8. enrichment given that the points a (5,7), b (8, 2) and c (1, 2) are the vertices of a triangle abc, calculate the length of altitude ah from vertex a to bc. 9. enrichment calculate the altitude ah in the triangle below. 10. at a local concert, two types of tickets were sold: one for $40 and the other for $60. the sale of 854 tickets generated an amount of $41,240. how many of each kind of tickets were sold? 11. find the solution to the following systems of linear equations by using the elimination method. $\begin{cases}9x + 4y = 9 \\ 8x + 3y = 8\end{cases}$ 12. find the solution to the following systems of linear equations by using the elimination method. $\begin{cases}3x + 5y = 17 \\ 2x + 9y = 17\end{cases}$ 13. find the solution to the following systems of linear equations by using the elimination method. $\begin{cases}3x - 2y = -1 \\ x + 6y = 27\end{cases}$

Question

Accepted Answer

### Question 10 Solution:
# Explanation:
## Step1: Define Variables
Let $ x $ be the number of \$40 tickets and $ y $ be the number of \$60 tickets.
We have two equations:
1. $ x + y = 854 $ (total number of tickets)
2. $ 40x + 60y = 41240 $ (total revenue)

## Step2: Simplify the Second Equation
Divide the second equation by 20: $ 2x + 3y = 2062 $

## Step3: Solve the First Equation for $ x $
$ x = 854 - y $

## Step4: Substitute $ x $ into the Simplified Second Equation
$ 2(854 - y) + 3y = 2062 $
$ 1708 - 2y + 3y = 2062 $
$ y = 2062 - 1708 = 354 $

## Step5: Find $ x $
$ x = 854 - 354 = 500 $

# Answer:
500 tickets of \$40 and 354 tickets of \$60 were sold.

### Question 11 Solution:
# Explanation:
## Step1: Label the Equations
Equation (1): $ 9x + 4y = 9 $
Equation (2): $ 8x + 3y = 8 $

## Step2: Eliminate $ y $
Multiply Equation (1) by 3: $ 27x + 12y = 27 $ (Equation 3)
Multiply Equation (2) by 4: $ 32x + 12y = 32 $ (Equation 4)
Subtract Equation 3 from Equation 4: $ 5x = 5 $ ⇒ $ x = 1 $

## Step3: Substitute $ x = 1 $ into Equation (1)
$ 9(1) + 4y = 9 $ ⇒ $ 4y = 0 $ ⇒ $ y = 0 $

# Answer:
$ x = 1 $, $ y = 0 $

### Question 12 Solution:
# Explanation:
## Step1: Label the Equations
Equation (1): $ 3x + 5y = 17 $
Equation (2): $ 2x + 9y = 17 $

## Step2: Eliminate $ x $
Multiply Equation (1) by 2: $ 6x + 10y = 34 $ (Equation 3)
Multiply Equation (2) by 3: $ 6x + 27y = 51 $ (Equation 4)
Subtract Equation 3 from Equation 4: $ 17y = 17 $ ⇒ $ y = 1 $

## Step3: Substitute $ y = 1 $ into Equation (1)
$ 3x + 5(1) = 17 $ ⇒ $ 3x = 12 $ ⇒ $ x = 4 $

# Answer:
$ x = 4 $, $ y = 1 $

### Question 13 Solution:
# Explanation:
## Step1: Label the Equations
Equation (1): $ 3x - 2y = -1 $
Equation (2): $ x + 6y = 27 $

## Step2: Eliminate $ x $
Multiply Equation (2) by 3: $ 3x + 18y = 81 $ (Equation 3)
Subtract Equation (1) from Equation 3: $ 20y = 82 $ ⇒ $ y = \frac{41}{10} = 4.1 $? Wait, no, let's recalculate:

Wait, Equation 3: $ 3x + 18y = 81 $
Equation 1: $ 3x - 2y = -1 $
Subtract Equation 1 from Equation 3: $ 20y = 82 $? No, 81 - (-1) = 82? Wait, 81 - (-1) is 82? Wait, 3x + 18y - (3x - 2y) = 81 - (-1) ⇒ 20y = 82 ⇒ $ y = \frac{82}{20} = \frac{41}{10} = 4.1 $? Wait, no, let's check again.

Wait, Equation (2) multiplied by 3: $ 3x + 18y = 81 $
Equation (1): $ 3x - 2y = -1 $
Subtract Equation (1) from Equation (3): $ (3x + 18y) - (3x - 2y) = 81 - (-1) $ ⇒ $ 20y = 82 $ ⇒ $ y = \frac{82}{20} = \frac{41}{10} = 4.1 $? Wait, that can't be. Wait, maybe I made a mistake. Let's do it again.

