Solve the following system of equations algebraically:

y = x^2+7x-4

y = - x-4

Y = x² + 7x - 4

y = - x - 4

x² + 7x - 4 = - x - 4

+ x + x

x² + 8x - 4 = - 4

+ 4 + 4

x² + 8x = 0

x (x) + x (8) = 0

x (x + 8) = 0

x = 0 or x + 8 = 0

- 8 - 8

x = - 8

y = - x - 4

y = - 0 - 4

y = 0 - 4

y = - 4

(x, y) = (0, - 4)

or

y = - x - 4

y = - (-8) - 4

y = 8 - 4

y = 4

(x, y) = (-8, 4)

The solutions are (0, - 4) and (8, - 4).

Y=x²+7x-4

y=-x-4

We can solve this system of equation by equalization method.

x²+7x-4=-x-4

x²+7x+x-4+4=0

x²+8x=0

x (x+8) = 0

we have two linear equations:

x=0 ⇒ x=0 ⇒ y=-x-4=-0-4=-4

(x+8) = 0 ⇒ x=-8 ⇒ y=-x-4 = - (-8) - 4=8-4=4

We have two solutions:

solution₁: x=0; y=-4

solution₂: x=-8; y=4