red} \text { Clear fractions by multiplying } \\ 23x &= 70 \quad \color {Red} \text { Simplify. Add } 30 \text { to both sides. } \\ x &= \dfrac{70}{23} \quad \color {Red} \text { Divide both sides by } 23 \end{aligned} \nonumber \]

To find \(y\), substitute \(70/23\) for \(x\) into equation \(y=-3+\dfrac{3}{2} x\). You could also substitute \(70/23\) for \(x\) in equations \ref{Eq4.2.5} or \ref{Eq4.2.6} and get the same result.

\[\begin{aligned} y &= -3+\dfrac{3}{2} x \\ y &= -3+\dfrac{3}{2}\left(\dfrac{70}{23}\right) \quad \color {Red} \text { Substitute } 70 / 23 \text { for } x \\ y &= -\dfrac{69}{23}+\dfrac{105}{23} \quad \color {Red} \text { Multiply. Make equivalent fractions. } \\ y &= \dfrac{36}{23} \quad \text { Simplify. } \end{aligned} \nonumber \]

Hence, \((x,y) = (70 /23,36/23)\) is the solution of the system.

Check: To check this solution, let’s use the graphing calculator to find the solution of the system. We already know that \(3x − 2y = 6\) is equivalent to \(y=-3+\dfrac{3}{2} x\). Let’s also solve Equation \ref{Eq4.2.6} for \(y\).

\[\begin{aligned} 4x+5y &= 20 \quad {\color {Red} \text { Equation }}\ref{Eq4.2.6} \\ 5y &= 20-4x \quad \color {Red} \text { Subtract } 4 x \text { from both sides. } \\ y &= \dfrac{20-4 x}{5} \quad \color {Red} \text { Divide both sides by } 5 \\ y &= 4-\dfrac{4}{5} x \quad \color {Red} \text { Divide both } 20 \text { and }-4 x \text { by } 5 \text { using the distributive property. } \end{aligned} \nonumber \]

Enter \(y=-3+\dfrac{3}{2} x\) and \(y=4-\dfrac{4}{5} x\) into the Y = menu of the graphing calculator (see Figure 4.32).

Press the ZOOM button and select 6:ZStandard . Press 2ND CALC to open the CALCULATE menu, select 5:intersect , then press the ENTER key three times in succession to enter “Yes” to the queries “First curve,” “Second curve,” and “Guess.” The result is shown in Figure \(\PageIndex{7}\).

At the bottom of the viewing window in Figure \(\PageIndex{7}\), note how the coordinates of the point of intersection are stored in the variables X and Y . Without moving the cursor, (the variables X and Y contain the coordinates of the cursor), quit the viewing window by pressing 2ND QUIT , which is located above the MODE key. Then press the CLEAR button to clear the calculator screen.

Now press the \(\mathrm{X}, \mathrm{T}, \theta, \mathrm{n}\) key, then the MATH button on your calculator:

Select 1:►Frac , then press the ENTER key to produce the fractional equivalent of the decimal content of the variable \(X\) (see Figure \(\PageIndex{9}\)).

Repeat the procedure for the variable \(Y\). Enter:

Exercise \(\PageIndex{3}\)

\[\begin{aligned}3 x-5 y&=3 \\ 5 x-6 y&=2\end{aligned} \nonumber \]

\((-8 / 7,-9 / 7)\)

Exceptional Cases Revisited

It is entirely possible that you might apply the substitution method to a system of equations that either have an infinite number of solutions or no solutions at all. Let’s see what happens should you do that.

Example \(\PageIndex{4}\)

\[2 x+3 y=6 \label{Eq4.2.7} \]

\[y=-\dfrac{2}{3} x+4 \label{Eq4.2.8} \]

Equation \ref{Eq4.2.8} is already solved for \(y\), so let’s substitute \(-\dfrac{2}{3} x+4\) for \(y\) in Equation \ref{Eq4.2.7}.

\[\begin{aligned} 2x+3y &= 6 \quad {\color {Red} \text { Equation }}\ref{Eq4.2.7} \\ 2x+3\left(-\dfrac{2}{3} x+4\right) &= 6 \quad \color {Red} \text { Substitute }-\dfrac{2}{3} x+4 \text { for } y \\ 2x-2x+12 &= 6 \quad \color {Red} \text { Distribute the } 3 \\ 12 &= 6\quad \color {Red} \text { Simplify. } \end{aligned} \nonumber \]

Goodness! What happened to the \(x\)? How are we supposed to solve for \(x\) in this situation? However, note that the resulting statement, \(12 = 6\), is false, no matter what we use for \(x\) and \(y\). This should give us a clue that there are no solutions. Perhaps we are dealing with parallel lines?

