Solve algebraically the simultaneous equations
y = 2x2 – 3x - 10
2x - y = -2

(-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)

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step-by-step explanation:

answer: the correct answer is -32

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