Solving equations with brackets

Example

Solve the equation 5(2c - 3) = 19.

The equation contains a set of brackets. The easiest way to solve equations with brackets is to expand the brackets.

5(2c - 3) = 19

Expand the bracket:

5 \times 2c - 5 \times 3 = 19

10c - 15 = 19

Isolate 10c by adding 15 to each side:

10c - 15 + 15 = 19 + 15

10c = 34

Isolate c by dividing by 10:

\frac{10c}{10} = \frac{34}{10}

c = \frac{34}{10} = \frac{17}{5} or 3.4

Question

The area of this rectangle is 56 cm2. Find the value of r.

Rectangle with sides labelled, (3r+2) cm and 7 cm

\text{Area of a rectangle} = \text{base} \times \text{height}. This means 3r + 2~cm will all be multiplied by 7. To show this in algebra, use a bracket for 3r + 2~cm to show that both terms are being multiplied by 7.

7 multiplied by (3r + 2)~cm can be written as 7(3r + 2)~cm as multiplication signs are not used in algebra.

\text{Area} = \text{base} \times \text{height}

Area = 56 cm2, base = (3r + 2) cm, height = 7 cm

So setting up an equation using this information gives:

56 = (3r + 2) \times 7

Multiply out the bracket.

56 = 21r + 14

Isolate 21r by subtracting 14 from both sides:

56 - 14 = 21r + 14 - 14

42 = 21r

Isolate r by dividing both sides by 21:

42 \div 21 = 21r \div 21

2 = r

Check that this answer is correct by substituting r = 2 into the original equation.

7(3r + 2) = 7 \times (3 \times 2 + 2) = 7 \times (6 + 2) = 7 \times 8 = 56

The equation balances, so r = 2 is the correct answer.