Question

Show your work for each question
Q1 Solve by substitution
y=4x-9
y=x-3
Q2 Solve by substitution
4x-y=20
y=-x+5
Q3 Solve the system by graphing.
y=-3x+4
y=3x-2

Q4 Sketch the solution to the system of inequalities.
y-52x-2
y<-12x+2

Q5 Solve by elimination.
8x-6y=-20
-16x+7y=30
Q6 The school that Stefan goes to is selling tickets to a choral performance. On the first day of ticket sales the school sold 3 senior citizen tickets and 1 child ticket for a total of $38. The school took in $52 on the second day by selling 3 senior citizen tickets and 2 child tickets. Find the price of a senior citizen ticket and the price of a child ticket.

Answer 1

Let's solve each question step-by-step:

### Q1: Solve by substitution
Given:
1. \( y = 4x - 9 \)
2. \( y = x - 3 \)

Since both equations equal \( y \), we can set them equal to each other:
\[
4x - 9 = x - 3
\]
Now, solve for \( x \):
\[
4x - x = -3 + 9
\]
\[
3x = 6 \\
x = 2
\]
Now, substitute \( x \) back into one of the original equations to solve for \( y \):
Using \( y = x - 3 \):
\[
y = 2 - 3 \\
y = -1
\]
Thus, the solution is \( (2, -1) \).

### Q2: Solve by substitution
Given:
1. \( 4x - y = 20 \)
2. \( y = -x + 5 \)

From the second equation, substitute \( y \) in the first equation:
\[
4x - (-x + 5) = 20
\]
This becomes:
\[
4x + x - 5 = 20 \\
5x - 5 = 20 \\
5x = 25 \\
x = 5
\]
Now substituting \( x \) back to find \( y \):
Using \( y = -x + 5 \):
\[
y = -5 + 5 \\
y = 0
\]
Thus, the solution is \( (5, 0) \).

### Q3: Solve the system by graphing
Given:
1. \( y = -3x + 4 \)
2. \( y = 3x - 2 \)

Let's find the points to graph both equations.

**For \( y = -3x + 4 \)**:
- If \( x=0 \): \( y = 4 \) (Point: (0, 4))
- If \( x=1 \): \( y = 1 \) (Point: (1, 1))

**For \( y = 3x - 2 \)**:
- If \( x=0 \): \( y = -2 \) (Point: (0, -2))
- If \( x=1 \): \( y = 1 \) (Point: (1, 1))

Now graph:
Plot the points and draw both lines. The intersection is at \( (1, 1) \), so the solution to the system is \( (1, 1) \).

### Q4: Sketch the solution to the system of inequalities
Given:
1. \( y \geq 5 - 2x \)
2. \( y < -12x + 2 \)

**For \( y \geq 5 - 2x \)**:
- The line \( y = 5 - 2x \) has a y-intercept of 5 and a slope of -2.
- You will shade above this line.

**For \( y < -12x + 2 \)**:
- The line \( y = -12x + 2 \) has a y-intercept of 2 and a slope of -12.
- You will shade below this line.

When shading the regions, the solution set is where the shading overlaps.

### Q5: Solve by elimination
Given:
1. \( 8x - 6y = -20 \)
2. \( -16x + 7y = 30 \)

First, multiply the first equation by 2:
\[
16x - 12y = -40
\]
Now we have:
1. \( 16x - 12y = -40 \)
2. \( -16x + 7y = 30 \)

Add both equations:
\[
(16x - 16x) + (-12y + 7y) = -40 + 30
\]
\[
-5y = -10 \\
y = 2
\]
Substituting \( y \) back into one of the original equations to solve for \( x \):
Using \( 8x - 6(2) = -20 \):
\[
8x - 12 = -20 \\
8x = -8 \\
x = -1
\]
Thus, the solution is \( (-1, 2) \).

### Q6: Interpret the scenario and create equations
Let \( s \) be the price of a senior citizen ticket and \( c \) be the price of a child ticket.

From Day 1:
\[
3s + 1c = 38 \quad \text{(Equation 1)}
\]
From Day 2:
\[
3s + 2c = 52 \quad \text{(Equation 2)}
\]

Now subtract Equation 1 from Equation 2:
\[
(3s + 2c) - (3s + 1c) = 52 - 38
\]
This simplifies to:
\[
c = 14
\]

Now substitute \( c \) back into Equation 1:
\[
3s + 1(14) = 38 \\
3s + 14 = 38 \\
3s = 24 \\
s = 8
\]
Thus, the prices are:
- Senior citizen ticket: $8
- Child ticket: $14.

Let me know if you need further explanations!