in exercises 1–4, solve the system using any method.
1. $y = 2x - 5$
$y = -4x + 1$
2. $y = x - 4$
$x + 2y = 4$
3. $y = \frac{1}{2}x + 1$
$y = -\frac{3}{4}x - 4$
4. $2x - 5y = 29$
$-2x + 7y = -35$

in exercises 5–8, use only the slopes and y-intercepts of the graphs of the equations to determine the number of solutions of the system. explain your reasoning.
5. $y = -6x - 13$
$6x + y = 13$
6. $3x + 4y = -12$
$\frac{3}{4}x + y = -3$
7. $6y = 3x + 39$
$x - 2y = 13$
8. $y = x + 3$
$y = -3x - 5$

9. the total cost for 5 adult tickets and 18 student tickets at a local theater is $98. the total cost was $98. the total cost for 4 adult tickets and 22 student tickets at the theater is $105. what is the cost for one adult ticket and one student ticket?

10. solve the literal equation $5xy - 2 = y + 2x$ for $x$.

Question

in exercises 1–4, solve the system using any method.
1. $y = 2x - 5$
   $y = -4x + 1$
2. $y = x - 4$
   $x + 2y = 4$
3. $y = \frac{1}{2}x + 1$
   $y = -\frac{3}{4}x - 4$
4. $2x - 5y = 29$
   $-2x + 7y = -35$

in exercises 5–8, use only the slopes and y-intercepts of the graphs of the equations to determine the number of solutions of the system. explain your reasoning.
5. $y = -6x - 13$
   $6x + y = 13$
6. $3x + 4y = -12$
   $\frac{3}{4}x + y = -3$
7. $6y = 3x + 39$
   $x - 2y = 13$
8. $y = x + 3$
   $y = -3x - 5$

9. the total cost for 5 adult tickets and 18 student tickets at a local theater is $98. the total cost was $98. the total cost for 4 adult tickets and 22 student tickets at the theater is $105. what is the cost for one adult ticket and one student ticket?

10. solve the literal equation $5xy - 2 = y + 2x$ for $x$.

Accepted Answer

### Problem 1: Solve the system $ \begin{cases} y = 2x - 5 \ y = -4x + 1 \end{cases} $
# Explanation:
## Step 1: Set the equations equal
Since both equal $ y $, set $ 2x - 5 = -4x + 1 $.
## Step 2: Solve for $ x $
Add $ 4x $ to both sides: $ 6x - 5 = 1 $.
Add 5 to both sides: $ 6x = 6 $.
Divide by 6: $ x = 1 $.
## Step 3: Find $ y $
Substitute $ x = 1 $ into $ y = 2x - 5 $: $ y = 2(1) - 5 = -3 $.
# Answer:
$ x = 1 $, $ y = -3 $ (or the ordered pair $ (1, -3) $)

### Problem 2: Solve the system $ \begin{cases} y = x - 4 \ x + 2y = 4 \end{cases} $
# Explanation:
## Step 1: Substitute $ y $ into the second equation
Substitute $ y = x - 4 $ into $ x + 2y = 4 $: $ x + 2(x - 4) = 4 $.
## Step 2: Solve for $ x $
Expand: $ x + 2x - 8 = 4 $.
Combine like terms: $ 3x - 8 = 4 $.
Add 8: $ 3x = 12 $.
Divide by 3: $ x = 4 $.
## Step 3: Find $ y $
Substitute $ x = 4 $ into $ y = x - 4 $: $ y = 4 - 4 = 0 $.
# Answer:
$ x = 4 $, $ y = 0 $ (or the ordered pair $ (4, 0) $)

### Problem 3: Solve the system $ \begin{cases} y = \frac{1}{2}x + 1 \ y = -\frac{3}{4}x - 4 \end{cases} $ (Note: Assumed typo, corrected $ y = -\frac{3}{4}x - 4 $)
# Explanation:
## Step 1: Set the equations equal
$ \frac{1}{2}x + 1 = -\frac{3}{4}x - 4 $.
## Step 2: Solve for $ x $
Multiply all terms by 4 to eliminate fractions: $ 2x + 4 = -3x - 16 $.
Add $ 3x $ to both sides: $ 5x + 4 = -16 $.
Subtract 4: $ 5x = -20 $.
Divide by 5: $ x = -4 $.
## Step 3: Find $ y $
Substitute $ x = -4 $ into $ y = \frac{1}{2}x + 1 $: $ y = \frac{1}{2}(-4) + 1 = -2 + 1 = -1 $.
# Answer:
$ x = -4 $, $ y = -1 $ (or the ordered pair $ (-4, -1) $)

### Problem 4: Solve the system $ \begin{cases} 2x - 5y = 29 \ -2x + 7y = -35 \end{cases} $
# Explanation:
## Step 1: Add the two equations
Add $ 2x - 5y = 29 $ and $ -2x + 7y = -35 $:
$ (2x - 2x) + (-5y + 7y) = 29 - 35 $.
Simplify: $ 2y = -6 $.
## Step 2: Solve for $ y $
Divide by 2: $ y = -3 $.
## Step 3: Find $ x $
Substitute $ y = -3 $ into $ 2x - 5y = 29 $:
$ 2x - 5(-3) = 29 $ → $ 2x + 15 = 29 $.
Subtract 15: $ 2x = 14 $.
Divide by 2: $ x = 7 $.
# Answer:
$ x = 7 $, $ y = -3 $ (or the ordered pair $ (7, -3) $)

