Mathematics • Year 8 • Unit 2 • Lesson 17
Simultaneous Equations — Mixed Challenge
Pull together verifying solutions, reading intersections, classifying systems (one / none / infinite), and setting up word problems. Six mixed problems, one "find the mistake", and one open-ended challenge.
1. Mixed problems — choose the right move
Each question uses a different combination of ideas from Lesson 17. Decide what's being asked BEFORE you write. Show working. 3 marks each
1.1 Check whether (1, 3) is a solution of y = 2x + 1 and x + y = 4. Show both checks and state your conclusion.
1.2 Two lines y = x + 2 and y = −x + 6 are drawn on the same axes. Where do they intersect, and what is the solution of the system? Solve by setting x + 2 = −x + 6.
1.3 Classify each system as one / none / infinite solutions. Give a one-line reason for each.
(a) y = 3x − 5 and y = 3x + 1.
(b) y = x and y = 2x.
(c) y = 4x + 2 and 2y = 8x + 4.
1.4 A point lies on BOTH y = 2x and x + y = 9. By setting 2x equal to (9 − x), find x and then y, and write the solution as (x, y).
1.5 The sum of two numbers is 20 and one number is three times the other. Define variables, write two equations and find the numbers.
1.6 A system has equations y = 2x − 1 and y = 2x + 5. (i) Without graphing, decide how many solutions it has. (ii) Explain what would have to change about the second equation for the system to have exactly ONE solution.
2. Find the mistake
Another student is asked: "Is (3, 2) a solution of the system y = 2x − 4 and x + y = 5?" Their working is below. Exactly one line contains a mistake. Spot it, explain why it's wrong, then redo correctly. 3 marks
Student's working:
Line 1: Test (3, 2) in Eq 1: y = 2x − 4.
Line 2: RHS = 2(3) − 4 = 2. LHS = 2. ✓
Line 3: Eq 1 passed, so (3, 2) MUST be the solution.
Line 4: Final answer: yes, (3, 2) is the solution.
(a) Which line contains the mistake (in reasoning)?
(b) Explain in one or two sentences why that reasoning is wrong.
(c) Re-do the check properly, including Eq 2, and state the correct final answer.
Stuck? A solution must satisfy BOTH equations. Passing Eq 1 alone tells you nothing about Eq 2.3. Open-ended challenge — design a system
This question has more than one valid answer. 4 marks
3.1 Design THREE different systems of two simultaneous equations whose solution is exactly (2, 5).
(a) Your first system must use the equation y = x + 3 as one of its equations. Provide a second equation that also passes through (2, 5) but is NOT y = x + 3.
(b) Your second system must contain at least one equation in the form ax + by = c with a, b, c whole numbers (e.g., 2x + 3y = ...).
(c) Your third system is your free choice — both equations made up by you. Both must pass through (2, 5) and not be the same line.
For each system: write the two equations clearly, and VERIFY that (2, 5) satisfies both.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Check (1, 3)
Eq 1: y = 2x + 1 → 3 = 2(1) + 1 = 3 ✓. Eq 2: x + y = 4 → 1 + 3 = 4 ✓. Both pass → YES, (1, 3) is the solution.
1.2 — Intersection of y = x + 2 and y = −x + 6
Set equal: x + 2 = −x + 6 → 2x = 4 → x = 2. Then y = 2 + 2 = 4. Intersection (2, 4), solution x = 2, y = 4.
1.3 — Classify the systems
(a) Same gradient (3), different intercepts (−5, +1) → parallel → NO solution.
(b) Different gradients (1, 2) → ONE solution (they meet at the origin (0, 0)).
(c) Divide the second by 2: y = 4x + 2 — same line as the first → INFINITE solutions.
1.4 — y = 2x and x + y = 9
Set 2x = 9 − x → 3x = 9 → x = 3. Then y = 2(3) = 6. Solution: (3, 6). Check: 3 + 6 = 9 ✓ and y = 2x → 6 = 6 ✓.
1.5 — Sum 20, one is three times the other
Let the smaller number = y and the bigger = x. Eq 1: x + y = 20. Eq 2: x = 3y. Substitute: 3y + y = 20 → 4y = 20 → y = 5. Then x = 15. Numbers: 15 and 5. Check: 15 + 5 = 20 ✓; 15 = 3(5) ✓.
1.6 — y = 2x − 1 and y = 2x + 5
(i) Same gradient, different intercepts → parallel → NO solution.
(ii) Change the GRADIENT of one of the equations to something other than 2 (e.g., y = 3x + 5 or y = −x + 5). Different gradients guarantee exactly one intersection.
2 — Find the mistake
(a) The mistake is on Line 3 (the reasoning, not the arithmetic).
(b) Passing Eq 1 only tells us that (3, 2) lies on the line y = 2x − 4. We need to check Eq 2 as well — a solution must satisfy BOTH equations.
(c) Test Eq 2: x + y = 5 → 3 + 2 = 5 ✓. Both equations pass, so (3, 2) IS the solution. Even though the student arrived at the right answer, their reasoning was unsound — they would have been wrong on a system where Eq 2 failed.
3 — Design a system with solution (2, 5)
(a) Use y = x + 3 (check: 5 = 2 + 3 ✓). Second equation: pick a different gradient, e.g., y = 2x + 1 (check: 5 = 2(2) + 1 = 5 ✓). System: y = x + 3 and y = 2x + 1.
(b) Need ax + by = c with 2a + 5b = c. Example: a = 1, b = 1 gives 1(2) + 1(5) = 7, so x + y = 7. Pair it with something else through (2, 5), e.g., y = 3x − 1 (check: 5 = 6 − 1 ✓). System: x + y = 7 and y = 3x − 1.
(c) Sample student answer: y = 5 (horizontal line through (2, 5)) and x = 2 (vertical line through (2, 5)). Both clearly contain (2, 5). System: y = 5 and x = 2. (Other valid answers: any two non-identical lines through (2, 5).)
Marking: 1 mark for each correctly designed system (3 marks total) — each system must have two DIFFERENT equations both passing through (2, 5), with verification shown. 1 extra mark for using the ax + by = c form correctly in (b).