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Algebra fluency: solving equations
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Every topic in Further Pure ends with solving something: a linear equation, a quadratic or a pair of simultaneous equations. These methods need to be quick and reliable.
Linear equations with fractions
Multiply both sides by the denominators to clear the fractions.
x + 34 = 2x − 13
Multiply by 12: 3(x + 3) = 4(2x − 1)
3x + 9 = 8x − 4, so 13 = 5x and x = 135
x + 34 = 2x − 13
Multiply by 12: 3(x + 3) = 4(2x − 1)
3x + 9 = 8x − 4, so 13 = 5x and x = 135
Quadratic equations
Rearrange to = 0 first.
Factorise: x2 − 5x − 14 = 0 gives (x − 7)(x + 2) = 0, so x = 7 or x = −2.
Formula when it will not factorise: for ax2 + bx + c = 0,
x = −b ± √(b2 − 4ac)2a
Never divide by x: x2 = 9x loses the solution x = 0. Write x2 − 9x = 0, x(x − 9) = 0.
Factorise: x2 − 5x − 14 = 0 gives (x − 7)(x + 2) = 0, so x = 7 or x = −2.
Formula when it will not factorise: for ax2 + bx + c = 0,
x = −b ± √(b2 − 4ac)2a
Never divide by x: x2 = 9x loses the solution x = 0. Write x2 − 9x = 0, x(x − 9) = 0.
Simultaneous linear equations
Elimination: make one variable's coefficients match, then add or subtract.
3x + 2y = 16 and 5x − 2y = 8
Add: 8x = 24, so x = 3. Then 9 + 2y = 16, so y = 3.5.
Substitution: rearrange one equation for x or y and substitute into the other. You will need this when one equation is a quadratic.
3x + 2y = 16 and 5x − 2y = 8
Add: 8x = 24, so x = 3. Then 9 + 2y = 16, so y = 3.5.
Substitution: rearrange one equation for x or y and substitute into the other. You will need this when one equation is a quadratic.
Solve 3x2 − 4x − 2 = 0, giving your answers to 3 significant figures.
- a = 3, b = −4, c = −2
- x = 4 ± √(16 + 24)6 = 4 ± √406
- x = 1.7208... or x = −0.3874...
Answer: x = 1.72 or x = −0.387
Clear fractions by multiplying through. Make a quadratic equal zero, then factorise or use the formula; never divide by x. For simultaneous equations, eliminate one variable or substitute.
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