Solving Systems of Equations by Substitution

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 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.

Example. Solve the system by substitution: $\begin{cases} 2x+y=7 \\ x-2y=6 \end{cases}$.

Both equations are true, so $(4,-1)$ is the solution to the system.

If one of the equations in the system is already given in slope-intercept form, Step 1 is done for us.

Example. Solve the system by substitution: $\begin{cases} x+y=-1 \\ y=x+5 \end{cases}$.

The second equation is already solved for $y$, so we substitute the expression $x+5$ in place of $y$ in the first equation: $x+(x+5)=-1$, so $2x+5=-1$, so $2x=-6$, and $x=-3$. Substitute $x=-3$ into $y=x+5$ to find $y$: $y=-3+5=2$. The ordered pair is $(-3,2)$.

Check: let $x=-3$ and $y=2$. Then $-3+2\overset{?}{=}-1$, so $-1=-1\ \checkmark$; and $2\overset{?}{=}-3+5$, so $2=2\ \checkmark$. Both equations are true, so $(-3,2)$ is the solution to the system.

If the equations are given in standard form, we’ll need to start by solving for one of the variables.

Example. Solve the system by substitution: $\begin{cases} 3x+y=5 \\ 2x+4y=-10 \end{cases}$.

We need to solve one equation for one variable. Solving the first equation for $y$ gives $y=-3x+5$. Substitute $-3x+5$ for $y$ in the second equation: $2x+4(-3x+5)=-10$, so $2x-12x+20=-10$, so $-10x=-30$, and $x=3$. Substitute $x=3$ into $y=-3x+5$ to find $y$: $y=-3(3)+5=-4$. The ordered pair is $(3,-4)$.

Check: let $x=3$ and $y=-4$. Then $3(3)+(-4)\overset{?}{=}5$, so $9-4\overset{?}{=}5$, and $5=5\ \checkmark$; and $2(3)+4(-4)\overset{?}{=}-10$, so $6-16\overset{?}{=}-10$, and $-10=-10\ \checkmark$. The solution is $(3,-4)$.

Sometimes it is easier to solve for $x$ rather than $y$ — look for a variable whose coefficient is $1$.

Example. Solve the system by substitution: $\begin{cases} x-2y=-2 \\ 3x+2y=34 \end{cases}$.

We will solve the first equation for $x$ and then substitute the expression into the second equation. Solving for $x$ gives $x=2y-2$. Substitute $2y-2$ for $x$ in the second equation: $3(2y-2)+2y=34$, so $6y-6+2y=34$, so $8y=40$, and $y=5$. Substitute $y=5$ into $x=2y-2$ to find $x$: $x=2(5)-2=8$. The ordered pair is $(8,5)$.

Check: let $x=8$ and $y=5$. Then $8-2(5)\overset{?}{=}-2$, so $8-10\overset{?}{=}-2$, and $-2=-2\ \checkmark$; and $3(8)+2(5)\overset{?}{=}34$, so $24+10\overset{?}{=}34$, and $34=34\ \checkmark$. The solution is $(8,5)$.

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

Example. Solve the system by substitution: $\begin{cases} y=-2x+5 \\ y=\tfrac{1}{2}x \end{cases}$.

Since both equations are solved for $y$, we can substitute one into the other: $\tfrac{1}{2}x=-2x+5$. Clear the fraction by multiplying both sides by $2$: $x=-4x+10$, so $5x=10$, and $x=2$. Substitute $x=2$ into $y=\tfrac{1}{2}x$ to find $y$: $y=\tfrac{1}{2}(2)=1$. The ordered pair is $(2,1)$.

Check: let $x=2$ and $y=1$. Then $1\overset{?}{=}\tfrac{1}{2}(2)$, and $1=1\ \checkmark$; and $1\overset{?}{=}-2(2)+5$, so $1\overset{?}{=}-4+5$, and $1=1\ \checkmark$. The solution is $(2,1)$.

Be very careful with the signs in the next example.

Example. Solve the system by substitution: $\begin{cases} 4x+2y=4 \\ 6x-y=8 \end{cases}$.

We need to solve one equation for one variable. Solving the first equation for $y$ gives $2y=-4x+4$, so $y=-2x+2$. Substitute $-2x+2$ for $y$ in the second equation: $6x-(-2x+2)=8$, so $6x+2x-2=8$, so $8x=10$, and $x=\tfrac{5}{4}$. Substitute $x=\tfrac{5}{4}$ into $4x+2y=4$ to find $y$: $4\left(\tfrac{5}{4}\right)+2y=4$, so $5+2y=4$, so $2y=-1$, and $y=-\tfrac{1}{2}$. The ordered pair is $\left(\tfrac{5}{4},-\tfrac{1}{2}\right)$.

Check: let $x=\tfrac{5}{4}$ and $y=-\tfrac{1}{2}$. Substituting back into both original equations gives $4=4\ \checkmark$ and $8=8\ \checkmark$. The solution is $\left(\tfrac{5}{4},-\tfrac{1}{2}\right)$.

Sometimes, when we solve the resulting equation, all the variable terms cancel out.

Example. Solve the system by substitution: $\begin{cases} 4x-3y=6 \\ 15y-20x=-30 \end{cases}$.

