**Given the following system of equations:** **$y=x^2+2x+7$,$2x+y=4$**

**Solution:**

Let's isolate one of the variables from one of the equations:

$2x+y=4$

$y=-2x+4$

**Now, let's substitute the value we isolated into the second equation. We will obtain:**

$-2x+4=x^2+2x+7$

Let's combine like terms and transpose members. We will obtain:

$-X^2-4x-3=0$

We will solve it with the help of the quadratic formula and will obtain:

$X=-1 ,-3$

Now we will gradually place an $X$ we found into one of the equations to find the complete points: Let's start with $X=-1$

$2\times(-1)+y=4$

$-2+y=4$

$y=6$

Let's write this solution $(-1,6)$

let's move to the second $X$ we found $X=-3$ and we will obtain:

$2*(-3)+y=4$

$-6+y=4$

$y=10$ let's write this solution $(-3,10)$

These solutions are the points of intersection of the parabola with the axis.