## Saturday, October 2, 2010

### 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 = 8x227y2 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 = 8x227 (16x2 – 120 x + 225)/9

0 = 8x23 ( 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