How do you work out the solutions for 3x-2y equals 1 and 3x square-2y square plus 5 equals 0?

Answer 1

How to solve:

3x - 2y = 1

3x2 - 2y2 + 5 = 0

Rearrange the first equation to make x or y the subject (that is x = something or y = something) and then substitute into the second equation and solve that:

3x - 2y = 1

=> y = (3x - 1)/2

3x2 - 2y2 + 5 = 0

=> 3x2 - 2((3x - 1)/2)2 + 5 = 0 [substitute for y]

=> 3x2 - 2(9x2 - 6x + 1)/4 + 5 = 0 [expand the square term]

=> 3x2 - (9x2 - 6x + 1)/2 + 5 = 0 [spot that 2w/4 is the same as w/2]

=> 6x2 - (9x2 - 6x + 1) + 10 = 0 [multiply equation by 2]

=> 6x2 - 9x2 + 6x - 1 + 10 = 0 [remove the brackets by multiplying by -1 as it is -1 x (..)]

=> -3x2 + 6x + 9 = 0 [collect together terms]

=> 3x2 - 6x - 9 = 0 [multiply whole equation by -1]

=> x2 - 2x - 3 = 0 [divide whole equation by 3]

=> (x - 3)(x + 1) =0 [factorize)

=> x = 3 or -1 [as one factor or the other must be zero]

Now use first equation to find corresponding y terms:

x = 3:y = (3 x (3) - 1) / 2

= 8 / 2

= 4

x = -1: y= (3 x (-1) - 1) /2

= -4 / 2

= -2

So the solution is the (x, y) pairs, or points, (3, 4) and (-1, -2).

The answer can be checked using the second equation:

(3, 4): 3(3)2 - 2(4)2 + 5 = 3 x 9 - 2 x 16 + 5

= 27 - 32 + 5

= 0

(-1, -2): 3(-1)2 - 2(-2)2 + 5 = 3 x 1 - 2 x 4 + 5

= 3 - 8 + 5

= 0