Loading please wait

The smart way to improve grades

Comprehensive & curriculum aligned

Try an activity or get started for free

Solve Simultaneous Equations both Linear and Quadratic

In this worksheet, students will practise solving linear, circle and quadratic simultaneous equations.

'Solve Simultaneous Equations both Linear and Quadratic' worksheet

Key stage:  KS 4

Year:  GCSE

GCSE Subjects:   Maths

GCSE Boards:   AQA, Eduqas, OCR, Pearson Edexcel,

Curriculum topic:   Algebra

Curriculum subtopic:   Solving Equations and Inequalities Algebraic Equations

Difficulty level:  

Worksheet Overview

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:

x2 + y2 = 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.

x2 + y2 = 25  →  (3 - 2y)2 + y2 = 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 + y2 = 25 is a quadratic, so we need to rearrange it into the form ay2 + by + c = 0

(3 - 2y)2 + y2 = 25

(3 – 2y)(3 – 2y) + y2 = 25

9 – 12y + 4y2 + y2 = 25

5y2 – 12y + 9 = 25

5y2 – 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.

 

the quadratic formula

 

Evaluating this gives the values: y1 = 3.35 and y2 = -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.

x1 = 3 – 2(y1) = 3 – 2(3.35) = -3.71

x2 = 3 – 2(y2) = 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.

What is EdPlace?

We're your National Curriculum aligned online education content provider helping each child succeed in English, maths and science from year 1 to GCSE. With an EdPlace account you’ll be able to track and measure progress, helping each child achieve their best. We build confidence and attainment by personalising each child’s learning at a level that suits them.

Get started
laptop

Try an activity or get started for free

  • National Tutoring Awards 2023 Shortlisted / Parents
    National Tutoring Awards 2023 Shortlisted
  • Private-Tutoring-WINNER-EducationInvestor-Awards / Parents
    Winner - Private Tutoring
  • Bett Awards Finalist / Parents
    Finalist
  • Winner - Best for Home Learning / Parents
    Winner - Best for Home Learning / Parents