Question

Find the solution set of this system:

y=−3x


x2−4=y


(4,12), (−1,3)


(−1,−3) only


(4,12) only


(−4,12), (1,−3)

Answer 1

(−4, 12), (1, −3)

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)

Answer 2

(−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)