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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.

Challenge

Finding Intersections with a Circle

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

a At what points does the line

y = \frac{1}{2} x + 1

intersect the circle?

b At what points does the parabola

y = \frac{1}{2} x^{2}

intersect the circle?

Discussion

Nonlinear System

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

{2 x + 4 y = 10 4 x^{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.

Identifying a nonlinear and linear system is a crucial step in figuring out which methods are best for solving any given system.

Pop Quiz

Is the System Nonlinear?

Determine if the given system of equations is nonlinear or linear.

Illustration

The Solutions of a Linear-Quadratic System

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?

Discussion

System of 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.

{y = x^{2} - 6 x + 3 y = - x^{2} + 2 x - 5

Similar to systems of linear equations, the solution to a system of quadratic equations are the values of the variables that make 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.

$y = x^{2} - 6 x + 3$	$y = - x^{2} + 2 x - 5$
$- 5 = 2^{2} - 6 (2) + 3$	$- 5 = - 2^{2} + 2 (2) - 5$
$- 5 = 4 - 12 + 3$	$- 5 = - 4 + 4 - 5$
$- 5 = - 5$	$- 5 = - 5$

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.

Pop Quiz

Is the System Quadratic?

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.

Illustration

The Solutions of a Quadratic-Quadratic System

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.

Discussion

Solving Nonlinear Systems Graphically

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.

{2 x - 2 = y 2 x^{2} - 8 x + 6 = y

To find the solutions, first each equation is graphed.

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.

Graphing the Second Equation

The second equation is a quadratic equation written in standard form.

a x^{2} + b x + c = y ⇓ 2 x^{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.

x = - \frac{b}{2 a}

SubstituteII

$a = 2$ , $b = - 8$

x = - \frac{- 8}{2 ( 2 )}

Solve for

x

Multiply

x = - \frac{- 8}{4}

CalcQuot

Calculate quotient

x = - (- 2)

NegNeg

$- (- a) = a$

x = 2

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 .

2 x^{2} - 8 x + 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

8 - 16 + 6 = y

AddTerms

Add terms

- 2 = y

RearrangeEqn

Rearrange equation

y = - 2

This means that the vertex lies on point

(2, - 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.

Finding the Points of Intersection

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

Linear Equation and Quadratic Equation Intersection

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

Example

Graphing a Linear-Quadratic 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.

{2 x - y = 0 x^{2} + 2 x - 1 = y (I) (II)

Along with Emily, solve the nonlinear system graphically.

Answer

Hint

Graph both equations and find the points of intersection.

Solution

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.

2 x - y = 0 \Rightarrow y = 2 x

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.

a x^{2} + b x + c = y ⇓ 1 x^{2} + 2 x + (- 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 = - \frac{b \pm 0}{2 a}

IdPropAdd

\IdPropAdd

x = - \frac{b}{2 a}

SubstituteII

$a = 1$ , $b = 2$

x = - \frac{2}{2 ( 1 )}

Multiply

x = - \frac{2}{2}

CalcQuot

Calculate quotient

x = - 1

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).

x^{2} + 2 x - 1 = y

Substitute

$x = - 1$

(- 1)^{2} + 2 (- 1) - 1 = y

Simplify

CalcPow

Calculate power

1 + 2 (- 1) - 1 = y

Multiply

1 - 2 - 1 = y

SubTerms

Subtract terms

- 2 = y

RearrangeEqn

Rearrange equation

y = - 2

This indicates that the vertex is at point

(- 1, - 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 $(1, 2) .$

Discussion

Solving Nonlinear Systems Using Elimination

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.

{- 2 x^{2} + 8 x + 16 = 2 y x + 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.

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.

{- 2 x^{2} + 8 x + 16 = 2 y x + 4 = y (I) (II)

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).

