Question 1: Solve the linear equation: 3x + 2 = 11. Then, verify your solution by substitution.

Answer: x=3

Question 2: Solve the quadratic equation: x^2 - 5x + 6 = 0. List both solutions.

Answer: x=2 or x=3

Question 3: A student attempts to solve the system: y = 2x + 1 and x^2 + y = 7. Their working shows they set x^2 + (2x + 1) = 7, then simplified to x^2 + 2x + 1 = 7 and concluded x= ±√6. Identify the mistake and correct the solution.

Answer: The mistake was treating the quadratic as if it was linear. The correct approach is to substitute y=2x+1 into the quadratic to get x^2 + (2x+1)=7, then x^2 + 2x + 1 = 7, leading to x^2 + 2x - 6=0, which factors to (x+3)(x-2)=0, so x=-3 or x=2.

Question 4: Plot the graph of y=2x and draw the quadratic y=x^2 - 4x + 3 on the same axes. (Sketch your graphs on the grid.)

Answer: Students draw two graphs correctly; y=2x is a straight line, y=x^2 - 4x + 3 is a parabola.

Question 5: Solve the system: y = x + 2 and y = x^2 - 1. Find all solutions graphically and algebraically.

Answer: Solutions are x=1, y=3 and x=-3, y=-1

Question 6: A student incorrectly solves the system: y=3x+1 and x^2 + y=10, stating the solutions x=2 and y=7. Is this correct? Explain and fix the solution if necessary.

Answer: The solution is incorrect because substituting y=3x+1 into x^2 + y=10 gives x^2 + 3x + 1=10, then x^2 + 3x - 9=0. Solving this quadratic, solutions are x=(-3±√45)/2, which are approximately -3.12 and 0.12. Corresponding y-values are y=3x+1.

Question 7: Identify the common misconception students have when solving simultaneous equations involving quadratics and linear equations and how to avoid it.

Answer: A common mistake is treating the quadratic as linear or substituting incorrectly. To avoid this, students should isolate one variable carefully and substitute correctly, then solve the resulting quadratic carefully.

Question 8: Solve the system: y = x^2 + 2x and y = 4x + 1. Show your working and verify the solutions.

Answer: Set x^2 + 2x = 4x + 1 ightarrow x^2 + 2x - 4x - 1=0 ightarrow x^2 - 2x -1=0. Solutions: x=1±√2. Then y=4(1±√2)+1.

Question 9: Challenge: For the quadratic y=x^2 + 4x + 4, find the points of intersection with the line y=6x+2. Then, explain why the solutions are what they are.

Answer: Set x^2 + 4x + 4=6x+2 ightarrow x^2 + 4x + 4 -6x - 2=0 ightarrow x^2 - 2x + 2=0. Discriminant = (-2)^2 - 4*1*2=4-8=-4, no real solutions. No real points of intersection, hence the parabola and line do not intersect.

Question 10: Error analysis: A student claims that solving y=2x+3 and y=x^2 - 1 by setting 2x+3=x^2 - 1 yields solutions x=-1 and x=2. Show the correct solution process and identify the mistake.

Answer: Correct method: 2x+3 = x^2 -1 ightarrow x^2 - 2x - 4=0. Discriminant = 4 - 4*1*-4=4+16=20. Solutions: x=(2±√20)/2. The student incorrectly simplified to x=-1 and x=2 without solving the quadratic correctly.