{{ stepNode.name }}

Proceed to next lesson

An error ocurred, try again later!

Chapter {{ article.chapter.number }}

{{ article.number }}. # {{ article.displayTitle }}

{{ article.introSlideInfo.summary }}

{{ 'ml-btn-show-less' | message }} {{ 'ml-btn-show-more' | message }} {{ 'ml-lesson-show-solutions' | message }}

{{ 'ml-lesson-show-hints' | message }}

| {{ 'ml-lesson-number-slides' | message : article.introSlideInfo.bblockCount}} |

| {{ 'ml-lesson-number-exercises' | message : article.introSlideInfo.exerciseCount}} |

| {{ 'ml-lesson-time-estimation' | message }} |

Image Credits *expand_more*

- {{ item.file.title }} {{ presentation }}

No file copyrights entries found

A system of equations does not always consist of only linear equations. The aim of this lesson is to show different ways to solve a system of equations if there are quadratic equations in the system.
### Catch-Up and Review

**Here are a few recommended readings before getting started with this lesson.**

Consider the equation of a circle centered at the origin with a radius of $22 .$

$x_{2}+y_{2}=(22 )_{2}⇓x_{2}+y_{2}=8 $

b At what points does the parabola $y=21 x_{2}$ intersect the circle?

A nonlinear system is a system of equations in which at least one of the equations is nonlinear. Consider the following system of equations.

${2x+4y=104x_{2}+y=-5 (I)(II) $

Even if the first equation in the system is linear, the system is a nonlinear system because the $x-$term of Equation (II) is elevated to the power of $2,$ making it nonlinear.Determine if the given system of equations is nonlinear or linear.

Some common nonlinear systems are linear-quadratic systems, which are systems of two equations that consist of a linear equation and a quadratic equation. These systems can have zero, one, or two different solutions. These solutions are the point of intersection of the graphs.

What happens if both equations in a system of two equations are quadratic equations?

A system of equations does not necessarily consist of linear equations. Different types of equations can be grouped into a system.

A system of quadratic equations is a system of equations that consists of only quadratic equations. **all** the equations true. In the example above, $x=2$ and $y=-5$ are a solution to the system. This can be verified by substituting the values into each equation.

In the examples, since the equations remain true, the values are a solution of the quadratic system. Quadratic systems can be solved graphically or algebraically. Since the equations in a quadratic system are not linear, these systems are nonlinear systems.

${y=x_{2}−6x+3y=-x_{2}+2x−5 $

Similar to systems of linear equations, the solution to a system of quadratic equations are the values of the variables that make $y=x_{2}−6x+3$ | $y=-x_{2}+2x−5$ |
---|---|

$-5=2_{2}−6(2)+3$ | $-5=-2_{2}+2(2)−5$ |

$-5=4−12+3$ | $-5=-4+4−5$ |

$-5=-5$ | $-5=-5$ |

For every given system of equations shown in the applet, determine if the system is a system of quadratic equations or not. Note that every equation in such a system needs to be a quadratic equation.

Another type of non-linear system is a quadratic-quadratic system, which consists of two quadratic equations. These systems can have zero, one, two, or infinitely many different solutions. These solutions are the points of intersection of the graphs.

Graphing the equations of a system of equations is helpful. Sometimes it is even possible to solve the system by referencing the graph only.

If the equations can be graphed, the solutions of a nonlinear system are the points of intersection of the graphs of every equation that makes the system. Therefore, by finding the points of intersection of the graphs, it is possible to solve a nonlinear system. As an example, consider a linear-quadratic system.
*expand_more*
*expand_more*
*expand_more*

${2x−2=y2x_{2}−8x+6=y $

To find the solutions, first each equation is graphed.
1

Graphing the First Equation

The first equation of the nonlinear system is a linear equation written in slope-intercept form. By using the slope of $2$ and the $y-$intercept of $-2,$ the equation is graphed in a coordinate plane.

Then, the second equation needs to be graphed.

