If you are aiming for the highest possible grades in GCSE maths, you need to be looking consistently at the 5/6 mark questions at the end of the paper.

One of the topics that frequently appears in this position is solving linear and non-linear simultaneous equations.

To solve non-linear simultaneous equations, you cannot use the elimination method as this only works for linear functions. Instead, you have to use the quadratic formula.

**Example: **

Solve the following simultaneous equations**:**

**x ^{2} + y^{2} = 25**

**x + 2y = 3**

**Give your answers to 3 significant figures.**

**Step 1:** **Rearrange one of the equations into the form x = or** **y =**

It doesn’t actually matter which one you do here, but there are a couple of guidelines:

Always rearrange the linear expression.

Try to avoid fractions.

Because of these rules, you need to rearrange the linear equation into the form **x = 3 – 2y**

**Step 2: Substitute this rearranged equation into the non-linear equation.**

x^{2} + y^{2} = 25 → **(3 - 2y) ^{2} + y^{2} = 25**

The reason behind this is that we now have an equation that only has one variable (y) that we can solve.

**Step 3: Expand and rearrange.**

We can see that the equation (3 - 2y)^{2} + y^{2} = 25 is a quadratic, so we need to** rearrange **it into the form **ay ^{2} + by + c = 0**

(3 - 2y)^{2} + y^{2} = 25

(3 – 2y)(3 – 2y) + y^{2} = 25

9 – 12y + 4y^{2} + y^{2} = 25

5y^{2} – 12y + 9 = 25

**5y ^{2} – 12y - 16 = 0**

**Step 4: Solve the quadratic to find the values of y.**

Sometimes we can factorise this but for this example, you are asked to give your answers to 3 significant figures. This means you have to use the quadratic formula.

Evaluating this gives the values:** y _{1} = 3.35 and y_{2} = -0.954**

**Step 5: Use the value of y to find the values of x.**

All we have to do to find the values of x is to substitute our values of y into the rearranged linear equation.

x_{1} = 3 – 2(y_{1}) = 3 – 2(3.35) = -3.71

x_{2} = 3 – 2(y_{2}) = 3 – 2(-0.954) = 4.91

So the solutions to our two simultaneous equations are (**-3.71,3.35)** and (**4.91,-0.954)**

Time for some questions.