GCSE: Further Simultaneous Equations

Dr J Frost (jfrost@tiffin.kingston.sch.uk)www.drfrostmaths.com

Last modified: 31st August 2015

StarterSolve the following simultaneous (linear) equations.

2x + 3y = 84x – y = -5

x = -0.5y = 3 ?

RECAP :: Equation of a circle

x

y

5-5

5

-5

The equation of this circle is:

x2 + y2 = 25

! The equation of a circle with centre at the origin and radius r is:

x2 + y2 = r2

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Quickfire Circles

1-1

1

-1

x2 + y2 = 1

3-3

3

-3

x2 + y2 = 9

4-4

4

-4

x2 + y2 = 16

8-8

8

-8

x2 + y2 = 64

10-10

10

-10

x2 + y2 = 100

6-6

6

-6

x2 + y2 = 36

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Motivation

x

y

1010

10

10

-2

x – y

= 2

x2 + y2 = 100

Given a circle and a line, we may wish to find the point(s) at which the circle and line intersect.How could we do this algebraically?

STEP 1: Rearrange linear equation to make x or y the subject.x = y + 2

STEP 2: Substitute into quadratic and solve.(y + 2)2 + y2 = 100y2 + 4y + 4 + y2 = 1002y2 + 4y – 96 = 0y2 + 2y – 48 = 0(y + 8)(y – 6) = 0y = -8 or y = 6

STEP 3: Use either equation to find the values of the other variable.When y = -8, x = -6When y = 6, x = 8

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Your Go

y = x2 – 3x + 4y – x = 1

STEP 1: Rearrange linear equation to make x or y the subject.

y = 1 + x

STEP 2: Substitute into quadratic and solve for one variable.

1 + x = x2 – 3x + 4x2 – 4x + 3 = 0(x – 1)(x – 3) = 0x = 1 or x = 3

STEP 3: Use either equation to find the values of the other variable.When x = 1, y = 2When x = 3, y = 4?

STEP 4 (OPTIONAL): Check that your pairs of values work.2 = 12 – (3 x 1) + 4 C4 = 32 – (3 x 3) + 4 C?

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Exercises

y = x2 + 7x – 2 y = 2x – 8

x2 + y2 = 8y = x + 4

y = x2

y = x + 2

x2 + y2 = 5x – 2y = 5

y = x2 – x – 2x + 2y = 11

y = x2 – 2x – 2x = 2y + 1

x = -3, y = -14x = -2, y = -12

x = -2, y = +2

x = -3, y = -14x = -2, y = -12

x = 1, y = -2

x + y = 1x2 + y2 = 1

x2 – y2 = 152x + 3y = 5

x2 – y2 = 152x + 3y = 5

y = x2 – 3xx = y – 9

x2 – 4y + 7 = 0y2 – 6z + 14 = 0z2 – 2x – 7 = 0[Source: BMO]

x = 2 – √13, y = 11 – √13X = 2 + √13, y = 11 + √13

x = 3, y = 4x = -5/2, y = 27/4

x = 1, y = 0x = 0, y = 1

x = -8, y = 7x = 4, y = -1

x = 1, y = 2,z = 3(Add equations, then complete the squares – you’ll end up with a sum of squares which must each be 0)

x = 5, y = -3x = 191/59, y = -255/59

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1

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3

4

5

7

8

9

10

N

6 x = 3, y = 1x = -1/2, y = -3/4?