Math Core

Graphing is a great way to see a solution, but reading a crossing point off a grid is slow and sometimes impossible to do exactly. Substitution turns a system of two equations into one equation with one variable, which you already know how to solve.

The idea

Look at this system:

\begin{aligned} y &= 2x \\ x + y &= 12 \end{aligned}

The first equation says $y$ and $2x$ are equal. So anywhere you see $y$ in the second equation, you can swap in $2x$ . That leaves an equation with only $x$ in it.

Solving by substitution

Solve one equation for one variable (skip this if it's already done).
Substitute that expression into the other equation.
Solve the new one-variable equation.
Plug that value back in to find the other variable.
Check the pair in both original equations.

Worked example: One variable is already alone

Solve the system $y = 2x$ and $x + y = 12$ .

Replace $y$ with $2x$ in the second equation:

\begin{aligned} x + 2x &= 12 \\ 3x &= 12 \\ x &= 4 \end{aligned}

Now find $y$ using $y = 2x$ : $y = 2(4) = 8$ .

Check: $8 = 2(4)$ ✓ and $4 + 8 = 12$ ✓. The solution is $(4, 8)$ .

Worked example: Substituting a binomial

Solve the system $y = x + 3$ and $2x + y = 15$ .

Replace $y$ with $x + 3$ . Use parentheses so the whole expression goes in:

\begin{aligned} 2x + (x + 3) &= 15 \\ 3x + 3 &= 15 \\ 3x &= 12 \\ x &= 4 \end{aligned}

Then $y = 4 + 3 = 7$ .

Check: $7 = 4 + 3$ ✓ and $2(4) + 7 = 15$ ✓. The solution is $(4, 7)$ .

When neither variable is alone

If no variable is by itself yet, pick the easiest one to isolate. A variable with a coefficient of $1$ or $-1$ is the best choice, because you won't create fractions.

Worked example: Isolate first

Solve the system $x - 2y = 1$ and $3x + y = 17$ .

The $x$ in the first equation has coefficient $1$ . Add $2y$ to both sides: $x = 2y + 1$ .

Substitute into the second equation and distribute:

\begin{aligned} 3(2y + 1) + y &= 17 \\ 6y + 3 + y &= 17 \\ 7y + 3 &= 17 \\ 7y &= 14 \\ y &= 2 \end{aligned}

Then $x = 2(2) + 1 = 5$ .

Check: $5 - 2(2) = 1$ ✓ and $3(5) + 2 = 17$ ✓. The solution is $(5, 2)$ .

Common mistake

When you substitute an expression like $2y + 1$ for a variable that has a coefficient, put the expression in parentheses and distribute. Writing $3 \cdot 2y + 1$ instead of $3(2y + 1)$ multiplies only the first term and gives the wrong answer.

When the variable disappears

Sometimes both variables cancel out when you substitute. What's left tells you what kind of system you have.

Worked example: No solution

Solve the system $y = 3x + 1$ and $6x - 2y = 5$ .

Substitute $3x + 1$ for $y$ :

\begin{aligned} 6x - 2(3x + 1) &= 5 \\ 6x - 6x - 2 &= 5 \\ -2 &= 5 \end{aligned}

The variable vanished and left a false statement. No value of $x$ can make $-2 = 5$ true, so the system has no solution. (The lines are parallel: both have slope $3$ .)

If the variable vanishes and leaves a true statement, like $4 = 4$ , every point on the line works. The system has infinitely many solutions.

Tip

After you find one variable, plug it into whichever equation is easiest, usually the one you solved for a variable in step 1. Then check in the other equation to catch mistakes.

Practice

Practice 1

Solve the system $y = 3x$ and $x + y = 20$ .

Enter a point like (2, -3)

Practice 2

Solve the system $y = x - 4$ and $3x + y = 16$ .

Enter a point like (2, -3)

Practice 3

Solve the system $x = 2y$ and $3x - y = 25$ .

Enter a point like (2, -3)

Practice 4

Solve the system $x + y = 10$ and $2x - y = 11$ .

Enter a point like (2, -3)

Practice 5

Solve the system $y = -2x + 5$ and $4x + 3y = 9$ .

Enter a point like (2, -3)

Practice 6

Solve the system $x - 3y = -2$ and $2x + 5y = 18$ .

Enter a point like (2, -3)

Practice 7

Use substitution to decide how many solutions the system $y = 2x - 1$ and $4x - 2y = 2$ has.

Practice 8

Solve the system $y = \dfrac{1}{2}x + 3$ and $y = 2x - 3$ .

Enter a point like (2, -3)