Equations with Brackets and the Unknown on Both Sides

Solve equations with the unknown on both sides, expand brackets using the distributive law, handle negative multipliers, and verify solutions by substitution.

Before you start

In Lesson 9.1, you solved one-step and two-step equations such as $3x - 4 = 11$ by applying inverse operations to both sides.
The distributive law (Lesson 8.2) states that $a(b + c) = ab + ac$ . Every term inside the bracket must be multiplied by the term outside.
Be especially careful with negative multipliers: $-3(x - 4) = -3x + 12$ . A negative times a negative gives a positive.
Always keep the equation balanced: whatever operation you do to the left-hand side, you must do to the right-hand side.

Equations with the Unknown on Both Sides

When the variable (the unknown) appears on both the left-hand side (LHS) and the right-hand side (RHS), our first goal is to have all the variable terms on one side and all the constant terms on the other. We do this as before, by adding or subtracting the same term on both sides.

It is usually easiest to remove the variable term with the smaller coefficient, so that the coefficient left is positive.

Collecting Variable Terms

1. Compare the variable terms on both sides.

2. Remove the variable term with the smaller coefficient from both sides: subtract it if it is positive, or add its opposite if it is negative (for $-3y$ , add $3y$ to both sides). The variable now appears on only one side.

3. Add or subtract the same number on both sides so that the constant terms are on the other side.

4. Divide both sides by the coefficient to isolate the variable.

Example 1: Variables on Both Sides

Solve each equation and check the solution:
(a) $5x + 3 = 2x + 18$
(b) $4 - 3y = 2y - 11$

Solution

(a) Solve $5x + 3 = 2x + 18$ :

Subtract $2x$ from both sides to collect variables on the left:

$5x - 2x + 3 = 2x - 2x + 18$

$3x + 3 = 18$

Subtract $3$ from both sides:

$3x = 18 - 3 = 15$

Divide both sides by $3$ :

$x = \frac{15}{3} = 5$

Check: $\text{LHS} = 5(5) + 3 = 28$ ; $\text{RHS} = 2(5) + 18 = 28 = \text{LHS} \quad \checkmark$

(b) Solve $4 - 3y = 2y - 11$ :

Add $3y$ to both sides so that the coefficient of $y$ is positive:

$4 - 3y + 3y = 2y - 11 + 3y$

$4 = 5y - 11$

Add $11$ to both sides:

$4 + 11 = 5y - 11 + 11$

$15 = 5y$

Divide both sides by $5$ :

$y = 3$

Check: $\text{LHS} = 4 - 3(3) = 4 - 9 = -5$ ; $\text{RHS} = 2(3) - 11 = 6 - 11 = -5 = \text{LHS} \quad \checkmark$

(a) $x = 5$ ; (b) $y = 3$ .

Equations with Single Brackets

When an equation contains brackets, we first remove the brackets by expanding using the distributive law: $a(b + c) = ab + ac$ .

After expanding, simplify any like terms on each side before rearranging.

Example 2: Equations with Brackets on One Side

Solve each equation:
(a) $4(x - 3) = 20$
(b) $3(2y + 5) - 7 = 26$

Solution

(a) Solve $4(x - 3) = 20$ :

Method 1: Expand the brackets first:

$4x - 12 = 20$

Add $12$ to both sides:

$4x = 20 + 12 = 32$

Divide by $4$ :

$x = 8$

Method 2: Since $20$ is divisible by $4$ , divide both sides by $4$ first:

$x - 3 = \frac{20}{4} = 5$

$x = 5 + 3 = 8$

Check: $\text{LHS} = 4(8 - 3) = 4(5) = 20 = \text{RHS} \quad \checkmark$

(b) Solve $3(2y + 5) - 7 = 26$ :

Expand the bracket:

$6y + 15 - 7 = 26$

Simplify constant terms on the LHS ( $15 - 7 = 8$ ):

$6y + 8 = 26$

Subtract $8$ from both sides:

$6y = 26 - 8 = 18$

Divide both sides by $6$ :

$y = 3$

Check: $\text{LHS} = 3(2(3) + 5) - 7 = 3(6 + 5) - 7 = 3(11) - 7 = 33 - 7 = 26 = \text{RHS} \quad \checkmark$

(a) $x = 8$ ; (b) $y = 3$ .

Negative Multipliers and Subtraction of Brackets

Extra care is required when a minus sign or negative number stands immediately in front of a bracket.

Remember that $-(b + c) = -b - c$ and $-(b - c) = -b + c$ . Every sign inside the bracket must be reversed.