2

Graphing the Second Equation

The second equation is a quadratic equation written in standard form.
This means that the vertex lies on point $(2,-2).$

$ax_{2}+bx+c=y⇓2x_{2}+(-8)x+6=y $

To graph a quadratic equation in standard form, first the axis of symmetry needs to be identified and graphed. To do so, the values are substituted into the formula.
This indicates that the axis of symmetry is a vertical line on $x=2.$
Then, to find the vertex of the parabola, the equation is evaluated to find the value of $y$ when $x=2.$
$2x_{2}−8x+6=y$

Substitute

$x=2$

$2(2)_{2}−8(2)+6=y$

Solve for $y$

CalcPow

Calculate power

$2(4)−8(2)+6=y$

Multiply

Multiply

$8−16+6=y$

AddTerms

Add terms

$-2=y$

RearrangeEqn

Rearrange equation

$y=-2$

Then, the $y-$intercept is the value of $c$ of the equation, which in this case is $6.$ Since the axis of symmetry divides the graph into two mirror images, the parabola also goes through point in $(4,6).$

Using these points, it is possible to draw the parabola.

3

Finding the Points of Intersection

Finally, finding the points in which the graph intersect, the solutions of the nonlinear system are found.

This means that the points $(1,0)$ and $(4,6)$ are the solutions of the nonlinear system.

Emily received a scholarship this summer and attended a camp for skydiving. She has always dreamed of feeling like flying in the sky. Surprisingly, her first task given by her instructor Sky Flyer is to solve a nonlinear system graphically.

Instructor Sky Flyer must be up to something by having Emily solve this system.${2x−y=0x_{2}+2x−1=y (I)(II) $

Along with Emily, solve the nonlinear system graphically. Graph both equations and find the points of intersection.

To solve this system graphically, the points of intersection of the graphs need to be found. To do so, each equation should be graphed. Equation (I) is a linear equation that can be written in slope-intercept form.
This indicates that the axis of symmetry is the vertical line $x=-1.$
Then, to find the vertex of the parabola, $-1$ will be substituted for $x$ in Equation (II).
This indicates that the vertex is at point $(-1,-2).$

$2x−y=0⇒y=2x $

The slope of $2$ and the $y-$intercept of $0$ can be used to graph this line in the coordinate plane.
Equation (II) is a quadratic equation written in standard form. $ax_{2}+bx+c=y⇓1x_{2}+2x+(-1)=y $

The first step to graph this equation is to find the axis of symmetry. That can be done by substituting the values of $a=1$ and $b=2$ into the Quadratic Formula to find the axis. Note that the root term is $0.$
$x=-2ab±0 $

IdPropAdd

Identity Property of Addition

$x=-2ab $

SubstituteII

$a=1$, $b=2$

$x=-2(1)2 $

Multiply

Multiply

$x=-22 $

CalcQuot

Calculate quotient

$x=-1$

$x_{2}+2x−1=y$

Substitute

$x=-1$

$(-1)_{2}+2(-1)−1=y$

Simplify

CalcPow

Calculate power

$1+2(-1)−1=y$

Multiply

Multiply

$1−2−1=y$

SubTerms

Subtract terms

$-2=y$

RearrangeEqn

Rearrange equation

$y=-2$

The value of $c$ on a quadratic equation written in standard form indicates the value of the $y-$intercept. In this case, the value of the $y-$intercept is $-1.$ Since the axis of symmetry divides the graph into mirror images, the points $(0,-1)$ and $(-2,-1)$ can be added.

These points can be used to graph the parabola.

Looking at the graph, the points of intersection can be located.

The solutions of the nonlinear system are the points $(-1,-2)$ and $(1,2).$

Instructor Sky Flyer continues teaching Emily about nonlinear systems. Emily is unsure but thinks they must be getting closer to applying this knowledge to something spectacular.

