Answer:
(-3/2, -1) and (4, 10)
Step-by-step explanation:
Given the simultaneous equations
y = 2x^2 – 3x - 10 ... 1
2x - y = -2 .... 2
From equation2;
-y = -2-2x
y = 2x+2 ..... 3
Substitute 3 into 1
Recall y = 2x^2 – 3x - 10
2x+2 = 2x^2 – 3x - 10
Collect like terms
2x^2 – 3x - 10 - 2x -2 = 0
2x^2 – 5x - 12 = 0
2x^2 – 8x+3x - 12 = 0
2x(x-4)+3(x-4) = 0
(2x+3)(x-4) = 0
2x+3 = 0 and x-4 = 0
2x = -3 and x = 4
x = -3/2 and x = 4
Recall that
y = 2x+2
when x = -3/2
y = 2(-3/2) + 2
y = -3+2
y = -1
when x = 4
y = 2(4)+2
y = 8+2
y = 10
Hence the solutions are (-3/2, -1) and (4, 10)
