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Year 8 Maths Equations

Equations with Brackets

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Theory

A bracket holds one quantity. In \(6(x - 4) = 42\) the whole bracket was multiplied by \(6\), so either divide both sides by \(6\) first (\(x - 4 = 7\)) or expand first (\(6x - 24 = 42\)). Both give \(x = 11\). This Year 8 maths page shows how.

Why it is two steps. In \(7(x - 2) = 35\), the \(-\,2\) happened inside the bracket first and the \(\times 7\) last. Undo the \(\times 7\) first.

Method 1: divide first. Divide both sides by the number in front, then undo the number inside. This is quicker whenever the division is tidy.

Method 2: expand first. Expanding means multiplying every term inside the bracket: \(3(x + 4) = 3x + 12\). Then collect like terms and solve. Use this when the pronumeral is on both sides.

The bracket may be on the right. \(45 = 9(k - 2)\) gives \(5 = k - 2\), so \(k = 7\). Write the answer with the pronumeral on the left.

The number in front may be negative. Divide both sides by it, sign and all; or expand with the sign: \(-2(4 - x) = -8 + 2x\).

Brackets on both sides. Expand both, take the smaller pronumeral term from both sides, then undo the number: \(2(x + 1) = 3(x - 4)\) gives \(2x + 2 = 3x - 12\), so \(x = 14\).

Modelling. When a group is repeated, put the group in a bracket: \(4\) bags each holding \(m\) apples and \(2\) pears hold \(4(m + 2)\) pieces of fruit.

Area model for 3(x + 5) A rectangle 3 high and x + 5 wide, split into a part x wide with area 3x and a part 5 wide with area 15. Together the area is 3x + 15. x3x 515 3 3(x + 5) = 3x + 15
The rectangle is \(3\) high and \(x + 5\) wide. Its area is \(3\) lots of \(x + 5\), or the two parts \(3x\) and \(15\) added, so the \(3\) multiplies the \(5\) as well as the \(x\).

Two routes to the same answer:

EquationSteps (both sides)Solution
\(4(x - 1) = 20\)\(\div 4\), then \(+\,1\)\(x = 6\)
\(-3(x + 2) = 9\)\(\div (-3)\), then \(-\,2\)\(x = -5\)
\(24 = 8(y - 1)\)\(\div 8\), then \(+\,1\)\(y = 4\)
\(5(x + 3) = 6x\)expand to \(5x + 15 = 6x\), then \(-\,5x\)\(x = 15\)
Expanding multiplies every term. \(4(y + 3) = 4y + 12\), never \(4y + 3\).
Do not touch the inside first. In \(9(y - 4) = 36\), adding \(4\) to both sides gives \(9(y - 4) + 4 = 40\): the bracket is unchanged. Divide by \(9\) first, or expand.

Divide first

  1. Divide both sides by the number in front of the bracket.
  2. Undo the number inside the bracket.
  3. Check in the original equation.

Expand first

  1. Expand every bracket: multiply each term inside.
  2. Collect the pronumeral terms on one side (take away the smaller one).
  3. Solve the two-step equation that is left and check.
Example 1 — Expanding the bracket
Solve \(3(x + 4) = 15\).
Solution
\(3 \times x + 3 \times 4\)\(=\)\(15\)
\(3x + 12\)\(=\)\(15\)
\(3x + 12 - 12\)\(=\)\(15 - 12\)
\(3x\)\(=\)\(3\)
\(\dfrac{3x}{3}\)\(=\)\(\dfrac{3}{3}\)
\(x\)\(=\)\(1\)
x =1

\(x = 1\). Dividing first is quicker here: \(x + 4 = 5\), so \(x = 1\).

Example 2 — Divide by the multiplier first
Solve \(6(x - 4) = 42\).
Solution
\(\dfrac{6(x - 4)}{6}\)\(=\)\(\dfrac{42}{6}\)
\(x - 4\)\(=\)\(7\)
\(x - 4 + 4\)\(=\)\(7 + 4\)
\(x\)\(=\)\(11\)
x =11

\(x = 11\). Check: \(6(11 - 4) = 6 \times 7 = 42\).