{- 2 x^{2} + 8 x + 16 = 2 y x + 4 = y

DivEqn

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

{- x^{2} + 4 x + 8 = y x + 4 = y

SysEqnSub

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

{- x^{2} + 4 x + 8 - (x + 4) = y - y x + 4 = y

SubTerms

$(I):$ Subtract terms

{- x^{2} + 3 x + 4 = 0 x + 4 = y

It can be noted that one of the equations only depends on

x

now.

Solving the Single Variable Equation

Equation (II) is a quadratic equation in terms of

x .

The equation is already written in standard form.

a x^{2} + b x + c = 0 ⇓ (- 1) x^{2} + (3) x + 4 = 0

This equation can be solved using the quadratic formula.

- x^{2} + 3 x + 4 = 0

UseQuadForm

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

x = \frac{- 3 \pm 3 ^{2} - 4 ( - 1 ) ( 4 )}{2 ( - 1 )}

CalcPow

Calculate power

x = \frac{- 3 \pm 9 - 4 ( - 1 ) ( 4 )}{2 ( - 1 )}

Multiply

x = \frac{- 3 \pm 9 + 1 6}{- 2}

AddTerms

Add terms

x = \frac{- 3 \pm 2 5}{- 2}

CalcRoot

Calculate root

x = \frac{- 3 \pm 5}{- 2}

There are two possible values for

x .

These can be found by adding or subtracting

5 .

$x = \frac{- 3 \pm 5}{- 2}$
$x = \frac{- 3 + 5}{- 2}$	$x = \frac{- 3 - 5}{- 2}$
$x = \frac{2}{- 2}$	$x = \frac{- 8}{- 2}$
$x = - 1$	$x = 4$

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

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 $(4, 8) .$

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

Example

Skydiving

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

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

h = - 4.9 t^{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 = - 5 t + 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.

Hint

Which variable is the easiest to eliminate?

Solution

First of all, the given equations can be combined to make a nonlinear system.

{h = - 4.9 t^{2} + 5450 h = - 5 t + 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 = - 4.9 t^{2} + 5450 h = - 5 t + 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 = - 4.9 t^{2} + 5450 h = - 5 t + 4450

SysEqnSub

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

{h - h = - 4.9 t^{2} + 5450 - (- 5 t + 4450) h = - 5 t + 4450

SubTerms

$(I):$ Subtract terms

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

Now a quadratic equation was obtained. The equation is already written in standard form.

a t^{2} + b t + c = 0 ⇓ - 4.9 t^{2} + 5 t + 1000 = 0

This equation is solved using the quadratic formula.

t = \frac{- b \pm b ^{2} - 4 a c}{2 a}

SubstituteValues

Substitute values

t = \frac{- ( 5 ) \pm 5 ^{2} - 4 ( - 4 . 9 ) ( 1 0 0 0 )}{2 ( - 4 . 9 )}

CalcPow

Calculate power

t = \frac{- 5 \pm 2 5 - 4 ( - 4 . 9 ) ( 1 0 0 0 )}{2 ( - 4 . 9 )}

Multiply

t = \frac{- 5 \pm 2 5 + 1 9 6 0 0}{- 9 . 8}

AddTerms

Add terms

t = \frac{- 5 \pm 1 9 6 2 5}{- 9 . 8}

The

\pm

sign indicates that there are two possible values for

t .

These can be found by either adding or subtracting the value of the root.

$t = \frac{- 5 \pm 1 9 6 2 5}{- 9 . 8}$
$t = \frac{- 5 + 1 9 6 2 5}{- 9 . 8}$	$t = \frac{- 5 - 1 9 6 2 5}{- 9 . 8}$
$t = - 13.7846$	$t = 14.8050$
$t \approx - 13.8$	$t \approx 14.8$

The two possible values for $t$ are about $14.8$ seconds and about $- 13.8$ seconds. Only positive numbers make sense in this case since the skydiver cannot wait a negative time before releasing the parachute. This means that the jumper waited about $14.8$ seconds to release the parachute.

