Unlocking Solutions: Solving Systems of Linear and Quadratic Equations

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

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

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.

$y = a x^{2} + b x + c$
$y = \frac{1}{6} x^{2} + 2 x + 5$	$y = - \frac{1}{2} x + (- 2) x + 5$

The graphs of both quadratic equations are parabolas. The values of

a

and

b

can be used to find the axis of symmetry of each graph. To do so, the values are substituted into a formula. This can be done for Equation (I).

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

SubstituteII

$a = \frac{1}{6}$ , $b = 2$

x = - \frac{2}{2 ( \frac{1}{6} )}

▼

Solve for

x

CancelCommonFac

Cancel out common factors

x = - \frac{2}{2 ( \frac{1}{6} )}

SimpQuot

Simplify quotient

x = - \frac{1}{\frac{1}{6}}

DivOneByFracD

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

x = - 6

This indicates that the axis of symmetry of the parabola of Equation (I) is the vertical line

x = - 6 .

To find the vertex of the parabola, the value

x = - 6

is substituted into Equation (I).

y = \frac{1}{6} x^{2} + 2 x + 5

Substitute

$x = - 6$

y = \frac{1}{6} (- 6)^{2} + 2 (- 6) + 5

▼

Solve for

y

CalcPow

Calculate power

y = \frac{1}{6} (36) + 2 (- 6) + 5

Multiply

y = \frac{1}{6} (36) - 12 + 5

MoveRightFacToNum

$\frac{a}{c} \cdot b = \frac{a \cdot b}{c}$

y = \frac{3 6}{6} - 12 + 5

CalcQuot

Calculate quotient

y = 6 - 12 + 5

AddTerms

Add terms

y = - 1

This indicates that the vertex of the first parabola lies in the point

(- 6, - 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 = - \frac{b}{2 a}

SubstituteII

$a = - \frac{1}{2}$ , $b = - 2$

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

▼

Solve for

x

MultPosNeg

$a (- b) = - a \cdot b$

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

MultFracByInverse

$\frac{a}{b} \cdot \frac{b}{a} = 1$

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

CalcQuot

Calculate quotient

x = - 2

This indicates that the axis of symmetry is the vertical line at

x = - 2 .

Also, to find the vertex of the second parabola,

x = - 2

is substituted into Equation (II).

y = - \frac{1}{2} x^{2} - 2 x + 5

Substitute

$x = - 2$

y = - \frac{1}{2} (- 2)^{2} - 2 (- 2) + 5

▼

Solve for

y

CalcPow

Calculate power

y = - \frac{1}{2} (4) - 2 (- 2) + 5

CalcQuot

Calculate quotient

y = - 2 - 2 (- 2) + 5

Multiply

y = - 2 + 4 + 5

AddTerms

Add terms

y = 7

Now it is known that the vertex of the second parabola lies at

(- 2, 7) .

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.

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

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

$x + 4 = y$
$- 1 + 4 = y$	$4 + 4 = y$
$y = 3$	$y = 8$

x + 4 = y

- 1 + 4 = y

4 + 4 = y

y = 3

y = 8

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Catch-Up and Review

Finding Intersections with a Circle

Nonlinear System

Is the System Nonlinear?

The Solutions of a Linear-Quadratic System

System of Quadratic Equations

Is the System Quadratic?

The Solutions of a Quadratic-Quadratic System

Solving Nonlinear Systems Graphically

Graphing a Linear-Quadratic System

Answer

Hint

Solution

Graphing a Quadratic-Quadratic System

Answer

Hint

Solution

Solving Nonlinear Systems Using Elimination

Skydiving

Hint

Solution

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