Example 3 — Decimal and fraction answers
Solve
(i) \(-4(x + 1) = 10\)
(ii) \(5(p - 2) = 4\)
Solution
\(x + 1\)\(=\)\(\dfrac{10}{-4} = -2.5\)
\(x\)\(=\)\(-2.5 - 1 = -3.5\)
\(p - 2\)\(=\)\(\dfrac{4}{5}\)
\(p\)\(=\)\(\dfrac{4}{5} + 2 = 2\dfrac{4}{5}\)
x =−3.5 p =24/5

(i) \(x = -3.5\). Check: \(-4(-3.5 + 1) = -4 \times (-2.5) = 10\).
(ii) \(p = 2\dfrac{4}{5}\). Check: \(5 \times \dfrac{4}{5} = 4\).

Example 4 — Expand, then collect
Solve
(i) \(2(x + 6) = 5x\)
(ii) \(2(4 - x) = 5 - 7x\)
Solution

(i) Expand to \(2x + 12 = 5x\), then subtract \(2x\) from both sides.
(ii) Expand to \(8 - 2x = 5 - 7x\). Add \(7x\) to both sides, then subtract \(8\).

\(12\)\(=\)\(3x\)
\(x\)\(=\)\(4\)
\(8 + 5x\)\(=\)\(5\)
\(5x\)\(=\)\(-3\)
\(x\)\(=\)\(-\dfrac{3}{5}\)
x =4 x =−3/5

(i) \(x = 4\). Check: \(2(4 + 6) = 20 = 5 \times 4\).
(ii) \(x = -\dfrac{3}{5}\). Check: both sides equal \(9\dfrac{1}{5}\).

Example 5 — Brackets on both sides
Solve \(2(x + 1) = 3(x - 4)\).
Solution
\(2x + 2\)\(=\)\(3x - 12\)
\(2x + 2 - 2x\)\(=\)\(3x - 12 - 2x\)
\(2\)\(=\)\(x - 12\)
\(14\)\(=\)\(x\)
x =14

\(x = 14\). Check: \(2(14 + 1) = 30\) and \(3(14 - 4) = 30\).

Example 6 — Writing the bracket
(i) A cinema ticket costs \(\$21\). Five friends each get the same discount, and together they pay \(\$80\). Find the discount.
(ii) Noor picks a number, adds \(4\), then multiplies the result by \(3\). She gets \(-6\). What was her number?
(iii) An equilateral triangle has sides of \((x + 5)\) cm and a square has sides of \(x\) cm. Their perimeters are equal. Find \(x\).
Solution
\(5(21 - d)\)\(=\)\(80\)
\(21 - d\)\(=\)\(16\)
\(d\)\(=\)\(5\)
\(3(n + 4)\)\(=\)\(-6\)
\(n + 4\)\(=\)\(-2\)
\(n\)\(=\)\(-6\)
\(3(x + 5)\)\(=\)\(4x\)
\(3x + 15\)\(=\)\(4x\)
\(x\)\(=\)\(15\)
d =5 n =−6 x =15

(i) Each friend gets \(\$5\) off, paying \(\$16\).
(ii) Her number was \(-6\). Check: \((-6 + 4) \times 3 = -6\).
(iii) \(x = 15\): both perimeters are \(60\) cm.

Common pitfalls

Multiplying only the first term. \(5(x - 3) = 20\) expands to \(5x - 15 = 20\), not \(5x - 3 = 20\). The correct answer is \(x = 7\).
Undoing the inside number first. In \(4(x + 2) = 28\), subtracting \(2\) first leaves the \(\times 4\) on the bracket. Divide by \(4\) first: \(x + 2 = 7\).
Losing the negative multiplier. For \(-5(x + 1) = 20\), divide by \(-5\): \(x + 1 = -4\), so \(x = -5\).
Dividing only one term. \(\dfrac{6x + 12}{6}\) is \(x + 2\), not \(x + 12\).

Frequently asked questions

How do you solve an equation with brackets?

Divide both sides by the number in front of the bracket, or expand the bracket first, then solve the equation that is left.

What does expand mean?

Multiply every term inside the bracket by the number outside: \(3(x + 4) = 3x + 12\).

When should I expand instead of dividing?

When the pronumeral is on both sides, or when the division would not come out tidily.

What if the number in front is negative?

Divide both sides by that negative number, or expand carefully with the sign: \(-2(x - 3) = -2x + 6\).

How do I solve brackets on both sides?

Expand both brackets, take the smaller pronumeral term from both sides, then finish as a two-step equation.