### Problem 5: Determine the number of solutions for $ \begin{cases} y = -6x - 13 \ 6x + y = 13 \end{cases} $
# Explanation:
## Step 1: Rewrite the second equation
Rewrite $ 6x + y = 13 $ in slope - intercept form ($ y = mx + b $): $ y = -6x + 13 $.
## Step 2: Compare slopes and y - intercepts
The first equation: $ y = -6x - 13 $ has slope $ m_1=-6 $, y - intercept $ b_1 = - 13 $.
The second equation: $ y = -6x + 13 $ has slope $ m_2=-6 $, y - intercept $ b_2 = 13 $.
Since $ m_1=m_2 $ (parallel lines) and $ b_1
eq b_2 $ (different y - intercepts), the lines are parallel and never intersect.
# Answer:
0 solutions (no solution)

### Problem 6: Determine the number of solutions for $ \begin{cases} 3x + 4y = -12 \ \frac{3}{4}x + y = -3 \end{cases} $
# Explanation:
## Step 1: Rewrite both equations in slope - intercept form
For $ 3x + 4y=-12 $:
Subtract $ 3x $: $ 4y=-3x - 12 $.
Divide by 4: $ y = -\frac{3}{4}x - 3 $.

For $ \frac{3}{4}x + y=-3 $:
Subtract $ \frac{3}{4}x $: $ y = -\frac{3}{4}x - 3 $.

## Step 2: Compare slopes and y - intercepts
Both equations have the same slope ($ m = -\frac{3}{4} $) and the same y - intercept ($ b=-3 $). This means the two equations represent the same line.
# Answer:
Infinitely many solutions

### Problem 7: Determine the number of solutions for $ \begin{cases} 6y = 3x + 39 \ x - 2y = 13 \end{cases} $
# Explanation:
## Step 1: Rewrite both equations in slope - intercept form
For $ 6y = 3x + 39 $:
Divide by 6: $ y=\frac{3}{6}x+\frac{39}{6}=\frac{1}{2}x + 6.5 $.

For $ x - 2y = 13 $:
Subtract $ x $: $ - 2y=-x + 13 $.
Divide by - 2: $ y=\frac{1}{2}x-6.5 $.

## Step 2: Compare slopes and y - intercepts
The first equation has slope $ m_1=\frac{1}{2} $, y - intercept $ b_1 = 6.5 $.
The second equation has slope $ m_2=\frac{1}{2} $, y - intercept $ b_2=-6.5 $.
Since $ m_1 = m_2 $ (parallel lines) and $ b_1
eq b_2 $ (different y - intercepts), the lines are parallel and never intersect.
# Answer:
0 solutions (no solution)

### Problem 8: Determine the number of solutions for $ \begin{cases} y = x + 3 \ y = -3x - 5 \end{cases} $
# Explanation:
## Step 1: Compare slopes
The first equation $ y = x + 3 $ has slope $ m_1 = 1 $.
The second equation $ y=-3x - 5 $ has slope $ m_2=-3 $.

## Step 2: Analyze the slopes
Since $ m_1
eq m_2 $, the two lines have different slopes, which means they are not parallel and will intersect at exactly one point.
# Answer:
1 solution

### Problem 9: Ticket Cost Problem
Let $ a $ = cost of adult ticket, $ s $ = cost of student ticket.
System: $ \begin{cases} 5a + 18s = 98 \ 4a + 22s = 105 \end{cases} $
# Explanation:
## Step 1: Eliminate $ a $
Multiply first equation by 4: $ 20a + 72s = 392 $.
Multiply second equation by 5: $ 20a + 110s = 525 $.
Subtract the first new equation from the second:
$ (20a - 20a)+(110s - 72s)=525 - 392 $ → $ 38s = 133 $.
Divide by 38: $ s = 3.5 $ (or $ \frac{7}{2} $).

## Step 2: Find $ a $
Substitute $ s = 3.5 $ into $ 5a + 18s = 98 $:
$ 5a + 18(3.5)=98 $ → $ 5a + 63 = 98 $.
Subtract 63: $ 5a = 35 $.
Divide by 5: $ a = 7 $.

## Step 3: Find total for one of each
One adult ($ \$7 $) + one student ($ \$3.50 $) = $ 7 + 3.5 = 10.5 $.
# Answer:
The cost for one adult ticket and one student ticket is $ \$10.50 $ (or $ \$10.5 $)

### Problem 10: Solve $ 5xy - 2 = y + 2x $ for $ x $
# Explanation:
## Step 1: Collect $ x $ terms
$ 5xy - 2x = y + 2 $.
Factor $ x $: $ x(5y - 2)=y + 2 $.

## Step 2: Solve for $ x $
Divide by $ 5y - 2 $ (assuming $ 5y - 2
eq0 $): $ x=\frac{y + 2}{5y - 2} $.
# Answer:
$ x=\frac{y + 2}{5y - 2} $ (for $ 5y
eq2 $)

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QUESTION IMAGE

Question

Explanation:

Problem 1: Solve the system \(

Step 1: Set the equations equal

Step 2: Solve for \( x \)

Step 3: Find \( y \)

Step 1: Substitute \( y \) into the second equation

Step 2: Solve for \( x \)

Step 3: Find \( y \)

Step 1: Set the equations equal

Step 2: Solve for \( x \)

Step 3: Find \( y \)

Answer:

Problem 2: Solve the system \(