Question

Solve the system of equations y=x^2-2 y=-2x+1

Answer 1

Answer:

D

Step-by-step explanation:

Given the 2 equations

y = x² - 2 → (1)

y = - 2x + 1 → (2)

Substitute y = x² - 2 into (2)

x² - 2 = - 2x + 1 ( subtract - 2x + 1 from both sides )

x² + 2x - 3 = 0 ← in standard form

(x + 3)(x - 1) = 0 ← in factored form

Equate each factor to zero and solve for x

x + 3 = 0 ⇒ x = - 3

x - 1 = 0 ⇒ x = 1

Substitute these values into (2) for corresponding values of y

x = - 3 : y = -2(- 3) + 1 = 6 + 1 = 7 ⇒ (- 3, 7 )

x = 1 : y = - 2(1) + 1 = - 2 + 1 = - 1 ⇒ (1, - 1 )

Answer 2

Answer:

D. (-3, 7) and (1, -1)

Step-by-step explanation:

$\left\{\begin{array}{ccc}y=x^2-2&(1)\\y=-2x+1&(2)\end{array}\right\\\\\text{substitute (1) to (2):}\\\\x^2-2=-2x+1\qquad\text{add 2x to both sides}\\x^2+2x-2=1\qquad\text{subtract 1 from both sides}\\x^2+2x-3=0\\x^2+3x-x-3=0\\x(x+3)-1(x+3)=0\\(x+3)(x-1)=0\iff x+3=0\ \vee\ x-1=0\\\\x+3=0\qquad\text{subtract 3 from both sides}\\x=-3\\\\x-1=0\qquad\text{add 1 to both sides}\\x=1\\\\\text{put the value of x to (1):}\\\\for\ x=-3\\y=(-3)^2-2=9-2=7\\\\for\ x=1\\y=1^2-2=1-2=-1$