Solving Linear Systems by Substitution

Graphing shows where two lines cross, but it can only estimate an answer like $\left(\tfrac{8}{7}, \tfrac{11}{7}\right)$. Substitution finds the exact solution with algebra. The idea is to use one equation to write one variable in terms of the other, so that the second equation has only one variable left, which you already know how to solve.

Key ideas

The method

1. Isolate one variable in one of the equations (for example, get $y = \ldots$ or $x = \ldots$).
2. Substitute that expression into the other equation. Use brackets!
3. Solve the new equation, which has only one variable.
4. Back-substitute that value into the isolated equation from step 1 to find the other variable.
5. Check the ordered pair in both original equations, and state the solution.

Why does this work? At the solution, $x$ and $y$ have the same values in both equations. So if the first equation says $y$ equals $2x - 7$, you can replace $y$ with $2x - 7$ in the second equation too.

You already used a special case in Grade 9: when both equations are solved for $y$, setting the right sides equal is substitution.

Choosing which variable to isolate

Look for a variable with a coefficient of $1$ or $-1$. Isolating it doesn’t create any fractions.

System	Best choice
$y = 2x - 7$ and $3x + 4y = 5$	$y$ is already isolated
$x - 3y = -1$ and $2x + 5y = 20$	isolate $x$ in the first: $x = 3y - 1$
$4x - y = 3$ and $2x + 3y = 7$	isolate $y$ in the first: $y = 4x - 3$

If no coefficient is $1$ or $-1$, substitution still works but creates fractions. In that case elimination is usually easier.

Systems with fractions or decimals

Clear them first. Multiply every term of an equation by the lowest common denominator (or by $10$ or $100$ for decimals). The new equation has the same solutions, and integer coefficients are much easier to work with.

When something strange happens

If the variable disappears and you’re left with a false statement such as $8 = 5$, the system has no solution (the lines are parallel). If you’re left with a true statement such as $0 = 0$, it has infinitely many solutions (the lines are the same). The elimination page looks at these cases in more detail.

Worked examples

Example 1: One variable already isolated

Solve the system.

$$\begin{aligned} y &= 2x - 7 \\ 3x + 4y &= 5 \end{aligned}$$

Solution. The first equation already gives $y$. Substitute $2x - 7$ for $y$ in the second equation:

$$\begin{aligned} 3x + 4(2x - 7) &= 5 \\ 3x + 8x - 28 &= 5 \\ 11x &= 33 \\ x &= 3 \end{aligned}$$

Back-substitute into $y = 2x - 7$: $y = 2(3) - 7 = -1$.

Check in the second equation: $3(3) + 4(-1) = 9 - 4 = 5$ ✓. (The first equation is true by construction: $2(3) - 7 = -1$ ✓.)

The solution is $(3, -1)$.

Example 2: Choosing what to isolate

Solve the system.

$$\begin{aligned} x - 3y &= -1 \\ 2x + 5y &= 20 \end{aligned}$$

Solution. In the first equation, $x$ has a coefficient of $1$, so isolate $x$:

$$x = 3y - 1$$

Substitute into the second equation:

$$\begin{aligned} 2(3y - 1) + 5y &= 20 \\ 6y - 2 + 5y &= 20 \\ 11y &= 22 \\ y &= 2 \end{aligned}$$

Back-substitute: $x = 3(2) - 1 = 5$.

Check: $5 - 3(2) = -1$ ✓ and $2(5) + 5(2) = 20$ ✓. The solution is $(5, 2)$.

Example 3: A system with fractions

Solve the system.

$$\begin{aligned} \frac{x}{2} + \frac{y}{3} &= 4 \\ \frac{x}{4} - y &= -5 \end{aligned}$$

Solution. Clear the fractions first. Multiply every term of the first equation by $6$, and every term of the second by $4$:

$$\begin{aligned} 3x + 2y &= 24 \\ x - 4y &= -20 \end{aligned}$$

Now isolate $x$ in the second equation: $x = 4y - 20$. Substitute into the first:

$$\begin{aligned} 3(4y - 20) + 2y &= 24 \\ 12y - 60 + 2y &= 24 \\ 14y &= 84 \\ y &= 6 \end{aligned}$$

Back-substitute: $x = 4(6) - 20 = 4$.