Let’s solve Equation \ref{Eq4.2.7} for \(y\), putting the equation into slope-intercept form, to help determine the situation.

\[\begin{aligned} 2x+3y &= 6 \quad {\color {Red} \text { Equation }}\ref{Eq4.2.7} \\ 3y &= -2x+6 \quad \color {Red} \text { Subtract } 2 x \text { from both sides. } \\ y &= -\dfrac{2}{3} x+2 \quad \color {Red} \text { Divide both sides by } 3 \end{aligned} \nonumber \]

Thus, our system is equivalent to the following two equations.

\[\begin{aligned} y &= -\dfrac{2}{3} x+2 \\ y &= -\dfrac{2}{3} x+4 \end{aligned} \nonumber \]

These lines have the same slope \(−2/3\), but different \(y\)-intercepts (one has \(y\)-intercept \((0,2)\), the other has \(y\)-intercept \((0,4)\)). Hence, these are two distinct parallel lines and the system has no solution.

Exercise \(\PageIndex{4}\)

\[\begin{aligned} x &=\dfrac{4}{3} y-7 \\ 6 x-8 y &=-3 \end{aligned} \nonumber \]

no solution

Example \(\PageIndex{5}\)

\[2x-6y=-8 \label{Eq4.2.9} \]

\[x=3y-4 \label{Eq4.2.10} \]

Equation \ref{Eq4.2.10} is already solved for \(x\), so let’s substitute \(3y−4\) for \(x\) in Equation \ref{Eq4.2.9}.

\[\begin{aligned} 2x-6y &= -8 \quad {\color {Red} \text { Equation }}\ref{Eq4.2.9} \\ 2(3y-4)-6y &= -8 \quad \color {Red} \text { Substitute } 3 y-4 \text { for } x \\ 6y-8-6y &= -8 \quad \color {Red} \text { Distribute the } 2 \\ -8 &= -8 \quad \color {Red}\text { Simplify. } \end{aligned} \nonumber \]

Goodness! What happened to the \(x\)? How are we supposed to solve for \(x\) in this situation? However, note that the resulting statement, \(−8=−8\), is a true statement this time. Perhaps this is an indication that we are dealing with the same line? Let’s put both equations \ref{Eq4.2.9} and \ref{Eq4.2.10} into slope-intercept form so that we can compare them.

Solve Equation \ref{Eq4.2.9} for \(y\):

\[\begin{aligned} 2 x-6 y &=-8 \\-6 y &=-2 x-8 \\ y &=\dfrac{-2 x-8}{-6} \\ y &=\dfrac{1}{3} x+\dfrac{4}{3} \end{aligned} \nonumber \]

Solve Equation \ref{Eq4.2.10} for \(y\):

\[\begin{aligned} x &=3 y-4 \\ x+4 &=3 y \\ \dfrac{x+4}{3} &=y \\ y &=\dfrac{1}{3} x+\dfrac{4}{3} \end{aligned} \nonumber \]

Hence, the lines have the same slope and the same \(y\)-intercept and they are exactly the same lines. Thus, there are an infinite number of solutions. Indeed, any point on either line is a solution. Examples of solution points are \((−4,0)\), \((−1,1)\), and \((2,2)\).

Exercise \(\PageIndex{5}\)

\[\begin{aligned}-28 x+14 y &=-126 \\ y &=2 x-9 \end{aligned} \nonumber \]

There are an infinite number of solutions. Examples of solution points are \((0,−9)\), \((5,1)\), and \((−3,−15)\).

When you substitute one equation into another and the variable disappears, consider:

• If the resulting statement is false, then you have two distinct parallel lines and there is no solution.
• If the resulting statement is true, then you have the same lines and there are an infinite number of solutions.

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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