SIMULATANEOUS EQUATIONS - ANSWERS

1.   x + 3y = 8
      x + 5y = 12

(1) x + 3y = 8
(2) x + 5y = 12

Subtract (1) from (2) to cancel x (because x - x = 0), giving:

5y - 3y = 12 - 8
2y = 4
y = 4/2
y = 2

Now we know y, just substitute y = 2 into either equation: (1) looks the easiest so, we get x + (3 times 2) = 8
x + 6 = 8
x = 8 - 6
x = 2
2.  2x - 3y = 4
     5x + 7y = 68

(1) 2x - 3y = 4
(2) 5x + 7y = 68

Multiply (1) by 5 and (2) by 2 to cancel the x's, giving:

10x - 15y = 20
10x + 14y = 136

Subtract (1) from (2) giving:

14y - (-15y) = 136 - 20
29y = 116
y = 4

Subtitute y = 4 into the original (1) giving:

2x - 12 = 4
2x = 16
x = 8
3.  4a - 3b = 6
     3a + b = 11 

(1) 4a - 3b = 6
(2) 3a + b = 11

Multiply (2) by 3 and add (1) and (2) to cancel b, giving:

13a = 39
a = 3

Subtitute to get b = 2
4.  4x + y = 3
     3x -5y = 31		 x = 2, y = -5
5.  3x + 2y = 13
     5x - 3y = 9

x = 3, y = 2
6.  2x + 7y = 31
     5x - 3y = 16 		 x = 5, y = 3
7.  3y - 2z = 5
     7y - 5z = 11		 y = 3, z = 2
8.  -y + 3x = 5
     5x - 2y = 6

x = 4, y = 7