Method: Comparison between quadratic equations

When we have a system of quadratic equations and the $Y$ is isolated in this way (with the same coefficient in both equations):

**$Y=ax^2+bx+c$**

**$Y=ax^2+bx+c$**

**We will proceed in the following order:**

- We will verify that the variable $Y$ is written in the same way in both equations
- We will compare the equations
- We will solve for the $X$s
- We will gradually substitute $X$s into one of the equations to solve for its $Y$
- We will neatly record the solutions we have found.

**Attention -** The other parameters do not necessarily have to be the same. Only the $Y$ needs to be isolated in the same way in order to equate the equations.

The solution of the system of quadratic equations represents the points of intersection of the parabolas. Therefore, we will be able to see the solution of the system of equations graphically as the points of intersection of the parabolas.

$y=5x^2-2x+6$

$y=-5x^2-2x+6$

- Let's verify that the $Y$ is really isolated in the same way.
- Let's compare the equations:

$5x^2-2x+6=-5x^2-2x+6$ - Let's solve for $X$s

$5x^2-2x+6=-5x^2-2x+6$

Let's transpose terms and we will get:

$10x^2=0$

$x^2=0$

$x=0$ - Let's find the $Y$ by substituting the $X$ we found into one of the equations:

$y=5x^2-2x+6$

$y=5*0^2-2*0+6$

$y=6$ - Let's note the solution we found: $(6, 0)$

at this point the functions intersect and that is the solution to the system of equations.