Higher only · simultaneous equations

A line meets
a curve: two
pairs of answers.

substitute,
then solve

when one equation has a square in it —

Substitute

put the line

into the curve

step 1 of 4

Substitute the line in

1. The line says y is the same as x + 1.

2. Replace y in the curve, in a bracket.

✓ Now there is only x

y = x + 1

x² + y² = 13

x² + (x + 1)² = 13

step 2 of 4

Expand and tidy

1. (x + 1)² = x² + 2x + 1, not x² + 1.

2. Make it equal 0, then divide by 2.

✓ A quadratic you can factorise

x² + x² + 2x + 1 = 13

2x² + 2x − 12 = 0

x² + x − 6 = 0

step 3 of 4

Solve for x

1. Multiply to −6, add to +1: +3 and −2.

2. Each bracket gives a value of x.

✓ Two x values, so two points

x² + x − 6 = 0

(x + 3)(x − 2) = 0

x = −3  or  x = 2

step 4 of 4

Find each y

1. Use the line: it is the easy one.

2. Keep each x with its own y.

✓ Check: (−3)² + (−2)² = 13

y = x + 1

x = −3  →  y = −2

x = 2  →  y = 3

(−3, −2) and (2, 3)

a line and a curve —

Substitute

the line into the curve

Solve

the quadratic: two x values

Pair up

each x with its own y

The answers are where
the two graphs cross.

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