Example

Acrobats on the Circus

Before having the chance to skydive, Instructor Sky Flyer wants to show how the students can train by practicing acrobatics. They all walk into a large gym to see a few pros practice. One acrobat jumped from a platform, and another dove from another higher platform just a split second later. Mid-air, they did a high-five while spinning!

The heights above the floor of both acrobats at which they did a high five can be modeled using a system of quadratic equations. These heights are measured from the moment the first acrobat jumps.

{h = - 16 (t - 1)^{2} + 16 h = - 16 (t - 2)^{2} + 16 (I) (II)

In these equations,

h

is the distance of the acrobats from the floor in feet, and

t

is the time in seconds.

a How long after jumping from their platforms did the acrobats high-five? Solve using elimination.

b At what height above the floor were the acrobats when they did the high-five?

Hint

a Which variable is present in the least amount of terms?

b Substitute the value from Part A into either equation.

Solution

a The first step to solve the given system using elimination is to find a variable to eliminate. The objective is to obtain a single equation that depends on a single variable. The variable

h

is present in only one term, so it can be eliminated to obtain an equation in terms of

t .

{h = - 16 (t - 1)^{2} + 16 h = - 16 (t - 2)^{2} + 16 (I) (II)

To eliminate the variable, Equation (II) is subtracted from Equation (I). To get started, it is practical to expand the perfect squares inside each parenthesis.

{h = - 16 (t - 1)^{2} + 16 h = - 16 (t - 2)^{2} + 16

ExpandNegPerfectSquare

$(I):$ $(a - b)^{2} = a^{2} - 2 a b + b^{2}$

{h = - 16 (t^{2} - 2 t + 1) + 16 h = - 16 (t - 2)^{2} + 16

ExpandNegPerfectSquare

$(II):$ $(a - b)^{2} = a^{2} - 2 a b + b^{2}$

{h = - 16 (t^{2} - 2 t + 1) + 16 h = - 16 (t^{2} - 4 t + 4) + 16

Distr

$(I):$ Distribute $- 16$

{h = - 16 t^{2} + 32 t - 16 + 16 h = - 16 (t^{2} - 4 t + 4) + 16

Distr

$(II):$ Distribute $- 16$

{h = - 16 t^{2} + 32 t - 16 + 16 h = - 16 t^{2} + 64 t - 64 + 16

AddTerms

Add terms

{h = - 16 t^{2} + 32 t h = - 16 t^{2} + 64 t - 48

SysEqnSub

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

{h - h = - 16 t^{2} + 32 t - (- 16 t^{2} + 64 t - 48) h = - 16 t^{2} + 64 t - 48

SubTerms

$(I):$ Subtract terms

{0 = - 32 t + 48 h = - 16 t^{2} + 64 t - 48

A linear equation that depends on

t

was obtained. Solving this equation gives the time at which the acrobats did the high-five.

0 = - 32 t + 48

SubEqn

$LHS - - 32 t = RHS - - 32 t$

32 t = 48

DivEqn

$LHS / 32 = RHS / 32$

t = \frac{4 8}{3 2}

ReduceFrac

$\frac{a}{b} = \frac{a / 1 6}{b / 1 6}$

t = \frac{3}{2}

This means that the acrobats did the high-five

\frac{3}{2},

1.5

seconds after the first one jumped.

b To find the height above the floor at which the acrobats did the high-five, the time found in Part A needs to be substituted into either equation of the given system. In this case, Equation (I) will be used.

h = - 16 (t - 1)^{2} + 16

Substitute

$t = 1.5$

h = - 16 (1.5 - 1)^{2} + 16

SubTerm

Subtract term

h = - 16 (0.5)^{2} + 16

CalcPow

Calculate power

h = - 16 (0.25) + 16

Multiply

h = - 4 + 16

AddTerms

Add terms

h = 12

The height at which the acrobats did the high-five is

12

feet.

Discussion

Solving Nonlinear Systems Using Substitution

In a similar way that a system of linear equations can be solved using the substitution method, there are nonlinear systems that can be solved by substituting. As an example, consider the following linear-quadratic system.

