Solutions for Simultaneous Equations

1. Solve the simultaneous equations:

x + y = 3 (1)

x² + y² = 29 (2)

Re-arrange (1) as x = 3 – y

Then y² = 9 – 3y – 3y + y² 

Therefore x² + y² = (9 – 6y + y²)+ y² = 29

2y² – 6y + 9 – 29 = 0

2y² – 6y – 20 = 0

y² – 3y – 10 = 0

(y – 5) (y + 2) = 0

y = 5 or y = - 2

When y = 5, x = 3 – y When y = 5, x = 3 – y

x = 3 – 5 x = 3 – - 2

x = -2 x = 5

The solutions are (-2, 5) and (5, -2).

Note: (1) is a line. (2) is a circle centred on the origin with radius √29. The solutions are reflections in y = x.

2. Solve the simultaneous equations:

y – x = 3 (1)

xy = 4 (2)

Multiply both sides of (1) by x: xy – x² = 3x

Substitute y = 4/x into (1), to get: 4(x/x) – x² = 3x

x² +3x – 4 = 0

(x + 4) (x – 1) = 0

x = -4 or x = 1

When x = -4, y = -1

When x= 1, y = 4

The solutions are (-4, -1) and (1, 4).

Note: (1) is a line. (2) is a hyperbola. The solutions are reflections in y = -x.

3. Solve the simultaneous equations:

2x + 3y = 14 (1)

xy = 4 (2)

Multiply both sides of (1) by x: 2x² +3xy = 14x

Substitute y = 4/x into (1), to get: 2x² +12(x/x) = 14x

2x² +12 – 14x = 0 (We can divide both sides by 2. Remember 0/2 = 0)

x² – 7x + 6 = 0

(x – 6) (x– 1) = 0

x = 6 or x = 1

When x = 6, y = 4/6 = 2/3

When = 1, y = 4

4. The line x – 3y = ^-1 intersects the ellipse 2(x – y) = 2x² – 11y² at A and B.

Find the co-ordinates of A and B.

Re-arrange the equation of the line to get, x = 3y – 1

Then x² = 9y² – 3y – 3y + 1

x² = 9y² – 6y + 1

Re-arrange the equation of the ellipse to get 2y – 2x + 2x² – 11y² = 0.

2y –2(3y – 1) + 2(3y – 1)² – 11y² = 0

2y – 6y + 2 + 2 (9y² – 6y + 1 ) – 11y² = 0

2y – 6y + 2 + 18y² – 12y + 2 – 11y² = 0

18y² – 11y² + 2y – 6y – 12y + 2 + 2 = 0

7y² – 16y + 4 = 0

(7y – 2) (y – 2) = 0

7y – 2 = 0 or y – 2 = 0

y = 2/7 or y = 2

x = 3y – 1 x = 3y – 1

x = 3(2/7) – 1 x = 3(2) – 1

x =– 1/7 x = 5

The points of intersection are A(-1/7, 2/7) and B (5, 2)

5. The line 4x – 3y = 15 intersects the curve 45 = 8x² – 27y² at A and B.

Find the co-ordinates of A and B.

Rearrange 4x – 3y = 15 to make y the subject of the equation:

3y = (4x – 15)

y = (4x – 15)/3

Therefore y² = (16x² – 120 x + 225)/9

Substitute into the curve:

45 = 8x² – 27 (16x² – 120 x + 225)/9



0 = 8x²– 3 ( 16 x² – 120 x + 225 ) – 45



0 = 8x² – 48x² – 675 – 360 x – 45



40x² + 360 x+ 720 = 0



x² + 9x + 18 = 0

(x + 6) (x + 3) = 0

x = -6 or x = -3

When x = - 6, y = -13

When x = -3, y = -9

6. The line y = 3x – 1 intersects the curve 11 = 2x² + 2y² – x + y at A and B.

Find the co-ordinates of A and B.

Square y = 3x – 1 to get y² = 9x² – 6x + 1, and substitute into the equation of the curve:

11 = 2x² + 2(9x² – 6x + 1) – x + 3x – 1

0 = 2x² + 18x² – 12x + 2 – x + 3x – 1 – 11

0 = 20x² – 10x – 10 (We can divide both sides by 20. )

0 = 2x² – 1x – 1

(2x + 1) (x – 1) = 0

x = - ½ or x = 1

When x = - ½, y = -2 ½

When x = 1, y = 2