Wait, Equation (2): $ x + 6y = 27 $ ⇒ $ x = 27 - 6y $
Substitute into Equation (1): $ 3(27 - 6y) - 2y = -1 $ ⇒ $ 81 - 18y - 2y = -1 $ ⇒ $ 81 - 20y = -1 $ ⇒ $ -20y = -82 $ ⇒ $ y = \frac{82}{20} = \frac{41}{10} = 4.1 $? Wait, no, 81 +1 = 82, so 20y = 82 ⇒ y = 4.1? Wait, but let's check with elimination again.

Wait, maybe I messed up the multiplication. Let's try to eliminate y instead.

Equation (1): $ 3x - 2y = -1 $ multiply by 3: $ 9x - 6y = -3 $ (Equation 3)
Equation (2): $ x + 6y = 27 $ (Equation 2)
Add Equation 3 and Equation 2: $ 10x = 24 $ ⇒ $ x = \frac{24}{10} = \frac{12}{5} = 2.4 $
Then substitute x into Equation (2): $ \frac{12}{5} + 6y = 27 $ ⇒ $ 6y = 27 - \frac{12}{5} = \frac{135 - 12}{5} = \frac{123}{5} $ ⇒ $ y = \frac{123}{30} = \frac{41}{10} = 4.1 $. Wait, so that's correct. So x = 12/5, y = 41/10.

Wait, but let's check with the original equations.

Equation (1): 3*(12/5) - 2*(41/10) = 36/5 - 82/10 = 72/10 - 82/10 = -10/10 = -1. Correct.

Equation (2): 12/5 + 6*(41/10) = 12/5 + 246/10 = 24/10 + 246/10 = 270/10 = 27. Correct. So the solution is x = 12/5, y = 41/10.

Wait, but let's do elimination properly.

Equation (1): 3x - 2y = -1

Equation (2): x + 6y = 27

Multiply Equation (1) by 3: 9x - 6y = -3 (Equation 3)

Add Equation (3) and Equation (2): 10x = 24 ⇒ x = 24/10 = 12/5 = 2.4

Then substitute x = 12/5 into Equation (2): 12/5 + 6y = 27 ⇒ 6y = 27 - 12/5 = (135 - 12)/5 = 123/5 ⇒ y = 123/(5*6) = 123/30 = 41/10 = 4.1

So the solution is x = 12/5, y = 41/10.

# Explanation:
## Step1: Label the Equations
Equation (1): $ 3x - 2y = -1 $
Equation (2): $ x + 6y = 27 $

## Step2: Eliminate $ y $
Multiply Equation (1) by 3: $ 9x - 6y = -3 $ (Equation 3)
Add Equation (3) and Equation (2): $ 10x = 24 $ ⇒ $ x = \frac{12}{5} $

## Step3: Substitute $ x = \frac{12}{5} $ into Equation (2)
$ \frac{12}{5} + 6y = 27 $ ⇒ $ 6y = 27 - \frac{12}{5} = \frac{123}{5} $ ⇒ $ y = \frac{41}{10} $

# Answer:
$ x = \frac{12}{5} $, $ y = \frac{41}{10} $ (or $ x = 2.4 $, $ y = 4.1 $)

### Question 8 Solution:
# Explanation:
## Step1: Find the Slope of BC
Points B(8,2) and C(1,2). Since the y-coordinates are equal, BC is horizontal. The equation of BC is $ y = 2 $.

## Step2: Find the Length of Altitude AH
The altitude from A(5,7) to BC is vertical? No, since BC is horizontal, the altitude is vertical? Wait, no. The altitude from A to BC is the vertical distance from A to BC because BC is horizontal. The y-coordinate of BC is 2, and the y-coordinate of A is 7. So the length is $ |7 - 2| = 5 $.

# Answer:
5

### Question 9 Solution:
# Explanation:
## Step1: Find the Slope of BC
Points B(9,3) and C(1,1). Slope $ m = \frac{3 - 1}{9 - 1} = \frac{2}{8} = \frac{1}{4} $.

## Step2: Equation of BC
Using point C(1,1): $ y - 1 = \frac{1}{4}(x - 1) $ ⇒ $ y = \frac{1}{4}x + \frac{3}{4} $.

## Step3: Slope of AH (Perpendicular to BC)
Slope of AH is $ -4 $ (negative reciprocal of $ \frac{1}{4} $).

## Step4: Equation of AH
Using point A(4,9): $ y - 9 = -4(x - 4) $ ⇒ $ y = -4x + 25 $.

## Step5: Find Intersection H of AH and BC
Solve $ \frac{1}{4}x + \frac{3}{4} = -4x + 25 $
Multiply by 4: $ x + 3 = -16x + 100 $ ⇒ $ 17x = 97 $ ⇒ $ x = \frac{97}{17} $? Wait, no, that can't be. Wait, maybe I made a mistake. Wait, looking at the graph, H is on BC, and A is (4,9). Let's check the coordinates of H. From the graph, BC goes from (1,1) to (9,3). The x-coordinate of H is around 6? Wait, maybe the altitude is vertical? No, wait, the graph shows A(4,9), B(9,3), C(1,1). Let's calculate the length of AH using the formula for distance from a point to a line.

