Systems of equations are simultaneous equations that share the same solutions. We can solve systems of two equations with two unknowns using the substitution method. This method consists of solving for one of the variables in one of the equations. Then we need to substitute that expression into the second equation to form a single equation with one unknown.
Here, we will learn how to solve systems of equations using the substitution method. We will solve several practice problems to understand the process used.
ALGEBRA
Relevant for…
Learning to solve systems of equations with the substitution method.
ALGEBRA
Relevant for…
Learning to solve systems of equations with the substitution method.
Steps to solve systems of equations by substitution
The substitution method for solving systems of equations consists of solving one of the equations for one of the variables. Then, we substitute the obtained expression in the second equation, and we will obtain an equation with a single variable, which can be solved easily.
In detail, we can follow the following steps to solve systems of equations by substitution:
1. Simplify both equations if possible.
This includes removing parentheses, combining like terms, and removing fractions.
2. Solve an equation for one variable.
We can choose any of the equations and solve for any of the variables.
3. Substitute the expression obtained in step 2 into the other equation.
By doing this, we will obtain an equation with only one unknown.
4. Solve the equation from step 3.
By doing this, we will get the value of one of the variables. If you need a revision, you can take a look at our article on how to solve linear equations with one variable.
5. Find the value of the other variable.
Substitute the value of the variable obtained in step 4 into any of the two equations and solve.
Systems of equations by substitution – Examples with answers
EXAMPLE 1
Find the solutions to the system of equations using the substitution method: $latex \begin{cases}x+2y=10 \\ 2x-y=5 \end{cases}$
Solution
Step 1: The equations are already simplified.
Step 2: Solving the first equation for x, we have:
$latex x+2y=10$
$latex x=10-2y$
Step 3: Substituting $latex for x=10-2y$ in the second equation, we have:
$latex 2x-y=5$
$latex 2(10-2y)-y=5$
$latex 20-4y-y=5$
Step 4: Solving for y, we have:
$latex 20-4y-y=5$
$latex -5y=-15$
$latex y=3$
Step 5: Substituting the value $latex y=3$ in the first equation, we have:
$latex x+2y=10$
$latex x+2(3)=10$
$latex x=4$
The solution to the system is $latex x=4,~~y=3$.
EXAMPLE 2
Solve the following system of equations using the substitution method: $latex \begin{cases}-2x-y=1 \\ 3x+4y=6 \end{cases}$
Solution
Step 1: The equations are already simplified.
Step 2: Solving the first equation for y, we have:
$latex -2x-y=1$
$latex -y=1+2x$
$latex y=-1-2x$
Step 3: Substituting $latex y=-1-2x$ into the second equation, we have:
$latex 3x+4y=6$
$latex 3x+4(-1-2x)=6$
$latex 3x-4-8x=6$
Step 4: Solving for x, we have:
$latex 3x-4-8x=6$
$latex -5x=10$
$latex x=-2$
Step 5: Substituting $latex x=-2$ into the first equation, we have:
$latex -2x-y=1$
$latex -2(-2)-y=1$
$latex 4-y=1$
$latex -y=-3$
$latex y=3$
The solution to the system is $latex x=-2, ~~y=3$.
EXAMPLE 3
Find the solution to the system of equations using the substitution method: $latex \begin{cases}2(2x-4)+y=3 \\ -x+2y=4 \end{cases}$
Solution
Step 1: We can simplify the first equation:
$latex \begin{cases}4x-8+y=3 \\ -x+2y=4 \end{cases}$
Step 2: Solving the first equation for y, we have:
$latex 4x-8+y=3$
$latex y=-4x+11$
Step 3: Substituting $latex y=-4x+11$ into the second equation, we have:
$latex -x+2y=4$
$latex -x+2(-4x+11)=4$
$latex -x-8x+22=4$
Step 4: Solving for x, we have:
$latex -x-8x+22=4$
$latex -9x=-18$
$latex x=2$
Step 5: Substituting $latex x=2$ into the second equation, we have:
$latex -x+2y=4$
$latex -2+2y=4$
$latex 2y=6$
$latex y=3$
The solution to the system is $latex x=2,~~y=3$.
