Harder Equations
Every question with a fully worked solution.
- Harder Equations - Video - Harder Equations - Example 1 Watch
Theory
When the pronumeral is on both sides, first take the smaller pronumeral term away from both sides. What is left is an ordinary one-step or two-step equation: \(5y - 5 = 4y + 20\) becomes \(y - 5 = 20\), so \(y = 25\). This Year 8 maths page shows how, with brackets and fractions too.
The balance stays balanced. Whatever you do to one side, do to the other. That includes taking away a whole term such as \(5x\).
Equivalent equations. Each move gives a shorter equation with the same solution. \(5x + 2 = 2x + 17\) and \(3x = 15\) are equivalent, because both give \(x = 5\).
Move the smaller term. In \(3x = 8x - 15\), subtract \(3x\): \(0 = 5x - 15\), so \(x = 3\). This keeps the coefficient positive.
A negative term moves by adding. In \(9 - 4c = -c\), add \(4c\) to both sides: \(9 = 3c\).
Collect like terms first if one side has two pronumeral terms (\(-3m - 5m = -8m\)), and simplify a side that has only numbers.
Decimal coefficients behave like any other number. To divide by a decimal, multiply the top and the bottom by \(10\): \(\dfrac{4}{0.5} = \dfrac{40}{5} = 8\).
All the types together. Expand the brackets, clear the fractions by multiplying every term by the LCD, then move all the pronumeral terms to one side and solve.
Answers need not be whole numbers. Leave a fraction simplified, as a mixed numeral or a decimal. Check every answer by working out the LHS and the RHS separately.
One move clears the pronumeral from a side:
| Equation | First move | Solution |
|---|---|---|
| \(8x = 5x + 21\) | \(-\,5x\): \(3x = 21\) | \(x = 7\) |
| \(4 - 6b = -2b\) | \(+\,6b\): \(4 = 4b\) | \(b = 1\) |
| \(6x = 10x - 9\) | \(-\,6x\): \(0 = 4x - 9\) | \(x = 2\dfrac{1}{4}\) |
| \(0.7t + 1 = 0.2t\) | \(-\,0.2t\): \(0.5t + 1 = 0\) | \(t = -2\) |
| \(5n - 8 = 6n + 1\) | \(-\,5n\): \(-8 = n + 1\) | \(n = -9\) |
How to solve a harder equation
- Expand any brackets and clear any fractions (multiply every term by the LCD).
- Collect like terms on each side.
- Move the smaller pronumeral term: subtract it, or add it if it is negative.
- Solve the one-step or two-step equation that is left.
- Check: LHS and RHS separately in the original equation.
| \(5y - 5 - 4y\) | \(=\) | \(4y + 20 - 4y\) |
| \(y - 5\) | \(=\) | \(20\) |
| \(y\) | \(=\) | \(25\) |
\(y = 25\). Check: LHS \(= 5 \times 25 - 5 = 120\), RHS \(= 4 \times 25 + 20 = 120\).
(i) \(9 - 4c = -c\)
(ii) \(-3m - 5m = 24\)
(i) Add \(4c\) to both sides.
(ii) Collect the two \(m\) terms first, then divide by \(-8\).
| \(9\) | \(=\) | \(3c\) |
| \(c\) | \(=\) | \(3\) |
| \(-8m\) | \(=\) | \(24\) |
| \(m\) | \(=\) | \(\dfrac{24}{-8} = -3\) |
(i) \(c = 3\). Check: \(9 - 12 = -3\).
(ii) \(m = -3\). Check: \(9 + 15 = 24\).
| \(0.8w - 0.3w\) | \(=\) | \(4\) |
| \(0.5w\) | \(=\) | \(4\) |
| \(w\) | \(=\) | \(\dfrac{4}{0.5} = \dfrac{40}{5} = 8\) |
\(w = 8\). Check: \(0.8 \times 8 = 6.4\) and \(0.3 \times 8 + 4 = 6.4\).
Expand the bracket: \(\dfrac{2x}{3} - 2 + x = \dfrac{x}{2} + 3\). The LCD of \(3\) and \(2\) is \(6\), so multiply every term by \(6\).
| \(4x - 12 + 6x\) | \(=\) | \(3x + 18\) |
| \(10x - 12\) | \(=\) | \(3x + 18\) |
| \(7x - 12\) | \(=\) | \(18\) |
| \(7x\) | \(=\) | \(30\) |
| \(x\) | \(=\) | \(\dfrac{30}{7} = 4\dfrac{2}{7}\) |
\(x = 4\dfrac{2}{7}\)
The LCD is \(6\). Multiply every term by \(6\); the right side becomes \(2(x + 1) = 2x + 2\).
| \(10x + 3x\) | \(=\) | \(2x + 2\) |
| \(13x - 2x\) | \(=\) | \(2\) |
| \(11x\) | \(=\) | \(2\) |
| \(\dfrac{11x}{11}\) | \(=\) | \(\dfrac{2}{11}\) |
| \(x\) | \(=\) | \(\dfrac{2}{11}\) |
\(x = \dfrac{2}{11}\)
Let \(w\) be the number of weeks. The \(w\) term on the left is negative, so add \(15w\) to both sides.
| \(200 - 15w\) | \(=\) | \(25w\) |
| \(200\) | \(=\) | \(40w\) |
| \(w\) | \(=\) | \(5\) |
They are equal after \(5\) weeks: \(200 - 75 = 125\) and \(25 \times 5 = 125\), so both have \(\$125\).
(i) Test Sam’s answer.
(ii) Solve the equation.
| \(\text{LHS}\) | \(=\) | \(2 \times 4 + 9 = 17\) |
| \(\text{RHS}\) | \(=\) | \(5 \times 4 - 1 = 19\) |
| \(10\) | \(=\) | \(3x\) |
| \(x\) | \(=\) | \(\dfrac{10}{3} = 3\dfrac{1}{3}\) |
(i) LHS \(\ne\) RHS, so \(x = 4\) is not the solution.
(ii) Subtract \(2x\) and add \(1\): \(x = 3\dfrac{1}{3}\).
Common pitfalls
Frequently asked questions
How do you solve an equation with x on both sides?
Subtract the smaller pronumeral term from both sides, then solve the equation that is left.
What if the pronumeral term is negative?
Add it to both sides instead: in \(9 - 4c = -c\), add \(4c\).
What are equivalent equations?
Equations with the same solution, such as \(5x + 2 = 2x + 17\) and \(3x = 15\).
How do I divide by a decimal?
Multiply the top and the bottom by \(10\) or \(100\) first: \(\dfrac{4}{0.5} = \dfrac{40}{5} = 8\).
How do I check a claimed answer?
Substitute it into each side separately. If the two values are equal it is the solution; if not, it is not.