Core Tier equations are solved with one fixed method. At Extended Tier, the challenge shifts to choosing the right tool for the job — clearing algebraic denominators, eliminating a variable between a linear and a quadratic equation, or picking between factorising, completing the square, and the quadratic formula depending on what a question actually asks for.

How do I solve an equation containing algebraic fractions?

Multiply every single term on both sides of the equation by the lowest common denominator. This clears every fraction at once, turning the equation into an ordinary linear or quadratic equation.

Multiplying every term by $(x+1)(x-2)$:

How do I solve simultaneous equations when one is linear and one is quadratic?

Rearrange the linear equation to make one variable the subject, substitute that expression into the quadratic equation, and solve the resulting quadratic for the remaining variable.

Substituting back into the linear equation gives the matching $y$-values: $(3,5)$ and $(-1,-3)$.

How can I rewrite the first two terms of a quadratic expression as the difference of two squares?

Halve the coefficient of $x$ and square it — this is the amount you need to add and then subtract, so the expression is unchanged overall:

How do I complete the square?

Do the step above, then combine the leftover constant with whatever number term the quadratic already had:

Setting this equal to zero and rearranging gives an alternative route to solving any quadratic, including ones that don't factorise:

How do I complete the square when there is a coefficient in front of the $x^2$ term?

Factor that coefficient out of the $x^2$ and $x$ terms only — leave any constant term outside the bracket — then complete the square inside the bracket as before.

How do I find the turning point by completing the square?

Once written as $a(x+p)^2+q$, the turning point of the graph is $(-p,\,q)$: the squared bracket is smallest (zero) when $x=-p$, so $y=q$ there. For $2(x-2)^2-5$, the turning point is $(2,-5)$, and since $a=2>0$, it's a minimum.

How do I decide whether to factorise, complete the square, or use the formula?

• Factorise when the question asks for exact values and the numbers look "nice" (integer roots).
• Complete the square when the question specifically asks for surd-form answers or the equation of a turning point.
• Use the quadratic formula when the equation doesn't factorise and the question allows decimal or surd answers:

How do I change the subject of a formula when the subject appears more than once, or has a power?

Collect every term containing the subject on one side of the equation and every other term on the other side, factorise out the subject if it appears more than once, then divide. If the subject is inside a power or root, undo that operation last.

Worked Example

Solve $\dfrac{3}{x+2}+\dfrac{2}{x-1}=2$, giving your answer in surd form.

1. Multiply every term by the common denominator $(x+2)(x-1)$:

1. Expand both sides:

1. Rearrange into $=0$ form:

1. Apply the quadratic formula with $a=2,\ b=-3,\ c=-5$:

1. This gives $x=\dfrac{10}{4}=2.5$ or $x=-1$ — both rational here, so state both exact solutions.