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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.

1 answer

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!