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

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

y = a x^{2} + b x + c

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

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

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

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

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

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

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

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

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

y = 5 x - 1

y_{1} = 5 (6) - 1

y_{2} = 5 (- 3) - 1

y_{1} = 29

y_{2} = - 16

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

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

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.

⎩ ⎪ ⎨ ⎪ ⎧ x^{2} + y^{2} = 8 y = \frac{1}{2} x - 1 (I) (II)

The

y

variable is already isolated in Equation (II), so this can be substituted into the

y

term in Equation (I).

⎩ ⎪ ⎨ ⎪ ⎧ x^{2} + y^{2} = 8 y = \frac{1}{2} x - 1 (I) (II)

Substitute

$(I):$ $y = \frac{1}{2} x - 1$

⎩ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎧ x^{2} + (\frac{1}{2} x - 1)^{2} = 8 y = \frac{1}{2} x - 1

ExpandNegPerfectSquare

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

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ x^{2} + \frac{1}{4} x^{2} - x + 1 = 8 y = \frac{1}{2} x - 1

AddTerms

$(I):$ Add terms

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ \frac{5}{4} x^{2} - x + 1 = 8 y = \frac{1}{2} x - 1

SubTerms

$(I):$ Subtract terms

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ \frac{5}{4} x^{2} - x - 7 = 0 y = \frac{1}{2} x - 1

It can be seen that the obtained quadratic equation is written in standard form.

a x^{2} + b x + c = 0 ⇓ \frac{5}{4} x^{2} + (- 1) x + (- 7) = 0

This equation can then be solved using the quadratic formula.

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

SubstituteValues

Substitute values

x = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 ( \frac{5}{4} ) ( - 7 )}{2 ( \frac{5}{4} )}

▼

Simplify right-hand side

CalcPow

Calculate power

x = \frac{- ( - 1 ) \pm 1 - 4 ( \frac{5}{4} ) ( - 7 )}{2 ( \frac{5}{4} )}

Multiply

x = \frac{- ( - 1 ) \pm 1 + 3 5}{\frac{5}{2}}

NegNeg

$- (- a) = a$

x = \frac{1 \pm 1 + 3 5}{\frac{5}{2}}

AddTerms

Add terms

x = \frac{1 \pm 3 6}{\frac{5}{2}}

CalcRoot

Calculate root

x = \frac{1 \pm 6}{\frac{5}{2}}

ExpandFrac

$\frac{a}{b} = \frac{a \cdot 2}{b \cdot 2}$

x = \frac{2 \pm 1 2}{5}

There are two possible values for

x .

These values change if adding or subtracting.

$x = \frac{2 \pm 1 2}{5}$
$x = \frac{2 + 1 2}{5}$	$x = \frac{2 - 1 2}{5}$
$2.8$	$- 2$

Finally, these values are substituted into Equation (II) to find the values of $y$ of the points of intersection.

$y = \frac{1}{2} x - 1$
$y = \frac{1}{2} (2.8) - 1$	$y = \frac{1}{2} (- 2) - 1$
$0.4$	$- 2$

The points of intersection are $(2.8, 0.4)$ and $(- 2, - 2) .$

b Finding the points of intersection between the circle and a parabola is achieved by finding the solutions of the nonlinear system.

⎩ ⎪ ⎨ ⎪ ⎧ x^{2} + y^{2} = 8 y = \frac{1}{2} x^{2} (I) (II)

The

x^{2}

term can be isolated on both equations.

⎩ ⎪ ⎨ ⎪ ⎧ x^{2} + y^{2} = 8 y = \frac{1}{2} x^{2} (I) (II)

SubEqn

$(I):$ $LHS - y^{2} = RHS - y^{2}$

⎩ ⎪ ⎨ ⎪ ⎧ x^{2} = 8 - y^{2} y = \frac{1}{2} x^{2}

RearrangeEqn

$(II):$ Rearrange equation

⎩ ⎪ ⎨ ⎪ ⎧ x^{2} = 8 - y^{2} \frac{1}{2} x^{2} = y

MultEqn

$(II):$ $LHS \cdot 2 = RHS \cdot 2$

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

The Equation (II) can be subtracted from Equation (I) to solve the system using elimination.

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

SysEqnSub

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

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

SubTerms

$(I):$ Subtract terms

{0 = 8 - y^{2} - 2 y x^{2} = 2 y

RearrangeEqn

$(I):$ Rearrange equation

{- y^{2} - 2 y + 8 = 0 x^{2} = 2 y

A quadratic equation of

y

is obtained. It can be noted that the equation is written in standard form.

a x^{2} + b x + c = 0 ⇓ (- 1) y^{2} + (- 2) y + 8 = 0

This equation can be solved using the quadratic formula.

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

SubstituteValues

Substitute values

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

CalcPow

Calculate power

y = \frac{- ( - 2 ) \pm 4 - 4 ( - 1 ) ( 8 )}{2 ( - 1 )}

Multiply

y = \frac{- ( - 2 ) \pm 4 + 3 2}{- 2}

NegNeg

$- (- a) = a$

y = \frac{2 \pm 4 + 3 2}{- 2}

AddTerms

Add terms

y = \frac{2 \pm 3 6}{- 2}

CalcRoot

Calculate root

y = \frac{2 \pm 6}{- 2}

Now the two possible values for

y

can be found by adding or subtracting.

$y = \frac{2 \pm 6}{- 2}$
$y = \frac{2 + 6}{- 2}$	$y = \frac{2 - 6}{- 2}$
$- 4$	$2$

Now two values of

y

were obtained, and it should be noted that the value of

y

in Equation (II) has to be greater than or equal to

0 .

Therefore, the valid solution is

2,

while

- 4

is an extraneous solution. The value of

2

can be substituted into either equation to find the possible values of

x .

In this case, it will be substituted into Equation (II).

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

Substitute

$y = 2$

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

MultEqn

$LHS \cdot 2 = RHS \cdot 2$

4 = x^{2}

RearrangeEqn

Rearrange equation

x^{2} = 4

CalcRoot

Calculate root

x = \pm 2

Here,

x

can be either

2

- 2

because the value of

x

is not limited by any equation. Therefore, the points of intersection are

(2, 2)

and

(- 2, 2) .

Solving Systems of Equations Including Quadratics

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

Acrobats on the Circus

Hint

Solution

Solving Nonlinear Systems Using Substitution

Skydive Recording

Hint

Solution

Planning New Formations

Hint

Solution

Finding the Intersections with a Circle

Hint

Solution

Solving Systems of Equations Including Quadratics

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