EXAMPLE 4
Solve the system of equations: $latex \begin{cases}3x+4y-27=0 \\ 5x+y-11=0 \end{cases}$
Solution
Step 1: The equations are already simplified.
Step 2: We can solve the second equation for y:
$latex 5x+y-11=0$
$latex y=-5x+11$
Step 3: Substituting $latex y=-5x+11$ into the first equation, we have:
$latex 3x+4y-27=0$
$latex 3x+4(-5x+11)-27=0$
$latex 3x-20x+44-27=0$
$latex -17x+17=0$
Step 4: Solving for x, we have:
$latex -17x+17=0$
$latex -17x=-17$
$latex x=1$
Step 5: Substituting $latex x=1$ into the second equation, we have:
$latex 5x+y-11=0$
$latex 5(1)+y-11=0$
$latex y-6=0$
$latex y=6$
The solution to the system is $latex x=1,~~y=6$.
EXAMPLE 5
Find the solution to the system of equations: $latex \begin{cases}2(-x+y)=-3x+y+9 \\ 2x+y=13 \end{cases}$
Solution
Step 1: Simplifying the first equation, we have:
$latex \begin{cases}x+y=9 \\ 2x+y=13 \end{cases}$
Step 2: Solving the first equation for y, we have:
$latex x+y=9$
$latex y=9-x$
Step 3: Substituting $latex y=9-x$ into the second equation, we have:
$latex 2x+y=13$
$latex 2x+9-x=13$
Step 4: Solving for x, we have:
$latex 2x+9-x=13$
$latex x=4$
Step 5: Substituting $latex x=4$ into the first equation, we have:
$latex x+y=9$
$latex 4+y=9$
$latex y=5$
The system solution is $latex x=4,~~y=5$.
EXAMPLE 6
Find the solution to the system of equations: $latex \begin{cases}2x-3y=7 \\ 2x+3y=1 \end{cases}$
Solution
Step 1: The equations are already simplified.
Step 2: Solving the first equation for x, we have:
$latex 2x-3y=7$
$latex 2x=3y+7$
$latex x=\frac{3y+7}{2}$
Step 3: Substituting $latex x=\frac{3y+7}{2}$ into the second equation, we have:
$latex 2x+3y=1$
$latex 2\left(\frac{3y+7}{2}\right)+3y=1$
$latex 3y+7+3y=1$
Step 4: Solving for y, we have:
$latex 3y+7+3y=1$
$latex 6y=-6$
$latex y=-1$
Step 5: Substituting $latex y=-1$ into the second equation, we have:
$latex 2x+3y=1$
$latex 2x+3(-1)=1$
$latex 2x-3=1$
$latex 2x=4$
$latex x=2$
The solution to the system is $latex x=2,~~y=-1$.
EXAMPLE 7
Solve the system of equations using the substitution method: $latex \begin{cases}3x-4y=5 \\ 6x-4y=2 \end{cases}$
Solution
Step 1: We can simplify the second equation by dividing it by 2:
$latex \begin{cases}3x-4y=5 \\ 3x-2y=1 \end{cases}$
Step 2: Solving the second equation for x, we have:
$latex 3x-2y=1$
$latex 3x=2y+1$
$latex x=\frac{2y+1}{3}$
Step 3: Substituting $latex x=\frac{2y+1}{3}$ into the first equation, we have:
$latex 3x-4y=5$
$latex 3\left(\frac{2y+1}{3}\right)-4y=5$
$latex 2y+1-4y=5$
Step 4: Solving for y, we have:
$latex 2y+1-4y=5$
$latex -2y=4$
$latex y=-2$
Step 5: Substituting $latex y=-2$ into the second equation, we have:
$latex 3x-2y=1$
$latex 3x-2(-2)=1$
$latex 3x+4=1$
$latex 3x=-3$
$latex x=-1$
The solution to the system is $latex x=-1,~~y=-2$.
Systems of equations by substitution – Practice problems
Solve the system of equations: $latex \begin{cases}x-2y=5 \\ 3x+y=8 \end{cases}$
Write the answer in the form x=?, y=?.
See also
Interested in learning more about systems of equations? Take a look at these pages:
Jefferson Huera Guzman
Jefferson is the lead author and administrator of Neurochispas.com. The interactive Mathematics and Physics content that I have created has helped many students.