Check in the original equations: $\dfrac{4}{2} + \dfrac{6}{3} = 2 + 2 = 4$ ✓ and $\dfrac{4}{4} - 6 = 1 - 6 = -5$ ✓. The solution is $(4, 6)$.

Example 4: An answer graphing couldn’t find

Solve the system.

$$\begin{aligned} 2x + 3y &= 7 \\ 4x - y &= 3 \end{aligned}$$

Solution. In the second equation $y$ has a coefficient of $-1$. Isolate it:

$$4x - y = 3 \quad\Rightarrow\quad -y = 3 - 4x \quad\Rightarrow\quad y = 4x - 3$$

Substitute into the first equation:

$$\begin{aligned} 2x + 3(4x - 3) &= 7 \\ 2x + 12x - 9 &= 7 \\ 14x &= 16 \\ x &= \frac{8}{7} \end{aligned}$$

Back-substitute:

$$y = 4\left(\frac{8}{7}\right) - 3 = \frac{32}{7} - \frac{21}{7} = \frac{11}{7}$$

Check in the first equation: $2\left(\dfrac{8}{7}\right) + 3\left(\dfrac{11}{7}\right) = \dfrac{16}{7} + \dfrac{33}{7} = \dfrac{49}{7} = 7$ ✓.

The solution is $\left(\dfrac{8}{7}, \dfrac{11}{7}\right)$. On a graph this would look like “about $(1.1, 1.6)$”, but substitution gives the exact answer.

Common mistakes

Forgetting the brackets. Substituting $y = 2x - 7$ into $3x + 4y = 5$ gives $3x + 4(2x - 7) = 5$. Without brackets you’d write $3x + 4 \cdot 2x - 7$ and multiply only the first term by $4$.

Substituting back into the same equation. If you isolate $x$ from the first equation, substitute it into the second. Putting it back into the first gives something like $0 = 0$ and tells you nothing.

Stopping after one variable. Finding $x = 3$ is only half the answer. A solution is an ordered pair: back-substitute to find $y$ too.

Sign errors when isolating. From $4x - y = 3$, the result is $y = 4x - 3$, not $y = 3 - 4x$. Check by substituting a value: with $x = 1$, both $4(1) - y = 3$ and $y = 4(1) - 3$ give $y = 1$.

Clearing fractions from only some terms. When you multiply an equation by $6$, every term gets multiplied, including the one on the right side.

Checking in only one equation. The equation you used to back-substitute will always check. The real test is the other one, so check both.

Practice

1. (Warm-up) Solve: $y = x + 1$ and $2x + y = 10$.

Solution

Substitute $x + 1$ for $y$:

$$2x + (x + 1) = 10 \quad\Rightarrow\quad 3x + 1 = 10 \quad\Rightarrow\quad x = 3$$

Then $y = 3 + 1 = 4$. Check: $2(3) + 4 = 10$ ✓. The solution is $(3, 4)$.

2. (Warm-up) Solve: $x = 2y$ and $3x - 4y = 6$.

Solution

Substitute $2y$ for $x$:

$$3(2y) - 4y = 6 \quad\Rightarrow\quad 2y = 6 \quad\Rightarrow\quad y = 3$$

Then $x = 2(3) = 6$. Check: $3(6) - 4(3) = 18 - 12 = 6$ ✓. The solution is $(6, 3)$.

3. (Core) Solve: $3x + y = 5$ and $2x - 3y = 18$.

Solution

Isolate $y$ in the first equation: $y = 5 - 3x$. Substitute into the second:

$$\begin{aligned} 2x - 3(5 - 3x) &= 18 \\ 2x - 15 + 9x &= 18 \\ 11x &= 33 \\ x &= 3 \end{aligned}$$

Then $y = 5 - 3(3) = -4$.

Check: $3(3) + (-4) = 5$ ✓ and $2(3) - 3(-4) = 6 + 12 = 18$ ✓. The solution is $(3, -4)$.

