Question

What is the solution of the linear-quadratic system of equations? y=x' + 5x + 7 y = x +4

Answer 1

Answer:

$\mathrm{Therefore,\:the\:final\:solutions\:for\:}y=x^2+5x+7,\:y=x+4\mathrm{\:are\:}$

$\begin{pmatrix}x=-1,\:&y=3\\ x=-3,\:&y=1\end{pmatrix}$

Step-by-step explanation:

Given the system of the equations

$y=x^2\:+\:5x\:+\:7$

$y\:=\:x\:+4$

steps to solve the system of the equations

$\begin{bmatrix}y=x^2+5x+7\\ y=x+4\end{bmatrix}$

subtract the equations

$y=x^2+5x+7$

$-$

$\underline{y=x+4}$

$y-y=x^2+5x+7-\left(x+4\right)$

$0=x^2+4x+3$

solving

$0=x^2+4x+3$

$\left(x+1\right)\left(x+3\right)=0$

$x+1=0\quad \mathrm{or}\quad \:x+3=0$

$x=-1,\:x=-3$

$\mathrm{Plug\:the\:solutions\:}x=-1,\:x=-3\mathrm{\:into\:}y=x^2+5x+7$

substitute $x=-1$ into $y=x^2+5x+7$

$y=x^2+5x+7$

$y=\left(-1\right)^2-5\cdot \:1+7$

$y=1-5+7$

$y=3$

Now substitute $x=-3$ into $y=x^2+5x+7$

$y=x^2+5x+7$

$y=\left(-3\right)^2+5\left(-3\right)+7$

$y=3^2-15+7$

$y=3^2-8$

$y=9-8$

$y=1$

$\mathrm{Therefore,\:the\:final\:solutions\:for\:}y=x^2+5x+7,\:y=x+4\mathrm{\:are\:}$

$\begin{pmatrix}x=-1,\:&y=3\\ x=-3,\:&y=1\end{pmatrix}$