Solution of a system of equations - one of them is linear and the other quadratic

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.

If you are interested in this article, you might also be interested in the following articles:

The functions y=x²

Family of parabolas y=x²+c: Vertical shift

Family of parabolas y=(x-p)²

Family of parabolas y=(x-p)²+k (combination of horizontal and vertical shift)

Vertex form of the quadratic function

Factored form of the quadratic function

Completing the square in a quadratic equation

Standard form of the quadratic function

System of quadratic equations - Algebraic and graphical solution

In the blog of Tutorela you will find a variety of articles about mathematics.