{y = x^{2} + 2 x - 19 y = 5 x - 1 (I) (II)

This system is solved using substitution following the steps below.

Finding a Term to Substitute

The first step to solve a nonlinear system using substitution is to identify which term is substituted from one equation to the other. The given system has the

y

variable already isolated so it is easy to select that term.

{y = x^{2} + 2 x - 19 y = 5 x - 1 (I) (II)

Substituting a Term

Now the value of

y

from Equation (II) is substituted into Equation (I). Then, the resulting equation is simplified as much as possible.

{y = x^{2} + 2 x - 19 y = 5 x - 1

Substitute

$(I):$ $y = 5 x - 1$

{5 x - 1 = x^{2} + 2 x - 19 y = 5 x - 1

SubEqn

$(I):$ $LHS - 5 x = RHS - 5 x$

{- 1 = x^{2} - 3 x - 19 y = 5 x - 1

AddEqn

$(I):$ $LHS + 1 = RHS + 1$

{0 = x^{2} - 3 x - 18 y = 5 x - 1

RearrangeEqn

$(I):$ Rearrange equation

{x^{2} - 3 x - 18 = 0 y = 5 x - 1

A quadratic equation that only depends on

x

was obtained.

Solving the Resulting Equation

A quadratic equation was obtained from the previous step. This equation is written in standard form.

a x^{2} + b x + c = 0 ⇓ 1 x^{2} + (- 3) x + (- 18) = 0

These values can be substituted into the quadratic formula.

x = \frac{- b \pm b ^{2} - 4 a c}{2 a}

SubstituteValues

Substitute values

x = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 ( 1 ) ( - 1 8 )}{2 ( 1 )}

CalcPow

Calculate power

x = \frac{- ( - 3 ) \pm 9 - 4 ( 1 ) ( - 1 8 )}{2 ( 1 )}

Multiply

x = \frac{- ( - 3 ) \pm 9 + 7 2}{2}

NegNeg

$- (- a) = a$

x = \frac{3 \pm 9 + 7 2}{2}

AddTerms

Add terms

x = \frac{3 \pm 8 1}{2}

CalcRoot

Calculate root

x = \frac{3 \pm 9}{2}

There are two possible values for

x

depending if there is a subtraction or an addition. These values are calculated individually.

$x = \frac{3 \pm 9}{2}$
$x_{1} = \frac{3 + 9}{2}$	$x_{2} = \frac{3 - 9}{2}$
$x_{1} = \frac{1 2}{2}$	$x_{2} = \frac{- 6}{2}$
$x_{1} = 6$	$x_{2} = - 3$

Substituting the Values

To find the values of $y,$ the values of $x_{1}$ and $x_{2}$ are substituted into either equation of the system. It is easier to substitute the values into Equation (II) because it is a linear equation, instead of quadratic.

$y = 5 x - 1$
$y_{1} = 5 (6) - 1$	$y_{2} = 5 (- 3) - 1$
$y_{1} = 29$	$y_{2} = - 16$

This indicates that the solutions of the system are the points $(6, 29)$ and $(- 3, - 16) .$

Example

Skydive Recording

Finally, all the math and practice is making sense. Emily is now ready for her first jump! Instructor Sky Flyer prepared an equation to record Emily's height above the ground.

Her height from the ground when she jumps from the plane, measured in feet, is given by the following quadratic equation.

h (t) = - 16 t^{2} + 100 t + 10000

Again,

h

is Emily's height from the floor, while

t

is the time in seconds since Emily jumped. Using a linear equation, the instructor calculated Emily's height above the ground after her parachute was released.

h (t) = - 16 t + 6700

a After how many seconds does Emily release her parachute? Use substitution to find the answer. Round the time to two decimal places.

b At what height does Emily release her parachute? Round the height to two decimal places.

Hint

a Which variable is most convenient to isolate?

b Substitute the answer from Part A into either equation.

