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


RELATED: