Question

question #6 solve each system of equations. -x^{2}-x + y+2 = 0 x + y+3 = 0 (1, - 4), (0, - 8) (-1, - 2) (1, - 4) no solution.

Explanation:

Step1: Isolate y in the second - equation

From $x + y+3 = 0$, we get $y=-x - 3$.

Step2: Substitute y into the first - equation

Substitute $y=-x - 3$ into $-x^{2}-x + y+2 = 0$. Then $-x^{2}-x+(-x - 3)+2 = 0$.
Simplify the equation:
$-x^{2}-x - x-3 + 2 = 0$.
$-x^{2}-2x - 1 = 0$.
Multiply through by - 1 to get $x^{2}+2x + 1 = 0$.

Step3: Solve the quadratic equation

The quadratic equation $x^{2}+2x + 1=(x + 1)^{2}=0$.
Using the zero - product property, if $(x + 1)^{2}=0$, then $x=-1$.

Step4: Find the value of y

Substitute $x=-1$ into $y=-x - 3$. So $y=-(-1)-3=1 - 3=-2$.

Answer:

$(-1,-2)$