Math Core

Graphing shows what a system means, but it can only estimate an intersection like $\left(\tfrac{2}{3}, \tfrac{5}{3}\right)$ . The substitution method finds the exact solution with algebra alone, by turning a system of two equations into one equation in one variable, which you already know how to solve.

The idea: replace a variable with its equal

Suppose a system tells you that $y = 3x - 4$ . Then $y$ and $3x - 4$ are the same number, so anywhere $y$ appears in the other equation, you may write $3x - 4$ instead. That swap removes $y$ entirely.

\begin{cases} y = 3x - 4 \\ 2x + y = 11 \end{cases} \quad\Longrightarrow\quad 2x + (3x - 4) = 11

The new equation has only $x$ in it. Solve it, then use the value of $x$ to find $y$ .

The substitution method

Isolate one variable in one of the equations. Choose a variable whose coefficient is $1$ or $-1$ if you can.
Substitute that expression into the other equation, in parentheses.
Solve the resulting one-variable equation.
Back-substitute the value into the isolated equation to find the other variable.
Check the ordered pair in both original equations.

Worked example: One equation is already solved

Solve the system.

\begin{cases} y = 3x - 4 \\ 2x + y = 11 \end{cases}

The first equation already isolates $y$ . Substitute $3x - 4$ for $y$ in the second:

\begin{aligned} 2x + (3x - 4) &= 11 \\ 5x - 4 &= 11 \\ 5x &= 15 \\ x &= 3 \end{aligned}

Back-substitute into $y = 3x - 4$ : $y = 3(3) - 4 = 5$ .

Check: $5 = 3(3) - 4$ is true, and $2(3) + 5 = 11$ is true. The solution is $(3, 5)$ .

Choosing what to isolate

When neither equation is solved for a variable, look for a coefficient of $1$ or $-1$ . Isolating that variable avoids fractions.

Worked example: Isolating a variable first

Solve the system.

\begin{cases} x - 2y = 1 \\ 3x + 4y = 23 \end{cases}

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

Substitute into the second equation. The parentheses are essential, because the $3$ must multiply the whole expression:

\begin{aligned} 3(2y + 1) + 4y &= 23 \\ 6y + 3 + 4y &= 23 \\ 10y + 3 &= 23 \\ 10y &= 20 \\ y &= 2 \end{aligned}

Back-substitute: $x = 2(2) + 1 = 5$ .

Check: $5 - 2(2) = 1$ and $3(5) + 4(2) = 15 + 8 = 23$ . The solution is $(5, 2)$ .

Common mistake

Two mistakes cause most wrong answers with substitution:

Dropping the parentheses. Writing $3 \cdot 2y + 1$ instead of $3(2y + 1)$ multiplies only part of the expression. Always wrap the substituted expression in parentheses, then distribute.
Substituting back into the same equation. If you isolate $x$ from the first equation, substitute into the second. Putting $x = 2y + 1$ back into $x - 2y = 1$ gives $2y + 1 - 2y = 1$ , or $1 = 1$ : true, but useless.

Substitution gives exact answers even when they are fractions, which is exactly where graphing fails.

Worked example: A fractional solution

Solve the system.

\begin{cases} 4x + y = 7 \\ 6x - 2y = 7 \end{cases}

Isolate $y$ in the first equation: $y = 7 - 4x$ . Substitute into the second:

\begin{aligned} 6x - 2(7 - 4x) &= 7 \\ 6x - 14 + 8x &= 7 && (-2)(-4x) = +8x \\ 14x &= 21 \\ x &= \tfrac{21}{14} = \tfrac{3}{2} \end{aligned}

Back-substitute: $y = 7 - 4\left(\tfrac{3}{2}\right) = 7 - 6 = 1$ .

Check: $4\left(\tfrac{3}{2}\right) + 1 = 6 + 1 = 7$ and $6\left(\tfrac{3}{2}\right) - 2(1) = 9 - 2 = 7$ . The solution is $\left(\tfrac{3}{2}, 1\right)$ .

When both equations are solved for the same variable

If both equations look like $y = \ldots$ , substitution simply means setting the two expressions equal. For $y = -2x + 7$ and $y = 3x - 8$ :

-2x + 7 = 3x - 8 \quad\Longrightarrow\quad 15 = 5x \quad\Longrightarrow\quad x = 3,

and $y = -2(3) + 7 = 1$ . This is the algebraic version of asking "where do the two lines have the same height?"

When the variable disappears

Sometimes, after you substitute and simplify, both variables cancel. What's left tells you which special case you're in.

A false statement such as $-6 = 5$ means no ordered pair can work: the system has no solution (parallel lines).
A true statement such as $4 = 4$ means every point on the line works: the system has infinitely many solutions (the same line).

Worked example: Two special cases

(a) Solve $y = 2x + 3$ and $4x - 2y = 5$ .

4x - 2(2x + 3) = 5 \quad\Longrightarrow\quad 4x - 4x - 6 = 5 \quad\Longrightarrow\quad -6 = 5.

That is false for every $x$ , so the system has no solution. (Indeed, $4x - 2y = 5$ rewrites as $y = 2x - \tfrac{5}{2}$ , which is parallel to $y = 2x + 3$ .)

(b) Solve $x = 4 - 2y$ and $3x + 6y = 12$ .

3(4 - 2y) + 6y = 12 \quad\Longrightarrow\quad 12 - 6y + 6y = 12 \quad\Longrightarrow\quad 12 = 12.

That is true for every $y$ , so there are infinitely many solutions: every point on the line $x + 2y = 4$ .

Tip

A false statement like $-6 = 5$ isn't a sign you made a mistake; it's the answer "no solution." But it's worth a quick slope check: if the two slopes are different, the lines must cross, and the false statement means an arithmetic slip somewhere.

Practice

Practice 1

Solve by substitution: $y = x + 3$ and $3x + y = 15$ .

Enter a point like (2, -3)

Practice 2

Solve by substitution: $x = 2y - 1$ and $2x + 3y = 12$ .

Enter a point like (2, -3)

Practice 3

Solve: $y = -2x + 7$ and $y = 3x - 8$ .

Enter a point like (2, -3)

Practice 4

Solve by substitution: $x + 4y = 10$ and $3x - 2y = 2$ .

Enter a point like (2, -3)

Practice 5

Solve by substitution: $2x - y = 5$ and $5x + 3y = 7$ .

Enter a point like (2, -3)

Practice 6

Solve the system $x - 3y = 4$ and $-2x + 6y = -8$ .

Practice 7

Solve by substitution: $x + 2y = 3$ and $3x - 4y = 4$ .

Enter a point like (2, -3)

Practice 8

Solve by substitution: $y = \dfrac{1}{2}x + 1$ and $3x - 4y = 2$ .

Enter a point like (2, -3)