The equation of BC: points B(9,3) and C(1,1). The slope is (3-1)/(9-1)=2/8=1/4. So equation is y - 1 = (1/4)(x - 1) ⇒ y = (1/4)x + 3/4.

The distance from A(4,9) to line BC: $ d = \frac{| \frac{1}{4}(4) - 9 + \frac{3}{4} |}{\sqrt{(\frac{1}{4})^2 + (-1)^2}} $

Calculate numerator: $ |1 - 9 + 0.75| = |-7.25| = 7.25 = \frac{29}{4} $

Denominator: $ \sqrt{\frac{1}{16} + 1} = \sqrt{\frac{17}{16}} = \frac{\sqrt{17}}{4} $

So distance $ d = \frac{\frac{29}{4}}{\frac{\sqrt{17}}{4}} = \frac{29}{\sqrt{17}} $? No, that's not right. Wait, maybe the graph is a grid, so let's count the vertical distance? Wait, A is (4,9), and BC is at y=1 to y=3. Wait, no, the y-coordinate of A is 9, and BC is between y=1 and y=3. Wait, maybe I misread the graph. Wait, the graph has A at (4,9), B at (9,3), C at (1,1). Let's calculate the length of AH by finding the vertical distance? No, because BC is not horizontal. Wait, maybe the problem is simpler. Let's use the formula for the length of the altitude: area of triangle divided by base length.

First, calculate the length of BC: distance between (1,1) and (9,3): $ \sqrt{(9-1)^2 + (3-1)^2} = \sqrt{64 + 4} = \sqrt{68} = 2\sqrt{17} $.

Area of triangle ABC: using coordinates A(4,9), B(9,3), C(1,1).

Area = $ \frac{1}{2} | (4(3 - 1) + 9(1 - 9) + 1(9 - 3)) | = \frac{1}{2} | 4(2) + 9(-8) + 1(6) | = \frac{1}{2} | 8 - 72 + 6 | = \frac{1}{2} | -58 | = 29 $.

Then altitude AH = $ \frac{2 \times \text{Area}}{\text{Length of BC}} = \frac{2 \times 29}{2\sqrt{17}} = \frac{29}{\sqrt{17}} $? No, that's not matching the graph. Wait, maybe the graph is drawn with integer coordinates for H. Looking at the graph, H is at (6,2)? Wait, no, the y-coordinate of BC is from 1 to 3. Wait, A is (4,9), and the vertical distance from A to BC: the y-coordinate of BC at x=4 is $ y = \frac{1}{4}(4) + \frac{3}{4} = \frac{7}{4} = 1.75 $. So the vertical distance is 9 - 1.75 = 7.25, which is 29/4. But the graph shows H at (6,2)? Wait, maybe the problem is simpler: since BC is from (1,1) to (9,3), the equation of BC is y = 0.25x + 0.75. At x=4, y=0.25*4 + 0.75=1.75. So the altitude length is 9 - 1.75=7.25=29/4. But maybe the graph is approximate. Wait, the problem says "Calculate the altitude AH". Maybe using the distance formula between A(4,9) and H. From the graph, H is at (6,2)? No, (6,2) is on BC? Let's check: (6,2) in BC: y = 0.25*6 + 0.75=1.5 + 0.75=2.25≠2. So no. Wait, maybe the slope of BC is (3-1)/(9-1)=2/8=1/4, so the equation is y= (1/4)x + 3/4. Then AH is perpendicular, so slope -4, equation y= -4x + 25. Intersection with BC: (1/4)x + 3/4 = -4x

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QUESTION IMAGE

Question

Explanation:

Question 10 Solution:

Step1: Define Variables

Step2: Simplify the Second Equation

Step3: Solve the First Equation for \( x \)

Step4: Substitute \( x \) into the Simplified Second Equation

Step5: Find \( x \)

Step1: Label the Equations

Step2: Eliminate \( y \)

Step3: Substitute \( x = 1 \) into Equation (1)

Step1: Label the Equations

Step2: Eliminate \( x \)

Step3: Substitute \( y = 1 \) into Equation (1)

Answer:

Question 11 Solution: