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A system of quadratic equations is a system of equations that consists of only quadratic equations.

${y=x_{2}−6x+3y=-x_{2}+2x−5 $

The graphs of quadratic systems with two equations can have $0,$ $1,$ $2,$ or infinitely many points of intersection. Therefore, the number of solutions of a quadratic system is also $0,$ $1,$ $2,$ or infinitely many.
Quadratic systems can be solved graphically or algebraically. Since the equations in a quadratic system are not linear, these systems are nonlinear systems.

Similar to solving linear systems and solving linear-quadratic systems, solving quadratic systems can be done by graphing both equations. The solution(s) can be found by identifying the coordinates of the point(s) of intersection.

Quadratic systems can be solved using substitution just like linear-quadratic systems. One equation is substituted into the other, resulting in either a quadratic or linear equation. The solution to the resulting equation gives the $x$-values for the solution to the entire system. Lastly, they are substituted into one of the original equations to find the corresponding $y$-values.

${y=x_{2}−4x+2y=-x_{2}+2x−2 $

Show Solution *expand_more*

We can solve the system by substituting one equation into the other.
From here, we can solve the equation using the quadratic formula.
The $x$-values for the solutions are $x=2$ and $x=1.$ Now, we substitute them into one of the original equations to find the corresponding $y$-values. We'll use the first equation, but it doesn't matter which.

$x_{2}−4x+2=-x_{2}+2x−2$

Now, we have a quadratic equation. Its solutions will give us the $x$-values for the solutions of the system. Let's rearrange the equation so that it is set equal to $0.$
$x_{2}−4x+2=-x_{2}+2x−2$

AddEqn

$LHS+x_{2}=RHS+x_{2}$

$2x_{2}−4x+2=2x−2$

SubEqn

$LHS−2x=RHS−2x$

$2x_{2}−6x+2=-2$

AddEqn

$LHS+2=RHS+2$

$2x_{2}−6x+4=0$

$2x_{2}−6x+4=0$

UseQuadForm

Use the Quadratic Formula: $a=2,b=-6,c=4$

$x=2⋅2-(-6)±(-6)_{2}−4⋅2⋅4 $

NegNeg

$-(-a)=a$

$x=2⋅26±(-6)_{2}−4⋅2⋅4 $

CalcPowProd

Calculate power and product

$x=46±36−32 $

SubTerm

Subtract term

$x=46±4 $

CalcRoot

Calculate root

$x=46±2 $

StateSol

State solutions

$x=46+2 x=46−2 $

AddSubTerms

Add and subtract terms

$x=8/4x=4/4 $

CalcQuot

Calculate quotient

$x_{1}=2x_{2}=1 $

$y=2_{2}−4⋅2+2=-2y=1_{2}−4⋅1+2=-1 $

Thus, the solutions to the quadratic system are
$(2,-2)and(1,-1). $

A quadratic system can also consist of quadratic inequalities, such as

${y<x_{2}+5x−4y≥2x_{2}−4x−10. $

The solution set to a system of quadratic inequalities is, similar to systems of linear inequalities, an entire region in the coordinate plane.
A system of quadratic inequalities can be solved graphically, similar to a system of linear inequalities. Specifically, the individual inequalities are graphed, and the intersection of the shaded regions is the solution to the system. For example,
*expand_more*
*expand_more*
*expand_more*

${y≥0.5x_{2}−4x+6y≤-x_{2}+8x−12 $

can be solved this way.
1

Graph one inequality

To find the solutions to the system, start by graphing one of the inequalities. Here, $y≥0.5x_{2}−4x+6$ has the boundary curve $y=0.5x_{2}−4x+6,$ and the region corresponding to the solution set lies inside the parabola.

2

Graph the other inequality

Next, graph the other inequality. Here, $y≤-x_{2}+8x−12$ has the boundary $y=-x_{2}+8x−12.$ The region corresponding to the solution set also lies inside the parabola.

3

Find the overlapping region

The solutions to the system are solutions to both individual inequalities. Meaning, these lie in the overlapping shaded regions. Here, that is the purple area.

Since the curves in their entirety are not part of the solution set, trim them down to only border the purple region. Now, the solution set of the system is shown.

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