4. (Core) Solve: $2x + 5y = -1$ and $x + 3y = 0$.

Solution

Isolate $x$ in the second equation: $x = -3y$. Substitute into the first:

$$2(-3y) + 5y = -1 \quad\Rightarrow\quad -y = -1 \quad\Rightarrow\quad y = 1$$

Then $x = -3(1) = -3$.

Check: $2(-3) + 5(1) = -1$ ✓ and $-3 + 3(1) = 0$ ✓. The solution is $(-3, 1)$.

5. (Core) Solve: $y = \dfrac{2}{3}x + 1$ and $x - 3y = -9$.

Solution

Substitute for $y$ in the second equation:

$$\begin{aligned} x - 3\left(\frac{2}{3}x + 1\right) &= -9 \\ x - 2x - 3 &= -9 \\ -x &= -6 \\ x &= 6 \end{aligned}$$

Then $y = \dfrac{2}{3}(6) + 1 = 5$.

Check: $6 - 3(5) = -9$ ✓. The solution is $(6, 5)$.

6. (Core) Solve: $\dfrac{x + 1}{2} + \dfrac{y}{4} = 3$ and $x - y = 2$.

Solution

Multiply every term of the first equation by $4$:

$$2(x + 1) + y = 12 \quad\Rightarrow\quad 2x + 2 + y = 12 \quad\Rightarrow\quad 2x + y = 10$$

From the second equation, $y = x - 2$. Substitute:

$$2x + (x - 2) = 10 \quad\Rightarrow\quad 3x = 12 \quad\Rightarrow\quad x = 4$$

Then $y = 4 - 2 = 2$.

Check in the original first equation: $\dfrac{4 + 1}{2} + \dfrac{2}{4} = \dfrac{5}{2} + \dfrac{1}{2} = 3$ ✓. The solution is $(4, 2)$.

7. (Core) A rectangular garden has a perimeter of $46$ m. Its length is $5$ m more than its width. Write a system of equations and solve it by substitution to find the dimensions.

Solution

Let $l$ be the length and $w$ the width, in metres.

$$\begin{aligned} l &= w + 5 \\ 2l + 2w &= 46 \end{aligned}$$

Substitute $w + 5$ for $l$:

$$2(w + 5) + 2w = 46 \quad\Rightarrow\quad 4w + 10 = 46 \quad\Rightarrow\quad w = 9$$

Then $l = 9 + 5 = 14$. The garden is $14$ m by $9$ m.

Check: $2(14) + 2(9) = 28 + 18 = 46$ ✓.

8. (Challenge) Try to solve $y = 3x - 4$ and $6x - 2y = 5$ by substitution. What happens, and what does it tell you about the graphs?

Solution

Substitute $3x - 4$ for $y$:

$$\begin{aligned} 6x - 2(3x - 4) &= 5 \\ 6x - 6x + 8 &= 5 \\ 8 &= 5 \end{aligned}$$

The $x$ disappears and the result is false. No values of $x$ and $y$ can make both equations true, so the system has no solution.

On a graph, the second equation is $y = 3x - \dfrac{5}{2}$. Both lines have slope $3$ but different $y$-intercepts ($-4$ and $-\dfrac{5}{2}$), so they are parallel and never meet.

9. (Challenge) Solve: $0.4x + 0.3y = 2.5$ and $0.2x - y = -2.2$.

Solution

Multiply both equations by $10$ to clear the decimals:

$$\begin{aligned} 4x + 3y &= 25 \\ 2x - 10y &= -22 \end{aligned}$$

Divide the second equation by $2$: $x - 5y = -11$, so $x = 5y - 11$. Substitute into the first:

$$\begin{aligned} 4(5y - 11) + 3y &= 25 \\ 20y - 44 + 3y &= 25 \\ 23y &= 69 \\ y &= 3 \end{aligned}$$

Then $x = 5(3) - 11 = 4$.

Check in the original equations: $0.4(4) + 0.3(3) = 1.6 + 0.9 = 2.5$ ✓ and $0.2(4) - 3 = 0.8 - 3 = -2.2$ ✓. The solution is $(4, 3)$.