Sign Rules for Expanding Brackets

$\bullet \; +a(b + c) = ab + ac$

$\bullet \; +a(b - c) = ab - ac$

$\bullet \; -a(b + c) = -ab - ac$

$\bullet \; -a(b - c) = -ab + ac$

Example 3: Negative Multipliers

Solve each equation and check the solution:
(a) $7 - (2x - 3) = 16$
(b) $5 - 3(x - 2) = 2$

Solution

(a) Solve $7 - (2x - 3) = 16$ :

Expand the minus sign across the bracket: $-(2x - 3) = -2x + 3$ :

$7 - 2x + 3 = 16$

Combine constant terms on LHS ( $7 + 3 = 10$ ):

$10 - 2x = 16$

Subtract $10$ from both sides:

$-2x = 16 - 10 = 6$

Divide both sides by $-2$ :

$x = \frac{6}{-2} = -3$

Check: $\text{LHS} = 7 - (2(-3) - 3) = 7 - (-6 - 3) = 7 - (-9) = 7 + 9 = 16 = \text{RHS} \quad \checkmark$

(b) Solve $5 - 3(x - 2) = 2$ :

Expand $-3(x - 2) = -3x + 6$ :

$5 - 3x + 6 = 2$

Combine constant terms ( $5 + 6 = 11$ ):

$11 - 3x = 2$

Subtract $11$ from both sides:

$-3x = 2 - 11 = -9$

Divide both sides by $-3$ :

$x = \frac{-9}{-3} = 3$

Check: $\text{LHS} = 5 - 3(3 - 2) = 5 - 3(1) = 5 - 3 = 2 = \text{RHS} \quad \checkmark$

(a) $x = -3$ ; (b) $x = 3$ .

The Multi-Step Solving Strategy (ESCS)

For complex equations with multiple brackets and terms on both sides, follow the four-step ESCS strategy:

The ESCS Method

1. E — Expand: Remove all brackets using the distributive law.

2. S — Simplify: Combine like terms on the LHS and like terms on the RHS separately.

3. C — Collect: Add or subtract the same term on both sides so that the variable terms are on one side and the numbers on the other.

4. S — Solve: Divide by the coefficient of the variable to find the solution, and check by substitution.

Example 4: Multi-Step Equations with Brackets on Both Sides

Solve $3(2x + 1) = 2(x + 7) - 3$ .

Solution

Step 1 (Expand): Expand brackets on both sides:

$6x + 3 = 2x + 14 - 3$

Step 2 (Simplify): Combine constants on the RHS ( $14 - 3 = 11$ ):

$6x + 3 = 2x + 11$

Step 3 (Collect): Subtract $2x$ from both sides:

$4x + 3 = 11$

Subtract $3$ from both sides:

$4x = 11 - 3 = 8$

Step 4 (Solve): Divide by $4$ :

$x = \frac{8}{4} = 2$

Check: $\text{LHS} = 3(2(2) + 1) = 3(4 + 1) = 3(5) = 15$ ; $\text{RHS} = 2(2 + 7) - 3 = 2(9) - 3 = 18 - 3 = 15 = \text{LHS} \quad \checkmark$

$x = 2$ .

Common Mistakes

Multiplying only the first term inside brackets: Writing $3(2x + 5)$ as $6x + 5$ instead of $6x + 15$ . The number outside must multiply every term inside the bracket.
Sign errors with negative multipliers: Writing $-4(x - 3)$ as $-4x - 12$ instead of $-4x + 12$ . A negative times a negative produces a positive.
Subtracting variable terms from only one side: In $5x + 3 = 2x + 18$ , subtracting $2x$ from the LHS without subtracting it from the RHS breaks the balance.
Doing arithmetic before expanding: In $7 - 2(x - 3) = 15$ , computing $7 - 2 = 5$ first breaks the order of operations. Multiplication comes before subtraction.

Practice Questions

Work each question on paper using the ESCS method, showing all steps clearly, then check your solution.

Question 1

Solve $8x - 3 = 2x + 1$ .

ShowHide solution

The variable term with the smaller coefficient is $2x$ . Subtract $2x$ from both sides:

$8x - 3 - 2x = 2x + 1 - 2x$

$6x - 3 = 1$

Add $3$ to both sides:

$6x - 3 + 3 = 1 + 3$

$6x = 4$

Divide both sides by $6$ :

$x = \frac{4}{6} = \frac{2}{3}$

Check: $\text{LHS} = 8\left(\frac{2}{3}\right) - 3 = \frac{16}{3} - \frac{9}{3} = \frac{7}{3}$ ; $\text{RHS} = 2\left(\frac{2}{3}\right) + 1 = \frac{4}{3} + \frac{3}{3} = \frac{7}{3} \quad \checkmark$

$x = \frac{2}{3}$

Question 2

Solve $7y - 4 = 9y + 6$ .

