Solving linear-quadratic systems algebraically

y+2=(x-3)^2
y+3=x

(2, - 1 ) and (5, 2 )

Step-by-step explanation:

y + 2 = (x - 3)² → (1)

y + 3 = x ( subtract 3 from both sides )

y = x - 3 → (2)

Substitute y = x - 3 into (1)

x - 3 + 2 = (x - 3)² ← expand using FOIL

x - 1 = x² - 6x + 9 ( subtract x - 1 from both sides )

0 = x² - 7x + 10 ← in standard form

0 = (x - 2)(x - 5) ← in factored form

Equate each factor to zero and solve for x

x - 2 = 0 ⇒ x = 2

x - 5 = 0 ⇒ x = 5

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

x = 2 : y = 2 - 3 = - 1 ⇒ (2, - 1 )

x = 5 : y = 5 - 3 = 2 ⇒ (5, 2 )

(4,5

look up a corrdinate plane and do what the thingy says and boom

a - base

b - exponent

so,