Question

question #2 solve each system of equations. -x^2 - 2x + 3y - 19 = 0 2x + y = 1 (-6, 9) (-4, 9), (-3, 5) (-3, 5) (-4, 9)

Explanation:

Step1: Isolate y in the linear - equation

From $2x + y=1$, we get $y = 1 - 2x$.

Step2: Substitute y into the quadratic - equation

Substitute $y = 1 - 2x$ into $-x^{2}-2x + 3y-19 = 0$.
\[

$$\begin{align*} -x^{2}-2x+3(1 - 2x)-19&=0\\ -x^{2}-2x + 3-6x-19&=0\\ -x^{2}-8x-16&=0\\ x^{2}+8x + 16&=0 \end{align*}$$

\]

Step3: Solve the quadratic equation

The quadratic equation $x^{2}+8x + 16=(x + 4)^{2}=0$.
So $x=-4$.

Step4: Find the value of y

Substitute $x = - 4$ into $y = 1 - 2x$.
$y=1-2\times(-4)=1 + 8=9$.

Answer:

$(-4,9)$