# Quadratic Equations System - Algebraic and Graphical Solution

In the system of quadratic equations, we must find the $X$ and $Y$ that satisfy the first equation as well as the second. The system can have one solution, two solutions, or even none. The concept of the solution of the system of equations highlights the points of intersection of the function. At the same points we find - the functions intersect. If one solution is found - the functions intersect once. If two solutions are found - the functions intersect twice. If no solution is found - the functions never intersect.

## Test yourself on system of quadratic equations!

Consider the following relationships between the variables x and y:

$$x^2+4=-6y$$

$$y^2+9=-4x$$

## Algebraic solution

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:

1. We will verify that the variable $Y$ is written in the same way in both equations
2. We will compare the equations
3. We will solve for the $X$s
4. We will gradually substitute $X$s into one of the equations to solve for its $Y$
5. 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.

## Graphical solution

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.

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### If two solutions are found - the functions intersect twice

Do you know what the answer is?

### Let's look at an example

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

1. Let's verify that the $Y$ is really isolated in the same way.
2. Let's compare the equations:
$5x^2-2x+6=-5x^2-2x+6$
3. Let's solve for $X$s
$5x^2-2x+6=-5x^2-2x+6$
Let's transpose terms and we will get:
$10x^2=0$
$x^2=0$
$x=0$
4. Let's find the $Y$ by substituting the $X$ we found into one of the equations:
$y=5x^2-2x+6$
$y=5*0^2-2*0+6$
$y=6$
5. 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.

If you are interested in this article, you may 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

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

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

## Examples and exercises with solutions of quadratic equation systems

### Exercise #1

Consider the following relationships between the variables x and y:

$x^2+4=-6y$

$y^2+9=-4x$

### Video Solution

$(x+2)^2+(y+3)^2=0$

### Exercise #2

Look at the rectangle in the figure.

x>0

The area of the rectangle is:

$x^2-13$.

Calculate x.

### Video Solution

$x=3$