We need to solve one equation for one variable. Solving the first equation for $x$ gives $x=\tfrac{3}{4}y+\tfrac{3}{2}$. Substitute this expression into the second equation: $15y-20\left(\tfrac{3}{4}y+\tfrac{3}{2}\right)=-30$, so $15y-15y-30=-30$, so $-30=-30$.

Since $-30=-30$ 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.

Look back at the equations in the last example — is there any way to recognize that they are the same line? Let’s see what happens when the lines are parallel instead.

Example. Solve the system by substitution: $\begin{cases} 5x-2y=-10 \\ y=\tfrac{5}{2}x \end{cases}$.

The second equation is already solved for $y$, so we substitute $\tfrac{5}{2}x$ for $y$ in the first equation: $5x-2\left(\tfrac{5}{2}x\right)=-10$, so $5x-5x=-10$, so $0=-10$.

Since $0=-10$ is a false statement, the equations are inconsistent. The graphs of the two equations would be parallel lines. The system has no solution.

Solve Applications of Systems of Equations by Substitution

We’ll use the same problem-solving strategy we used for solving systems of equations by graphing. Now that we know how to solve systems by substitution, that’s what we’ll do in Step 5.

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

Let $n$ be the first number and $m$ be the second number. The sum of the two numbers is zero, so $n+m=0$. One number is nine less than the other, so $n=m-9$. The system is $\begin{cases} n+m=0 \\ n=m-9 \end{cases}$.

We’ll solve by substitution, since the second equation is already solved for $n$. Substitute $m-9$ for $n$ in the first equation: $(m-9)+m=0$, so $2m-9=0$, so $2m=9$, and $m=\tfrac{9}{2}$. Substitute $m=\tfrac{9}{2}$ into $n=m-9$ to find $n$: $n=\tfrac{9}{2}-9=-\tfrac{9}{2}$.

The numbers are $\tfrac{9}{2}$ and $-\tfrac{9}{2}$.

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

Let $L$ be the length and $W$ be the width. The perimeter of a rectangle is $2L+2W=P$, so $2L+2W=88$. The length is five more than twice the width, so $L=2W+5$. The system is $\begin{cases} 2L+2W=88 \\ L=2W+5 \end{cases}$.

Substitute $2W+5$ for $L$ in the first equation: $2(2W+5)+2W=88$, so $4W+10+2W=88$, so $6W+10=88$, so $6W=78$, and $W=13$. Substitute $W=13$ into $L=2W+5$ to find $L$: $L=2(13)+5=31$.

Does a rectangle with length $31$ and width $13$ have perimeter $88$? Yes. The length is $31$ and the width is $13$.

Example. 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.

Let $a$ be the measure of the first angle and $b$ be the measure of the second angle. One small angle is ten more than three times the other, so $a=3b+10$. The sum of the measures of the angles of a triangle is $180$, and one angle of a right triangle already measures $90°$, so $a+b+90=180$. The system is $\begin{cases} a=3b+10 \\ a+b+90=180 \end{cases}$.

Substitute $3b+10$ for $a$ in the second equation: $(3b+10)+b+90=180$, so $4b+100=180$, so $4b=80$, and $b=20$. Substitute $b=20$ into $a=3b+10$ to find $a$: $a=3(20)+10=70$.

The measures of the small angles are $20°$ and $70°$.

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

Let $s$ be Heather’s salary and $n$ be the number of training sessions. Option A gives $s=25{,}000+15n$. Option B gives $s=10{,}000+40n$. The system is $\begin{cases} s=25{,}000+15n \\ s=10{,}000+40n \end{cases}$.

Substitute $25{,}000+15n$ for $s$ in the second equation: $25{,}000+15n=10{,}000+40n$, so $25{,}000=10{,}000+25n$, so $15{,}000=25n$, and $n=600$.

Are 600 training sessions a reasonable number, and are the two options equal when $n=600$? Yes. The salary options would be equal for $600$ training sessions.

Key terms

substitution method — solving one equation of a system for a variable, then replacing that variable with the resulting expression in the other equation so it becomes an equation in one variable. consistent, dependent system — a system whose equations, after substitution, reduce to a true numeric statement (like $-30=-30$); the two equations describe the same line and the system has infinitely many solutions. inconsistent system — a system whose equations, after substitution, reduce to a false numeric statement (like $0=-10$); the two equations describe parallel lines and the system has no solution.

This section is adapted from Elementary Algebra 2e, Section 5.2: Solving Systems of Equations by Substitution by Lynn Marecek, MaryAnne Anthony-Smith, and Andrea Honeycutt Mathis, © OpenStax, licensed under CC BY-NC-SA 4.0. Access the original for free at openstax.org. Changes: recast the first worked example’s step-by-step breakdown as a markdown table and condensed the remaining worked examples into prose with inline checks; omitted the Be Prepared quiz, Self Check checklist, media links, and end-of-section exercises; and converted the practice problems (“Try Its”) into interactive exercises with instant feedback — including turning the dependent/inconsistent-system Try Its into a multiple-choice classification, since the math checker cannot reliably grade a bare true/false numeric statement as an answer.