Question

What is the solution of the linear-quadratic system of equations?

y=x' + 5x + 7
y = x +4

Answer 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}`

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}`