# Solving Simultaneous Equations – A Level Maths Revision

Having trouble cracking those simultaneous equations questions? No problem, just sit back and watch us spit out those x and y‘s – whilst you avoid the zzzzzzz’s! – as we take you through step-by-step methodology on a range of example content.

We combine simultaneous equations questions with a healthy set of guidance, tips and model solutions to provide you with an equation that’s perfect for AS Level revision practice. Just read on below…

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Simultaneous equations are equations with more than one algebraic term. A single equation with more than one variable is impossible to solve – it has an infinite number of solutions. Consider this simple example: $a + b = 10$

Possible pairs of values include: $a = 1, b = 9$ $a = 12, b = -2$ $a = 0.08, b = 0.92$ $a = \sqrt{3}, b = 10 - \sqrt{3}$

As soon as we have a second equation, the pair of equations becomes solveable: $a + b = 10$ $a - b = 2$

We can solve these two equations to give: $a = 6, b = 4$

Because we have solved the equations at the same time, or simultaneously, they are called simultaneous equations.

The simultaneous equations question examples below are based on the requirements of the A Level Maths curriculum. For questions and answers on A level simultaneous equations, have a look at this resource.

To revise GCSE simultaneous equations, try this preparation for A Level resource.

If you’re still learning simultaneous equations at GCSE, there are a range of foundation resources here or higher resources here.

At A level you are expected to use your calculator to solve both linear simultaneous equations and simultaneous equations using polynomials, wherever possible. This sheet explains how to solve simultaneous equations, both with and without a calculator. The instructions are for the Casio CLASSWIZ fx-991EX, which is recommended by all exam boards.

Solving Simultaneous Equation Questions – Example Content

Example Question 1

Solve the simultaneous equations: $2x + 3y = 7$ $5x - 4y = 6$

It can help to label the equations (1) and (2) to keep track of them: $2x + 3y = 7 \qquad (1)$ $5x - 4y = 6 \qquad (2)$

We will solve this pair of equations through elimination – removing one of the algebraic terms. To do this, we need to find a common coefficient for either x or y. In this case, there is no advantage in choosing to eliminate either x or y, so we will remove the x coefficient.

Multiply equation (1) by 5 and (2) by 2. We will call our new equations (3) and (4): $10x + 15y = 35 \qquad (1) \times 5 = (3)$ $10x - 8y = 12 \qquad (2) \times 2 = (4)$

We can now eliminate x by calculating (3) – (4): $23y = 23 \qquad (3) - (4)$ $y = 1$

To find x, we will substitute our value for y into equation (1) (although we could choose to substitute into any of our equations here): $2x + 3 \times 1 = 7$ $2x + 3 = 7$ $2x = 4$ $x = 2$

Our solution is: x = 2 and y = 1.

This also means that the graphs of 2x + 3y = 7 and 5x – 4y = 6 intersect at the point (2, 1).

Example Question 2

Use your calculator to solve the simultaneous equations: $4x + 2y = 1$ $2x + 4y = 5$

While you are expected to know how to solve a pair of linear simultaneous equations by elimination, you will usually use your calculator to do this. These instructions are for the Casio CLASSWIZ fx-991EX; other calculators may solve these in a different way. Note that a scientific calculator used at GCSE will not be able to solve these automatically.

Start by pressing [MENU] then choose A: Equation/Func. You should have two options, we want the first one 1: Simul Equation.

Next, we need to input the number of unknowns, this is simply the number of letters in our pair of equations, 2.

Finally, we need to input the coefficients of our equations. Your equations should be in the form
ax + by = c. Type each coefficient, then hit [=] to move to the next. If you need to navigate back through the coefficients, use the arrows.

When you have done this, press [=] again to see the solution for x, then hit [=] again to see the solution for y. $x = -\frac{1}{2}, y = \frac{3}{2}$

Example Question 3

Solve the simultaneous equations: $y = x^2 + 2x + 3 \qquad (1)$ $3x + 2y = 8 \qquad (2)$

The calculator can only solve linear simultaneous equations, so we can’t use it directly in this case; instead we will use substitution. In this case, y is already the subject of our linear equation, so we can replace y in equation (2) by substituting in equation (1). If y was not the subject of the linear equation, we would rearrange, then substitute. $3x + 2(x^2 - 2x + 3) = 8$ $3x + 2x^2 - 4x + 6 = 8$ $2x^2 - x - 2 = 0$

While your calculator cannot solve quadratic simultaneous equations, it can solve quadratic equations with a single variable. As before, hit [MENU] and option A: Equation/Func, but this time choose
2: Polynomial. Input the polynomial degree as 2 (as the highest power is 2) and then the coefficients in the same way as before (make sure you type -1 as the coefficient for x).

Your calculator will give two solutions: $x = \frac{1+\sqrt{17}}{4} \: \text{and} \: x = \frac{1-\sqrt{17}}{4}$

Now we have each x solution, we need to find the corresponding values of y. We can do this by substituting each value of x into equation (1), or rearranging equation (2) then substituting there. In either case, our results will be the same: $\text{When} \: x = \frac{1+\sqrt{17}}{4}, \: y = \frac{29 - 3 \sqrt{17}}{8}$ $\text{When} \: x = \frac{1-\sqrt{17}}{4}, \: y = \frac{29 + 3 \sqrt{17}}{8}$

So our two pairs of answers, correct to 3 s.f., are: $x = 1.28, y = 2.08 \: \text{and} \: x = -0.781, y = 5.17$

Remember, these results are in pairs and are not interchangeable. y only equals 2.08 when x equals 1.28, and y only equals 5.17 when x is -0.781. These answers could also be written as two pairs of coordinates, where equations (1) and (2) intersect: (1.28, 2.08) and (-0.781, 5.17).

For this example, we have two distinct solutions. However this is not always the case. When working with quadratic simultaneous equations, sometimes there is only one solution and sometimes there are no solutions. This can be easily demonstrated by sketching the three options:

Example Question 4

Find the points of intersection of: $x^2 + y^2 = 36 \: \text{and} \: 3x + 2y = 8$

This is slightly different to examples 2 and 3, the first equation is a circle. However, we can still solve by substitution and we can still have 2, 1 or 0 pairs of solutions.

We can rearrange the linear equation to make y the subject: $y = 4 - \frac{3}{2}x \qquad (1)$ $x^2 + y^2 = 36 \qquad (2)$

We can now substitute (1) into (2): $x^2 + (4 - \frac{3}{2}x)^2 = 36$ $x^2 + 16 - 12x + \frac{9}{4}x^2 = 36$ $\frac{13}{4}x^2 -12x - 20 = 0$

Using our calculator to solve this, we get: $x = -1.25 \: \text{and} \: x = 4.94$

We can substitute these back into either equation to find the corresponding values of y: $\text{When}\: x = -1.25, y =5.87$ $\text{When}\: x = 4.95, y = -3.41$

This gives us our two points of intersection:

(-1.25, 5.87) and (4.94, -3.41).

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