Solving Linear Equations

An equation says that two expressions are equal, like $4x - 7 = 21$. Solving it means finding the value of the variable that makes the equation true. Linear equations show up everywhere: comparing phone plans, finding the side of a garden from its perimeter, or working out how many weeks it takes to save for something. This page builds up from one-step equations to equations with brackets and fractions.

Key ideas

Keep the equation balanced

Think of an equation as a balance scale. The two sides are equal, and they stay equal as long as you do the same thing to both sides: add, subtract, multiply or divide by the same number (but never divide by $0$).

To get the variable by itself, use inverse operations. They undo each other:

To undo…	…do this to both sides
$+\,5$	subtract $5$
$-\,5$	add $5$
$\times\,3$	divide by $3$
$\div\,3$	multiply by $3$

For a two-step equation like $4x - 7 = 21$, undo the steps in reverse order. $x$ was multiplied by $4$ and then $7$ was subtracted, so first add $7$, then divide by $4$.

A general plan

1. Expand any brackets (distributive property).
2. Clear fractions by multiplying every term on both sides by the lowest common denominator (LCD).
3. Collect the variable terms on one side and the constants on the other.
4. Isolate the variable by dividing by its coefficient.
5. Check by substituting your answer into the original equation.

Variables on both sides

When the variable appears on both sides, move all the variable terms to one side first. It’s usually easiest to move the smaller one, so the coefficient stays positive. For $7x + 2 = 3x + 14$, subtract $3x$ from both sides to get $4x + 2 = 14$.

Equations with fractions

Fractions make equations messy. Multiply every term on both sides by the LCD of the denominators, and the fractions disappear. Put brackets around numerators with more than one term, so the multiplication reaches all of it.

No solution, or infinitely many

Sometimes the variable disappears completely:

• $2(x + 3) = 2x + 1$ becomes $2x + 6 = 2x + 1$, then $6 = 1$. That’s false, so there is no solution. No value of $x$ works.
• $2(x + 3) = 2x + 6$ becomes $6 = 6$. That’s always true, so there are infinitely many solutions. Every value of $x$ works, because the two sides are equivalent expressions.

Writing equations from word problems

1. Let a variable stand for the unknown, and say what it means (with units).
2. Write the other quantities in terms of that variable.
3. Write an equation using the information in the problem.
4. Solve it.
5. Answer the question in a sentence, and check that the answer makes sense in the story.

Worked examples

Example 1: One and two steps

Solve.

• (a) $x - 8 = -3$
• (b) $4x - 7 = 21$

Solution.

(a) Undo “subtract $8$” by adding $8$ to both sides:

$$x - 8 + 8 = -3 + 8 \quad\Rightarrow\quad x = 5$$

(b) Undo the subtraction first, then the multiplication:

$$\begin{aligned} 4x - 7 &= 21 \\ 4x &= 28 && \text{add } 7 \text{ to both sides} \\ x &= 7 && \text{divide both sides by } 4 \end{aligned}$$

Check: $4(7) - 7 = 28 - 7 = 21$. ✓

Example 2: Brackets and variables on both sides

Solve $5(x - 2) = 3x + 8$.

Solution.

$$\begin{aligned} 5(x - 2) &= 3x + 8 \\ 5x - 10 &= 3x + 8 && \text{expand} \\ 2x - 10 &= 8 && \text{subtract } 3x \\ 2x &= 18 && \text{add } 10 \\ x &= 9 && \text{divide by } 2 \end{aligned}$$

Check: left side $5(9 - 2) = 5(7) = 35$; right side $3(9) + 8 = 27 + 8 = 35$. ✓

Example 3: Fractions

Solve $\dfrac{x}{3} + \dfrac{x - 1}{4} = 5$.

Solution. The LCD of $3$ and $4$ is $12$. Multiply every term by $12$:

$$\begin{aligned} 12\left(\frac{x}{3}\right) + 12\left(\frac{x - 1}{4}\right) &= 12(5) \\ 4x + 3(x - 1) &= 60 \\ 4x + 3x - 3 &= 60 \\ 7x - 3 &= 60 \\ 7x &= 63 \\ x &= 9 \end{aligned}$$

Check: $\dfrac{9}{3} + \dfrac{9 - 1}{4} = 3 + 2 = 5$. ✓

Notice the brackets around $x - 1$. Without them, you’d only multiply the $x$ by $3$ and get the wrong answer.

Example 4: Comparing phone plans

Plan A costs $20 per month plus $5 per GB of data. Plan B costs $35 per month plus $2 per GB. For how many GB do the two plans cost the same? What is that cost?

Solution. Let $g$ be the number of GB used in a month. In dollars:

• Plan A costs $20 + 5g$.
• Plan B costs $35 + 2g$.

The costs are equal when

$$\begin{aligned} 20 + 5g &= 35 + 2g \\ 20 + 3g &= 35 && \text{subtract } 2g \\ 3g &= 15 && \text{subtract } 20 \\ g &= 5 && \text{divide by } 3 \end{aligned}$$

At $g = 5$, Plan A costs $20 + 5(5) = 45$ dollars.

Answer: the plans cost the same, $45, when you use 5 GB.

Check: Plan B costs $35 + 2(5) = 45$ dollars too. ✓ (For less than 5 GB, Plan A is cheaper; for more, Plan B is. You’ll compare relations like these again in comparing linear relations.)

Common mistakes

Doing something to only one side. If you subtract $3x$ on the left, you must subtract $3x$ on the right too. Otherwise the equation is no longer balanced. Writing each step as a new line helps.

