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:
\(10c \div 10 = 34 \div 10\)
\(c = \frac{34}{10} = \frac{17}{5}\) or 3.4
Question
The area of this rectangle is 56 cm2. Find the value of \(r\).
\(\text{Area of a rectangle} = \text{base} \times \text{height}\). This means \(3r + 2\) will all be multiplied by 7. To show this in algebra, use a bracket for \(3r + 2\) to show that both terms are being multiplied by 7.
7 multiplied by \((3r + 2)\) can be written as \(7(3r + 2)\) as multiplication signs are not used in algebra.
\(\text{Area} = \text{base} \times \text{height}\)
Area = \(7(3r + 2)\)
The area of the rectangle has been given in the question as 56 cm2:
\(56 = 7(3r + 2)\)
Expand the bracket:
\(56 = 7 \times 3r + 7 \times 2\)
\(56 = 21r + 14\)
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.
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