Question 1: Calculate the value of y when x = 3 for the equation y = 2x + 1.

Answer: 7

Question 2: Complete the table for y = 3x - 2 by calculating y for x = 0, 2, and 4.

Answer: {"x=0":"-2","x=2":"4","x=4":"10"}

Question 3: Choose three x-values that will produce y-values between 0 and 10 for y = x + 4.

Answer: ["x=1 (y=5)","x=2 (y=6)","x=6 (y=10)"]

Question 4: Plot points for y = 2x on the grid when x = -2, 0, and 2. Show the (x, y) pairs.

Answer: ["(-2, -4)","(0, 0)","(2, 4)"]

Question 5: Identify which x-values from the list [-3, 0, 2, 5] will give y-values greater than 4 for y = 2x + 1.

Answer: ["x=2 (y=5)","x=5 (y=11)"]

Question 6: Construct a table of values for y = -x + 3, choosing x-values from -2 to 2 at 1-unit intervals.

Answer: {"x=0":"3","x=1":"2","x=2":"1","x=-1":"4","x=-2":"5"}

Question 7: Explain why choosing x = 0 is always a safe starting point when plotting y = 2x + 1.

Answer: Because substituting x=0 simplifies the equation to y=1, providing an easy initial point for plotting.

Question 8: Using the equation y = 4x - 3, select x-values to plot so that y-values are positive and less than 10.

Answer: ["x=1 (y=1)","x=2 (y=5)","x=3 (y=9)"]

Question 9: Identify and correct the mistake: The student chooses x = -1, 0, 1 for y= x^2 + 2x. The table shows y-values of 1, 2, 3 respectively.

Answer: The mistake is in calculating y-values. Correct y-values are: x=-1 (y= -1), x=0 (y=0), x=1 (y=3).

Question 10: Challenge: For y = 3x + 2, choose x-values such that the y-values are multiples of 3.

Answer: ["x=1 (y=5)","x=0 (y=2)","x=2 (y=8)"]

Question 11: Determine x-values where y = x - 5 equals 0, 3, and -4. Show your working.

Answer: {"x=0 (y= -5)":"x=5","x=3 (y= -2)":"x=8","x=-4 (y= -9)":"x= -4"}