Question

3x+10, x^2+4x-2 solve the system.

Answer 1

Answer:

(-4,-2) and (3,19)

Step-by-step explanation:

We want to solve the system:

$y = 3x + 10$

$y = {x}^{2} + 4x - 2$

We equate the right hand sides:

${x}^{2} + 4x - 2 = 3x + 10$

Rewrite is standard quadratic form.

${x}^{2} + 4x - 3 x - 2 - 10 = 0$

This simplifies to:

${x}^{2} + x - 12= 0$

We factor:

$(x - 3)(x + 4) = 0$

$x = 3 \: or \: x = - 4$

When x=3,

$y = 3 \times 3 + 10 = 19$

Hence a solution is (3,19)

When x=-4,

$y = 3 \times - 4 + 10 = - 2$

Another solution is (-4,-2)