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

How to solve: 3x - 2y = 1 3x 2 2y 2 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 3x 2 2y 2 5 = 0 3x 2 2((3x - 1)/2) 2 5 = 0 substitute for y 3x 2 2(9x 2 6x + 1)/4 + 5 = 0 expand the square term 3x 2 (9x 2 6x + 1)/2 + 5 = 0 spot that 2w/4 is the same as w/2 6x 2 (9x 2 6x + 1) + 10 = 0 multiply equation by 2 6x 2 9x 2 6x - 1 + 10 = 0 remove the brackets by multiplying by -1 as it is -1 x (..) 3x 2 6x + 9 = 0 collect together terms 3x 2 6x - 9 = 0 multiply whole equation by -1 x 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.

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