Undoing steps in the wrong order. In $4x - 7 = 21$, dividing by $4$ first means you must divide every term: $x - \dfrac{7}{4} = \dfrac{21}{4}$. That works but is messy. Undo addition and subtraction first, then multiplication and division.

Not multiplying every term by the LCD. In $\dfrac{x}{2} + 3 = 7$, multiplying by $2$ gives $x + 6 = 14$, not $x + 3 = 14$. The $3$ and the $7$ must be multiplied too.

Sign errors with brackets. $-2(x - 4)$ is $-2x + 8$. And when you move a term across the equals sign, you’re really subtracting (or adding) it on both sides, so its sign flips.

Skipping the check. Substitute your answer into the original equation, not a later line, because a mistake in an early step would carry through to the later lines.

Panicking when the variable disappears. If you end up with a false statement like $6 = 1$, there is no solution. If you end up with a true statement like $6 = 6$, every number is a solution. Both are real answers.

Practice

1. (Warm-up) Solve.

• (a) $x + 9 = 4$
• (b) $-3y = 27$
• (c) $\dfrac{m}{5} = -2$

Solution

(a) Subtract $9$ from both sides: $x = 4 - 9 = -5$.

(b) Divide both sides by $-3$: $y = \dfrac{27}{-3} = -9$.

(c) Multiply both sides by $5$: $m = -10$.

2. (Warm-up) Solve $6x + 5 = -13$ and check your answer.

Solution

$$\begin{aligned} 6x + 5 &= -13 \\ 6x &= -18 && \text{subtract } 5 \\ x &= -3 && \text{divide by } 6 \end{aligned}$$

Check: $6(-3) + 5 = -18 + 5 = -13$. ✓

3. (Core) Solve $7x - 4 = 3x + 20$.

Solution

$$\begin{aligned} 7x - 4 &= 3x + 20 \\ 4x - 4 &= 20 && \text{subtract } 3x \\ 4x &= 24 && \text{add } 4 \\ x &= 6 \end{aligned}$$

Check: $7(6) - 4 = 38$ and $3(6) + 20 = 38$. ✓

4. (Core) Solve $3(2x - 1) - 4 = 2(x + 5) + 3$.

Solution

$$\begin{aligned} 3(2x - 1) - 4 &= 2(x + 5) + 3 \\ 6x - 3 - 4 &= 2x + 10 + 3 && \text{expand} \\ 6x - 7 &= 2x + 13 && \text{collect like terms} \\ 4x - 7 &= 13 && \text{subtract } 2x \\ 4x &= 20 && \text{add } 7 \\ x &= 5 \end{aligned}$$

Check: left side $3(9) - 4 = 23$; right side $2(10) + 3 = 23$. ✓

5. (Core) Solve $\dfrac{2x}{3} - 1 = \dfrac{x + 4}{2}$.

Solution

The LCD of $3$ and $2$ is $6$. Multiply every term by $6$:

$$\begin{aligned} 6\left(\frac{2x}{3}\right) - 6(1) &= 6\left(\frac{x + 4}{2}\right) \\ 4x - 6 &= 3(x + 4) \\ 4x - 6 &= 3x + 12 \\ x - 6 &= 12 \\ x &= 18 \end{aligned}$$

Check: left side $\dfrac{36}{3} - 1 = 11$; right side $\dfrac{22}{2} = 11$. ✓

6. (Core) A rectangular vegetable garden is 4 m longer than it is wide. Its perimeter is 56 m. Find its width and length.

Solution

Let $w$ be the width in metres. Then the length is $w + 4$. Perimeter is $2(\text{length}) + 2(\text{width})$:

$$\begin{aligned} 2(w + 4) + 2w &= 56 \\ 2w + 8 + 2w &= 56 \\ 4w + 8 &= 56 \\ 4w &= 48 \\ w &= 12 \end{aligned}$$

The garden is $12$ m wide and $16$ m long.

Check: $2(16) + 2(12) = 32 + 24 = 56$. ✓

7. (Core) Three consecutive even numbers add to 150. What are they?

Solution

Consecutive even numbers go up by 2. Let the smallest be $n$. Then the numbers are $n$, $n + 2$ and $n + 4$:

$$\begin{aligned} n + (n + 2) + (n + 4) &= 150 \\ 3n + 6 &= 150 \\ 3n &= 144 \\ n &= 48 \end{aligned}$$

The numbers are $48$, $50$ and $52$.

Check: $48 + 50 + 52 = 150$, and all three are even. ✓

8. (Challenge) Solve each equation, or explain why it has no solution or infinitely many solutions.

• (a) $4(x - 1) + 2 = 4x - 2$
• (b) $3(x + 2) = 3x - 5$

Solution

(a) Expand the left side: $4x - 4 + 2 = 4x - 2$, so $4x - 2 = 4x - 2$. Subtract $4x$: $-2 = -2$. That’s always true, so there are infinitely many solutions: every value of $x$ works. (The two sides are equivalent expressions.)

(b) Expand: $3x + 6 = 3x - 5$. Subtract $3x$: $6 = -5$. That’s false, so there is no solution.

9. (Challenge) For what value of $k$ does the equation $2(3x - 1) = kx + 5$ have no solution? Can you choose $k$ so it has infinitely many solutions?

Solution

Expand the left side: $6x - 2 = kx + 5$.

If $k = 6$, subtracting $6x$ from both sides gives $-2 = 5$, which is false. So $k = 6$ gives no solution.

For any other $k$, the $x$ terms don’t cancel, and you can solve: $(6 - k)x = 7$, so $x = \dfrac{7}{6 - k}$. That’s exactly one solution.

Infinitely many solutions would need both sides to be identical, which would need $-2 = 5$ as well as $k = 6$. That never happens, so no value of $k$ gives infinitely many solutions.