Solution

a When Emily releases her parachute, both heights have to be equal. Therefore, to find the time is the same as finding the solution of the following nonlinear system.

{h = - 16 t^{2} + 100 t + 10000 h = - 16 t + 6700 (I) (II)

The first step to solve the system using substitution is to find a variable to isolate in one equation. In this case, the variable

h

is already isolated in both equations, so it is a convenient choice.

{h = - 16 t^{2} + 100 t + 10000 h = - 16 t + 6700 (I) (II)

The value of

h

from Equation (II) can be substituted into Equation (I).

{h = - 16 t^{2} + 100 t + 10000 h = - 16 t + 6700 (I) (II)

Substitute

$(I):$ $h = - 16 t + 6700$

{- 16 t + 6700 = - 16 t^{2} + 100 t + 10000 h = - 16 t + 6700

SubEqn

$(I):$ $LHS - - 16 t + 6700 = RHS - - 16 t + 6700$

{0 = - 16 t^{2} + 100 t + 10000 - (- 16 t + 6700) h = - 16 t + 6700

SubTerms

$(I):$ Subtract terms

{0 = - 16 t^{2} + 116 t + 3300 h = - 16 t + 6700

The quadratic equation obtained only depends on

t .

This quadratic equation is already written in standard form.

a t^{2} + b t + c = 0 ⇓ - 16 t^{2} + 116 t + 3300 = 0

This equation can be solved using the quadratic formula.

t = \frac{- b \pm b ^{2} - 4 a c}{2 a}

SubstituteValues

Substitute values

t = \frac{- 1 1 6 \pm 1 1 6 ^{2} - 4 ( - 1 6 ) ( 3 3 0 0 )}{2 ( - 1 6 )}

CalcPow

Calculate power

t = \frac{- 1 1 6 \pm 1 3 4 5 6 - 4 ( - 1 6 ) ( 3 3 0 0 )}{2 ( - 1 6 )}

Multiply

t = \frac{- 1 1 6 \pm 1 3 4 5 6 + 2 1 1 2 0 0}{- 3 2}

AddTerms

Add terms

t = \frac{- 1 1 6 \pm 2 2 4 6 5 6}{- 3 2}

There are two possible values for the time

t .

$t = \frac{- 1 1 6 \pm 2 2 4 6 5 6}{- 3 2}$
$t = \frac{- 1 1 6 + 2 2 4 6 5 6}{- 3 2}$	$t = \frac{- 1 1 6 - 2 2 4 6 5 6}{- 3 2}$
$t \approx - 11.17$	$t \approx 18.44$

The time after Emily jumps has to be positive. Because of this, the only solution that makes sense is the positive one. Therefore, Emily released her parachute about $18.44$ seconds after she jumped.

b To find the height at which Emily released her parachute, the value from Part A can be substituted into either equation. In this case, the value will be substituted into Equation (II) because it is linear. Since the value of

t

is rounded, the equality is substituted with an approximation symbol.

h = - 16 t + 6700

Substitute

$t = 18.44$

h \approx - 16 \cdot 18.44 + 6700

AddTerms

Add terms

h \approx 6404.96

Therefore, the height at which Emily released her parachute is about

6404.96

feet.

Closure

Finding the Intersections with a Circle

In this lesson's challenge, an equation of a circle centered at the origin with a radius of

22

was given.

x^{2} + y^{2} = 8

The challenge was to find the points of intersection of that circle.

a Given the line

y = \frac{1}{2} x - 1,

at what points does it intersect with the circle? Round the answer to one decimal place if needed.

b Given the parabola

y = \frac{1}{2} x^{2},

determine at what points it intersects with the circle. Round the answer to one decimal place if needed.

Hint

a Consider using substitution.

b Isolate the

x^{2}

terms on both equations.

Solution

a The method used to find the points of intersection of the circle and the line is the same as finding the solution of the nonlinear system made by combining these equations.