Question

Find the solution set of this system:

Answer 1

ANSWER

(−4,12), (1,−3)

EXPLANATION

The system has equations:

y=−3x

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

We equate the two equations to get:

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

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

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

$x(x + 4) - 1(x + 4) = 0$

$(x + 4)(x - 1) = 0$

$x = - 4 \: or \: x = 1$

When x=1,

y=-3(1)=-3

when x=-4, y=-3(-4)=12

The solution is (1,-3) and (-4,12)

Answer 2

Answer:

(−4, 12), (1, −3)

Step-by-step explanation:

y = −3x

x^2 −4 = y

so

x^2 −4 = −3x

x^2 + 3x - 4 = 0

(x + 4)(x - 1) = 0

x + 4 = 0; x = -4

x - 1 = 0; x = 1

when x = 1, y = -3(1) = -3

when x = -4, y = -3(-4) = 12

Answer

(-4 , 12), (1 , -3)