Question

Solve the system
y = x^2 + 2x - 2
y = x + 10

Answer 1

x1 = -4,

y1 = 6,

x2 = 3,

y2 = 13

Explanation:

{y = x^2 + 2x - 2,

{y = x + 10;

Since both equations are equal to y, they are equal to one another aswell:

`{x}^(2) + 2x - 2 = x + 10`

Collect like-terms:

`{x}^(2) + x - 12 = 0`

a = 1, b = 1, c = -12

Let's solve this quadratic equation:

`d = {b}^(2) - 4ac = {1}^(2) - 4 * 1 * ( - 12) = 1 + 48 = 49 > 0`

`x1 = ( - b - √(d) )/(2a) = ( - 1 - 7)/(2 * 1) = ( - 8)/(2) = - 4`

`x2 = ( - b + √(d) )/(2a) = ( - 1 + 7)/(2 * 1) = (6)/(2) = 3`

{x1 = -4,

{y1 = -4 + 10 = 6;

{x2 = 3,

{y2 = 3 + 10 = 13

Answer 2

(- 4, 6 ) , (3, 13 )

Explanation:

y = x² + 2x - 2 → (1)

y = x + 10 → (2)

substitute y = x² + 2x - 2 into (2)

x² + 2x - 2 = x + 10 ← subtract x + 10 from both sides

x² + x - 12 = 0 ← in standard form

(x + 4)(x - 3) = 0 ← in factored form

equate each factor to zero and solve for x

x + 4 = 0 ⇒ x = - 4

x - 3 = 0 ⇒ x = 3

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

x = - 4 : y = - 4 + 10 = 6 ⇒ (- 4, 6 )

x = 3 : y = 3 + 10 = 13 ⇒ (3, 13 )