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

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

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

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

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

2.8

- 2

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

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

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

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

0.4

- 2

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

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

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

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

- 4

2

Solve the nonlinear system by graphing.

{y = 2 x^{2} - 8 x - 3 y = x + 2 (I) (II)

Select the solution or solutions to the system of equations.

We are asked to solve the nonlinear system by graphing. This means we need to graph both equations in the system and find their points of intersection, which will be the solutions to the system. Let's first graph Equation I.

Equation I

Let's write the quadratic equation in standard form to identify the values of a, b, and c. y=2x^2-8x-3 ⇓ y= 2x^2+( -8)x+( -3) Next, we can determine the axis of symmetry of the parabola. To do so, we will substitute a= 2 and b= -8 into the formula for the axis of a parabola.

The axis of symmetry is the vertical line x=2. Let's graph it on a coordinate plane.

The vertex of the parabola can be determined by evaluating Equation I when x= 2.

This indicates that the vertex of the parabola is at the point (2,-11).

In a quadratic equation in standard form, the value of c indicates the y-intercept. In this case, c= -3. Because the axis of symmetry divides the graph into mirror images, the points (0, -3) and (4, -3) are on the graph of the parabola.

We can now connect these points with a smooth curve to obtain the graph of the parabola.

Equation II

Equation II is a linear equation with slope 1 and y-intecept 2. We can use this information to add the graph of Equation II to our diagram.

Finding the Solutions

We can now examine the graphs of the equations to locate the points of intersection.

Therefore, the solutions to the nonlinear system are (-0.5,1.5) and (5,7).

Use the Substitution Method to solve the nonlinear system of equations.

{y = - 3 x^{2} + 6 x + 9 y = 6 x - 3 (I) (II)

Select the solution or solutions to the system of equations.

We will determine the solutions to the nonlinear system by using the Substitution Method. y=-3x^2+6x+9 & (I) y=6x-3 & (II) We can note that the y-variable is isolated in Equation II. This means we can substitute its value, 6x-3, for y in Equation I.

We have obtained a simple quadratic equation in Equation I. -3x^2+12=0 ⇔ 3x^2=12 This means that we can apply square roots to find the values of x.

We can now substitute the values of x into either equation and simplify to find the corresponding values of y. Let's use Equation II since it is a simpler equation. y=6x-3 We will first substitute x= 2.

We have found that y=9 when x=2. This means that one solution to the system is the point (2,9). Now we will evaluate Equation II when x= -2.

When x=-2, the value of y is -15. This means that (-2,-15) is also a solution to the system. Therefore, the solutions to the system are (2,9) and (-2,-15).

Use the Elimination Method to determine the solution or solutions to the nonlinear system.

{y = 3 x + 7.8 y = 5 x^{2} + 3 x - 2 (I) (II)

We will use the Elimination Method to solve the nonlinear system of equations. To do so, one of the variable terms needs to be eliminated by adding or subtracting one equation from the other equation. This means that either the x-terms or the y-terms must cancel each other out. y=3x+7.8 & (I) y=5x^2+3x-2 & (II) Note that if we subtract Equation I from Equation II, we will eliminate the y-terms.

We have written a simple quadratic equation for Equation II. 5x^2-9.8=0 ⇔ 5x^2=9.8 We can find the value or values of x by isolating the x-variable and applying square roots.

We can now use Equation I to find the corresponding y-value for each value of x. Let's first substitute 1.4 for x.

We have found that y=12 when x=1.4. This means that one solution to the system is the point (1.4,12). Now we will substitute -1.4 for x and simplify.

When x=-1.4, the value of y is 3.6. This means that (-1.4,3.6) is also a solution to the equation. Therefore, the solutions to the system are (1.4,12) and (-1.4,3.6).

Determine the values of

A

and

B

such that one of the solutions to the nonlinear system is

(2, - 10) .

{y = A x^{2} + B x - 2 y = - \frac{5}{3} B x + 10 (I) (II)

We will begin by substituting the given solution to the nonlinear system, x=2 and y=-10, into the system of equations and simplifying as much as possible.

We can see that our nonlinear system has turned out into a new system of linear equations with only two unknowns, A and B. Furthermore, we only have the B-variable in Equation II. This means we can determine the value of B by solving Equation II.

We can now substitute the value of B back into Equation I to find the value of A.

These values allow the nonlinear system to have (2,-10) as one of its solutions. Let's substitute these values into the original system to write the complete equations.

We have determined the values of A and B that make (2,-10) a solution to the given system of equations. We can use the simplified form of the resulting system to find any other solutions to the system. To do so, we will substitute the value of y from Equation II into Equation I.

We can see that the quadratic equation obtained only depends on x. This equation is already in standard form. ax^2+bx+c=0 ⇓ -5x^2+ 16x+( -12)=0 We can substitute these values into the quadratic formula.

There are two possible values for x. Let's simplify to find them.

x = -16 ± 4/-10
x=-16+4/-10	x=-16-4/-10
x=-12/-10	x=-20/-10
x=1.2	x=2

We can now substitute these values of x into Equation II to find the solutions to the system. However, we already know that (2,-10) is a solution to the system. This means that we only need to find the value of y when x=1.2. Let's do it!

We now know the two solutions to the system, (1.2,-2) and (2,-10).

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

Equation I

Equation II

Finding the Solutions

Solving Systems of Equations Including Quadratics

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	18 Theory slides
	9 Exercises - Grade E - A
	Each lesson is meant to take 1-2 classroom sessions