Sky Flyer gives the following quadratic-quadratic system to Emily and the other students learning to skydive.${y=61 x_{2}+2x+5y=-21 x−2x+5 (I)(II) $

Along with Emily, solve the nonlinear system graphically.
How do you find the axis of symmetry of a parabola if the quadratic equation is written in standard form?

To solve a quadratic-quadratic system graphically, both quadratic equations are graphed to identify the points of intersection. Looking at the system, it can be noted that both equations of the system are written in standard form.

$y=ax_{2}+bx+c$ | |
---|---|

$y=61 x_{2}+2x+5$ | $y=-21 x+(-2)x+5$ |

$x=-2ab $

SubstituteII

$a=61 $, $b=2$

$x=-2(61 )2 $

Solve for $x$

CancelCommonFac

Cancel out common factors

$x=-2 (61 )2 $

SimpQuot

Simplify quotient

$x=-61 1 $

DivOneByFracD

$a/b1 =ab $

$x=-6$

$y=61 x_{2}+2x+5$

Substitute

$x=-6$

$y=61 (-6)_{2}+2(-6)+5$

Solve for $y$

CalcPow

Calculate power

$y=61 (36)+2(-6)+5$

Multiply

Multiply

$y=61 (36)−12+5$

MoveRightFacToNum

$ca ⋅b=ca⋅b $

$y=636 −12+5$

CalcQuot

Calculate quotient

$y=6−12+5$

AddTerms

Add terms

$y=-1$

It is possible to find the $y-$intercept of the parabola with the value of $c.$ In Equation (I), this value is $5.$ Also, since the axis of symmetry mirrors the image of the parabola, there is another known point at $(-12,5).$ These points can be used to graph the parabola.

The same can be done to graph Equation (II). The axis of symmetry can be found by substituting the values of $a$ and $b$ in the formula.$x=-2ab $

SubstituteII

$a=-21 $, $b=-2$

$x=-2(-21 )-2 $

Solve for $x$

MultPosNeg

$a(-b)=-a⋅b$

$x=--2(21 )-2 $

MultFracByInverse

$ba ⋅ab =1$

$x=--1-2 $

CalcQuot

Calculate quotient

$x=-2$

The value $c=5$ on Equation (II) indicates that the $y-$intercept of the second parabola is $y=5.$ Since the axis of symmetry mirrors the image, there is another known point $(−4,5).$

Now that the two equations are graphed, the solutions of the nonlinear system can be determined. They are the points of intersection of the graphs.

The points $(0,5)$ and $(-6,-1)$ are the solutions of the nonlinear system.

Just like in a system of linear equations, there are many methods that can be used to solve a nonlinear system.

Some nonlinear systems can be solved by eliminating one of the variables of the equations involved, similar to the elimination method used in systems of linear equations. Consider the following example.
*expand_more*
*expand_more*
*expand_more*
*expand_more*

It should be noted that not every nonlinear system can be solved using this method.

${-2x_{2}+8x+16=2yx+4=y (I)(II) $

Since Equation (I) is a quadratic equation, the system is nonlinear. The system can be solved by elimination following the steps below.
1

Find a Variable to Eliminate

There are two different variables in the nonlinear system, $x$ and $y.$ The equations have different powers for the terms of $x,$ as one is quadratic and the other is linear. An the other hand, both equations have linear terms of $y.$ Therefore, variable $y$ will be eliminated.

${-2x_{2}+8x+16=2yx+4=y (I)(II) $

2

Eliminating the Variable

For a variable to be eliminated, its coefficient has to be the same in both equations. This can be achieved by dividing Equation (I) by $2$ and then subtracting Equation (II) from Equation (I).
It can be noted that one of the equations only depends on $x$ now.