ShowHide solution

Subtract $7y$ from both sides:

$-4 = 9y - 7y + 6$

$-4 = 2y + 6$

Subtract $6$ from both sides:

$-4 - 6 = 2y$

$-10 = 2y$

Divide both sides by $2$ :

$y = \frac{-10}{2} = -5$

Check: $\text{LHS} = 7(-5) - 4 = -35 - 4 = -39$ ; $\text{RHS} = 9(-5) + 6 = -45 + 6 = -39 \quad \checkmark$

$y = -5$

Question 3

Solve $3(2x - 4) = 18$ .

ShowHide solution

Expand the brackets:

$6x - 12 = 18$

Add $12$ to both sides:

$6x = 18 + 12 = 30$

Divide both sides by $6$ :

$x = \frac{30}{6} = 5$

Check: $\text{LHS} = 3(2(5) - 4) = 3(10 - 4) = 3(6) = 18 = \text{RHS} \quad \checkmark$

$x = 5$

Question 4

Solve $12 - (3x - 4) = 1$ .

ShowHide solution

Expand the minus sign across the bracket: $-(3x - 4) = -3x + 4$ (both signs inside change):

$12 - 3x + 4 = 1$

Combine constant terms on LHS ( $12 + 4 = 16$ ):

$16 - 3x = 1$

Subtract $16$ from both sides:

$16 - 3x - 16 = 1 - 16$

$-3x = -15$

Divide both sides by $-3$ :

$x = \frac{-15}{-3} = 5$

Check: $\text{LHS} = 12 - (3(5) - 4) = 12 - (15 - 4) = 12 - 11 = 1 = \text{RHS} \quad \checkmark$

$x = 5$

Question 5

Solve $10 - 2(3x - 1) = 24$ .

ShowHide solution

Expand $-2(3x - 1) = -6x + 2$ :

$10 - 6x + 2 = 24$

Combine constants on LHS ( $10 + 2 = 12$ ):

$12 - 6x = 24$

Subtract $12$ from both sides:

$-6x = 24 - 12 = 12$

Divide both sides by $-6$ :

$x = \frac{12}{-6} = -2$

Check: $\text{LHS} = 10 - 2(3(-2) - 1) = 10 - 2(-6 - 1) = 10 - 2(-7) = 10 + 14 = 24 = \text{RHS} \quad \checkmark$

$x = -2$

Question 6

Solve $4(2x - 3) = 3x + 8$ .

ShowHide solution

Expand the brackets on LHS:

$8x - 12 = 3x + 8$

Subtract $3x$ from both sides:

$5x - 12 = 8$

Add $12$ to both sides:

$5x = 8 + 12 = 20$

Divide both sides by $5$ :

$x = \frac{20}{5} = 4$

Check: $\text{LHS} = 4(2(4) - 3) = 4(8 - 3) = 4(5) = 20$ ; $\text{RHS} = 3(4) + 8 = 12 + 8 = 20 \quad \checkmark$

$x = 4$

Question 7

Solve $3(2x + 5) = 2(x - 2) + 11$ .

ShowHide solution

Expand brackets on both sides:

$6x + 15 = 2x - 4 + 11$

Simplify constants on RHS ( $-4 + 11 = 7$ ):

$6x + 15 = 2x + 7$

Subtract $2x$ from both sides:

$4x + 15 = 7$

Subtract $15$ from both sides:

$4x = 7 - 15 = -8$

Divide both sides by $4$ :

$x = \frac{-8}{4} = -2$

Check: $\text{LHS} = 3(2(-2) + 5) = 3(-4 + 5) = 3(1) = 3$ ; $\text{RHS} = 2(-2 - 2) + 11 = 2(-4) + 11 = -8 + 11 = 3 \quad \checkmark$

$x = -2$

Question 8

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

ShowHide solution

Expand brackets on both sides:

$5x - 5 = 6x + 3$

Subtract $5x$ from both sides:

$-5 = 6x - 5x + 3$

$-5 = x + 3$

Subtract $3$ from both sides:

$x = -5 - 3 = -8$

Check: $\text{LHS} = 5(-8 - 1) = 5(-9) = -45$ ; $\text{RHS} = 3(2(-8) + 1) = 3(-16 + 1) = 3(-15) = -45 \quad \checkmark$

$x = -8$

Question 9

Solve $2(3x + 4) - 3(x - 1) = 20$ .

