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How do you solve linear equations, including with brackets, fractions and the unknown on both sides?

Solve linear equations in one unknown, including those with brackets, fractions and the unknown on both sides, and form linear equations from worded and geometric contexts.

A focused answer to the Eduqas GCSE Mathematics algebra content on solving linear equations, covering brackets, fractions, the unknown on both sides, and forming equations from worded and geometric situations.

Generated by Claude Opus 4.810 min answer

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  1. What this dot point is asking
  2. The balance method
  3. Brackets and fractions
  4. The unknown on both sides
  5. Forming equations
  6. Checking your solution

What this dot point is asking

The Eduqas algebra content requires you to solve linear equations in one unknown, however they are dressed up: with brackets to expand, fractions to clear, or the unknown appearing on both sides. It also asks you to form an equation from a worded or geometric situation and then solve it. Linear equations are foundational, appearing on both components and at both tiers, and the skill of forming an equation is exactly where Eduqas tests AO2 and AO3 reasoning, so it carries marks beyond the routine solving.

The balance method

An equation is a statement that two sides are equal. Whatever you do to one side you must do to the other to keep the balance, and you undo the operations applied to the unknown in reverse order.

The reverse order matters: in 3x+73x + 7 the unknown was multiplied by 33 then 77 was added, so you undo the +7+7 first, then the Γ—3\times 3.

Brackets and fractions

When an equation contains brackets, expand them before collecting terms. So 4(x+3)=204(x + 3) = 20 becomes 4x+12=204x + 12 = 20, then 4x=84x = 8 and x=2x = 2.

When an equation contains fractions, clear them by multiplying every term by the lowest common denominator.

The unknown on both sides

When the unknown appears on both sides, collect the unknown terms on one side (usually the side that keeps the coefficient positive) and the numbers on the other. For 7xβˆ’4=3x+127x - 4 = 3x + 12, subtract 3x3x from both sides to get 4xβˆ’4=124x - 4 = 12, add 44 to get 4x=164x = 16, and divide to get x=4x = 4.

Forming equations

Many marks come from turning a situation into an equation. A worded problem ("I think of a number, multiply it by 44 and subtract 33 to get 1717") becomes 4nβˆ’3=174n - 3 = 17. A geometric fact provides the equation: angles on a straight line sum to 180∘180^\circ, angles round a point to 360∘360^\circ, and the angles of a triangle to 180∘180^\circ. Set the relevant sum equal to the total, then solve. Because forming the equation is the reasoning step, always state the fact you are using ("angles in a triangle sum to 180∘180^\circ") so the method mark is secure.

A perimeter or money context works the same way. If a rectangle has length x+5x + 5 and width xx, and its perimeter is 3434 cm, then 2(x+5)+2x=342(x + 5) + 2x = 34, which gives 4x+10=344x + 10 = 34, so x=6x = 6 and the rectangle is 1111 cm by 66 cm. The discipline is always the same: name the quantity with a letter, write the relationship as an equation, solve it, then translate the answer back into the context (and check it makes sense, since a negative length would signal an error).

Checking your solution

Substituting the answer back into the original equation is the fastest way to catch a slip and is worth doing every time. For 7xβˆ’4=3x+127x - 4 = 3x + 12 with solution x=4x = 4, the left side is 7(4)βˆ’4=247(4) - 4 = 24 and the right side is 3(4)+12=243(4) + 12 = 24, so the two sides match and the answer is confirmed. On the non-calculator Component 1 this self-check costs only a few seconds and protects an otherwise hard-won mark, which is why examiners' reports repeatedly recommend it.

Exam-style practice questions

Practice questions written in the style of WJEC Eduqas exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Eduqas 20183 marksSolve 5(xβˆ’2)=2x+75(x - 2) = 2x + 7. (Foundation, Component 1, non-calculator.)
Show worked answer β†’

Expand the bracket first: 5xβˆ’10=2x+75x - 10 = 2x + 7.

Collect the unknowns on the left by subtracting 2x2x: 3xβˆ’10=73x - 10 = 7.

Add 1010 to both sides: 3x=173x = 17. Divide by 33: x=173x = \dfrac{17}{3} (or 5235\tfrac{2}{3}).

Markers award a mark for expanding, a mark for collecting terms correctly, and a mark for the final value. Forgetting to multiply the βˆ’2-2 inside the bracket by 55 (writing 5xβˆ’25x - 2) is the most common slip.

Eduqas 20224 marksThe angles of a triangle are x+20x + 20, 2x2x and 3xβˆ’83x - 8 degrees. Form an equation and solve it to find the size of each angle. (Higher, Component 2, calculator.)
Show worked answer β†’

The angles of a triangle sum to 180∘180^\circ, so form the equation (x+20)+2x+(3xβˆ’8)=180(x + 20) + 2x + (3x - 8) = 180.

Collect like terms: 6x+12=1806x + 12 = 180.

Subtract 1212: 6x=1686x = 168. Divide by 66: x=28x = 28.

Substitute back: the angles are 48∘48^\circ, 56∘56^\circ and 76∘76^\circ, which check by summing to 180∘180^\circ.

Markers give marks for a correct equation, for solving to x=28x = 28, and for the three angles. The reasoning mark depends on stating the angle sum, and a final check that the angles total 180∘180^\circ secures full marks.

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