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{{ printedBook.courseTrack.name }} {{ printedBook.name }} A system of quadratic equations is a system where all equations are quadratic. They can be expressed as follows. ${y=x_{2}−6x+3y=-x_{2}+2x−5 $ The graphs to quadratic systems can have $0,$ $1,$ $2,$ or infinitely many points of intersection. Thus, the number of solutions of a quadratic system is also $0,$ $1,$ $2,$ or infinitely many.

Quadratic systems can be solved graphically or algebraically.

Similar to solving linear systems and linear-quadratic systems, a quadratic system can be solved 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.

Solve the system using substitution. ${y=x_{2}−4x+2y=-x_{2}+2x−2 $

Show Solution

We can solve the system by substituting one equation into the other.
$x_{2}−4x+2=-x_{2}+2x−2$
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 $0.$
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.
$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). $

$x_{2}−4x+2=-x_{2}+2x−2$

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

$2x_{2}−4x+2=2x−2$

SubEqn$LHS−2x=RHS−2x$

$2x_{2}−6x+2=-2$

AddEqn$LHS+2=RHS+2$

$2x_{2}−6x+4=0$

$2x_{2}−6x+4=0$

UseQuadFormUse 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 $

CalcPowProdCalculate power and product

$x=46±36−32 $

SubTermSubtract term

$x=46±4 $

CalcRootCalculate root

$x=46±2 $

StateSolState solutions

$x=46+2 x=46−2 $

AddSubTermsAdd and subtract terms

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

CalcQuotCalculate quotient

$x_{1}=2x_{2}=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,
${y≥0.5x_{2}−4x+6y≤-x_{2}+8x−12 $
can be solved this way.
### 1

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

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

Graph one inequality

Graph the other inequality

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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