ShowHide solution

Expand both sets of brackets on LHS:

$6x + 8 - 3x + 3 = 20$

Combine like terms on LHS ( $6x - 3x = 3x$ and $8 + 3 = 11$ ):

$3x + 11 = 20$

Subtract $11$ from both sides:

$3x = 20 - 11 = 9$

Divide both sides by $3$ :

$x = \frac{9}{3} = 3$

Check: $\text{LHS} = 2(3(3) + 4) - 3(3 - 1) = 2(9 + 4) - 3(2) = 2(13) - 6 = 26 - 6 = 20 = \text{RHS} \quad \checkmark$

$x = 3$

Question 10 (Setting: School Hall Seating)

In a secondary school hall, there are two types of seating rows. A standard row has $x$ chairs, and an extended row has $(2x - 5)$ chairs. If 5 standard rows have the same total number of chairs as 2 extended rows plus 46 chairs, form an equation in $x$ and determine how many chairs are in a standard row.

ShowHide solution

Total chairs in 5 standard rows: $5x$ chairs.

Total chairs in 2 extended rows: $2(2x - 5)$ chairs.

Form the equation:

$5x = 2(2x - 5) + 46$

Expand the brackets on RHS:

$5x = 4x - 10 + 46$

Simplify constants on RHS ( $-10 + 46 = 36$ ):

$5x = 4x + 36$

Subtract $4x$ from both sides:

$5x - 4x = 36$

$x = 36$

Check: 5 standard rows $= 5(36) = 180$ chairs; 2 extended rows $+ 46 = 2(2(36) - 5) + 46 = 2(67) + 46 = 134 + 46 = 180$ chairs $\quad \checkmark$

The equation is $5x = 2(2x - 5) + 46$ ; there are 36 chairs in a standard row.

Question 11 (Setting: Supermarket Shopping)

At a supermarket, a box of biscuits costs $x$ dollars and a tin of chocolates costs $(2x + 3)$ dollars. Mrs Tan bought 3 boxes of biscuits and 2 tins of chocolates, and paid a total of $48. Form an equation in $x$ and find the cost of one box of biscuits.

ShowHide solution

Cost of 3 boxes of biscuits: $3x$ dollars.

Cost of 2 tins of chocolates: $2(2x + 3)$ dollars.

Total cost: $(3x + 2(2x + 3))$ dollars.

Form the equation:

$3x + 2(2x + 3) = 48$

Expand the brackets:

$3x + 4x + 6 = 48$

Combine like terms on LHS:

$7x + 6 = 48$

Subtract $6$ from both sides:

$7x = 48 - 6 = 42$

Divide both sides by $7$ :

$x = \frac{42}{7} = 6$

Check: 3 boxes of biscuits cost $3(6) = 18$ dollars; 2 tins of chocolates cost $2(2(6) + 3) = 2(15) = 30$ dollars; total is $18 + 30 = 48$ dollars $\quad \checkmark$

The equation is $3x + 2(2x + 3) = 48$ ; each box of biscuits costs $6.

Question 12 (Setting: Park Connector Cycling)

On a cycling trip along a park connector, Ken cycles for 3 hours at $(2d + 1)$ km/h and Leo cycles for 4 hours at $(d + 2)$ km/h. Ken cycles 5 km further than Leo. Form an equation in $d$ and find Ken's speed.

ShowHide solution

Distance = speed $\times$ time.

Ken's distance: $3(2d + 1)$ km.

Leo's distance: $4(d + 2)$ km.

Ken cycles 5 km further than Leo:

$3(2d + 1) - 4(d + 2) = 5$

Expand both sets of brackets:

$6d + 3 - 4d - 8 = 5$

Combine like terms on LHS ( $6d - 4d = 2d$ and $3 - 8 = -5$ ):

$2d - 5 = 5$

Add $5$ to both sides:

$2d - 5 + 5 = 5 + 5$

$2d = 10$

Divide both sides by $2$ :

$d = \frac{10}{2} = 5$

Ken's speed is $2(5) + 1 = 11$ km/h and Leo's speed is $5 + 2 = 7$ km/h.

Check: Ken cycles $3(11) = 33$ km; Leo cycles $4(7) = 28$ km; $33 - 28 = 5$ km $\quad \checkmark$

The equation is $3(2d + 1) - 4(d + 2) = 5$ , so $d = 5$ and Ken's speed was 11 km/h.

Key Points

To solve an equation with the unknown on both sides, add or subtract the same terms on both sides until the variable terms are on one side and the numbers on the other.
Removing the variable term with the smaller coefficient from both sides keeps the coefficient left positive.
Expand brackets using the distributive law: $a(b + c) = ab + ac$ and $a(b - c) = ab - ac$ .
Take special care with negative signs: $-a(b - c) = -ab + ac$ .
Follow the ESCS method: Expand, Simplify, Collect, Solve.
Always verify your answer by substituting it back into both sides of the original equation.