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

Ignacio and his brother are flying kites at the beach. The path of Ignacio's kite can be modeled by the equation $y = - x^{2} - 4 x + 12$ and his brother kite's path by the equation $y = 4 x + 28 .$ Determine the point where the paths of the kites meet.

We will find the point where the brothers' kites meet. To do so, we will write a system of nonlinear equations using the given models. y=- x^2-4x+12 & (I) y=4x+28 & (II) If this system has a solution, we can state that the kite paths cross. If there are no solutions, we know that the paths of the kites do not cross. To determine if the system has a solution, we will solve the system using the Substitution Method. Let's do it by substituting 4x+28 for y in Equation I.

Note that Equation I has only the x-variable. Additionally, it is a perfect square trinomial. This means there is only one solution for this equation and the paths of the kites cross each other at one point. Let's factor the trinomial.

We have determined the x-coordinate of the point where the paths cross. To find its y-coordinate, we can substitute this value into either of the equations. Let's substitute x=-4 into Equation II.

This means that the paths of the kites cross each other at (-4,12).

The daily number of customers

y

at Restaurant A is given by the following function.

y = 0.5 x^{2} - 7 x + 70

In this equation,

x

is the number of days since the beginning of the month. The number of customers for Restaurant B is modeled by a linear function. On days

10

and

14,

both restaurants have the same number of customers. What is the function that models the number of customers for Restaurant B?

We are asked to find a linear function that models the daily number of customers at Restaurant B. To do so, we will begin by finding the number of customers the restaurants have on days 10 and 14. This information can be used to determine the equation for the linear function.

Number of Customers on Days 10 and 14

We know that the number of customers on days 10 and 14 is the same for both restaurants. We will evaluate the function for Restaurant A to determine the number of customers for the restaurants on these days. Let's first evaluate the function when x= 10.

On day 10, both restaurants have 50 customers. We can write this information as the ordered pair (10,50). We will calculate the number of customers on day 14 in a similar fashion.

On day 14, the number of customers in each restaurant is 70. This can be written as (14,70).

Writing the Function for Restaurant B

We just found two points that are on the graphs of both of the lines. This is because we know the number of customers on days 10 and 14 are equal for both restaurants. ( 10, 50) ( 14, 70) We can use these points to determine the slope of the line that models the customer count at Restaurant B. To do so, we will use the Slope Formula.

We can write a partial equation for the line in the slope-intercept form using this slope. y= 5x+ b We are missing the value of b. To determine the y-intercept b, we can substitute either of the points into the equation and solve it for b. Let's use ( 10, 50).

We can finally complete the equation to obtain the function for Restaurant B. y=5x+ 0 ⇔ y=5x

Use the graph to find the requested information.

Determine the coordinates of point

P .

Determine the coordinates of point

Q .

We can see in the graph that point P is on the y-axis. It is the y-intercept of both functions.

This means that the x-coordinate of point P is 0. Additionally, we can evaluate the quadratic equation when x= 0 to determine the y-coordinate of P. Let's do it!

Therefore, the coordinates of P are (0,2).

Let's now determine the coordinates of Q. To do so, we will first write an equation for the line. Since we know the line passes through (0,2) and (1,1.5), we can find its slope using the Slope Formula.

We can now use the slope-intercept form to write the equation for the line. Recall that we found in the previous part that the y-intercept is b= 2. y= -0.5x+ 2 We can now use this equation and the given equation of the parabola to write a nonlinear system. y=-3x^2+7x+2 & (I) y=-0.5x+2 & (II) Solving this system will give us the points of intersection of the graphs. Let's use the Substitution Method.

Next, we will solve Equation I by factoring it and using the Zero Product Property.

Remember that 0 is the x-coordinate of point P, so the other solution to the equation is the x-coordinate of point Q. We can now substitute x=2.5 into either of the equations to find the y-coordinate of Q. Let's substitute into the linear equation.

Therefore, the coordinates of point Q are (2.5,0.75).

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

Number of Customers on Days 10 and 14

Writing the Function for Restaurant B

Solving Systems of Equations Including Quadratics

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	18 Theory slides
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