${-2x_{2}+8x+16=2yx+4=y $

DivEqn

$(I):$ $LHS/2=RHS/2$

${-x_{2}+4x+8=yx+4=y $

SysEqnSub

$(I):$ Subtract $(II)$

${-x_{2}+4x+8−(x+4)=y−yx+4=y $

SubTerms

$(I):$ Subtract terms

${-x_{2}+3x+4=0x+4=y $

3

Solving the Single Variable Equation

Equation (II) is a quadratic equation in terms of $x.$ The equation is already written in standard form.
There are two possible values for $x.$ These can be found by adding or subtracting $5.$

$ax_{2}+bx+c=0⇓(-1)x_{2}+(3)x+4=0 $

This equation can be solved using the quadratic formula.
$-x_{2}+3x+4=0$

UseQuadForm

Use the Quadratic Formula: $a=-1,b=3,c=4$

$x=2(-1)-3±3_{2}−4(-1)(4) $

CalcPow

Calculate power

$x=2(-1)-3±9−4(-1)(4) $

Multiply

Multiply

$x=-2-3±9+16 $

AddTerms

Add terms

$x=-2-3±25 $

CalcRoot

Calculate root

$x=-2-3±5 $

$x=-2-3±5 $ | |
---|---|

$x=-2-3+5 $ | $x=-2-3−5 $ |

$x=-22 $ | $x=-2-8 $ |

$x=-1$ | $x=4$ |

Therefore, the values of $x$ that solve the nonlinear system are $-1$ and $4.$

4

Substituting the Values

Now that there are two known values for $x,$ the solution values for $y$ can be found by substituting the values of $x$ into either equation. Since Equation (II) is linear, it is easier to do so in this equation.

$x+4=y$ | |
---|---|

$-1+4=y$ | $4+4=y$ |

$y=3$ | $y=8$ |

Combining each $x-$value value with its correspondent $y-$value, it can be seen that the solutions can be written as the points $(-1,3)$ and $(4,8).$

Instructor Sky Flyer takes the students out to the yard. Emily, super excited, looks up and sees someone free falling with a parachute! ### Hint

### Solution

The height above the ground of the skydiver can be modeled by a quadratic equation.

$h=-4.9t_{2}+5450 $

In this equation, $t$ is the time passed in seconds after the jump and $h$ is the skydiver's height above the ground in meters. After releasing her parachute, the rate at which she is falling slows. The model for that height is given by the following linear equation.
$h=-5t+4450 $

How long after jumping did the jumper release their parachute? Use the elimination method to find the solution, and round that time to one decimal place. {"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":["x"],"constants":["PI"]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":null,"formTextAfter":"seconds","answer":{"text":["14.8"]}}

Which variable is the easiest to eliminate?

First of all, the given equations can be combined to make a nonlinear system.
Now a quadratic equation was obtained. The equation is already written in standard form.

${h=-4.9t_{2}+5450h=-5t+4450 (I)(II) $

Finding how long after jumping did the jumper release his parachute is the same as finding the time $t$ where the model changed from Equation (I) to Equation (II). This can be done by solving the nonlinear system. The first step to solve the nonlinear system by elimination is to find a variable to eliminate.
${h=-4.9t_{2}+5450h=-5t+4450 $

The variable $h$ is chosen because there are only linear terms of that variable. Then, the elimination is done by subtracting Equation (II) from Equation (I).
${h=-4.9t_{2}+5450h=-5t+4450 $

SysEqnSub

$(I):$ Subtract $(II)$

${h−h=-4.9t_{2}+5450−(-5t+4450)h=-5t+4450 $

SubTerms

$(I):$ Subtract terms

${0=-4.9t_{2}+5t+1000h=-13t+450 $

$at_{2}+bt+c=0⇓-4.9t_{2}+5t+1000=0 $

This equation is solved using the quadratic formula.
$t=2a-b±b_{2}−4ac $

SubstituteValues

Substitute values

$t=2(-4.9)-(5)±5_{2}−4(-4.9)(1000) $

CalcPow

Calculate power

$t=2(-4.9)-5±25−4(-4.9)(1000) $

Multiply

Multiply

$t=-9.8-5±25+19600 $

AddTerms

Add terms

$t=-9.8-5±19625$