Mathematics • Year 8 • Unit 2 • Lesson 17
Introduction to Simultaneous Equations
Two equations, two unknowns, one shared answer. Build fluency with checking solutions, reading the solution from an intersection point, classifying systems (one / none / infinite), and setting up two equations from a word problem.
1. I do — fully worked example
Read every line. The pair (x, y) is a solution of a system only if it satisfies BOTH equations. Checking is just substituting and comparing LHS with RHS.
Problem. Is (3, 2) the solution of the system below?
Eq 1: x + y = 5
Eq 2: x − y = 1
Step 1 — Substitute into Eq 1: replace x with 3, y with 2.
LHS = 3 + 2 = 5. RHS = 5. LHS = RHS ✓
Reason: the pair must satisfy Eq 1, so we test it there first.
Step 2 — Substitute into Eq 2 with the SAME pair.
LHS = 3 − 2 = 1. RHS = 1. LHS = RHS ✓
Reason: the pair must ALSO satisfy Eq 2. Passing both is the test.
Step 3 — State the conclusion.
Both checks pass → (3, 2) IS the solution.
Graphically: the two lines y = 5 − x and y = x − 1 cross exactly at (3, 2).
Answer: Yes — (3, 2) is the solution.
2. We do — fill in the missing steps
Same shape as Section 1, with blanks. Fill every blank. 5 marks
Problem. Is (4, 1) the solution of the system below?
Eq 1: x + y = 5
Eq 2: 2x − y = 7
Step 1 — Substitute into Eq 1.
LHS = ______ + ______ = ______. RHS = 5. LHS ___ RHS ( ✓ or ✗ )
Step 2 — Substitute into Eq 2 with the SAME pair.
LHS = 2(______) − ______ = ______ − ______ = ______. RHS = 7. LHS ___ RHS ( ✓ or ✗ )
Step 3 — Conclusion: Because the pair satisfies ______ equation(s), it ______ (is / is not) the solution of the system.
3. You do — independent practice
Show every check. 3.1–3.3 are foundation (verify pairs). 3.4–3.6 are standard (reading intersection points and classifying systems). 3.7–3.8 are extension (setting up two equations from words).
Foundation — verify the pair
3.1 Is (2, 3) a solution of x + y = 5 and y = 2x − 1? Show both checks. 2 marks
3.2 Is (5, 0) a solution of x + y = 5 and x − y = 4? 2 marks
3.3 Two lines on a graph intersect at the point (−1, 4). State the solution of the simultaneous equations they represent. 1 mark
Standard — classify the system
3.4 Lines y = 2x + 1 and y = 2x − 3 are graphed on the same axes. State how many solutions the system has and explain in one sentence. 2 marks
3.5 Lines y = x + 2 and 2y = 2x + 4 are graphed on the same axes. State how many solutions the system has and explain why. 2 marks
3.6 Lines y = 3x − 2 and y = −x + 6 have different gradients. Without graphing, state how many solutions the system has and explain in one sentence. 2 marks
Extension — set up two equations from words
3.7 At the canteen, 1 sandwich plus 2 drinks costs $9, and 1 sandwich plus 1 drink costs $7. Let s = price of a sandwich and d = price of a drink. Write the TWO equations — do not solve. 2 marks
3.8 The sum of two numbers is 14 and their difference is 4. Let the bigger number be x and the smaller be y. Write two equations relating x and y, then check that (9, 5) satisfies both. 3 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2 — We do (4, 1)
Step 1: LHS = 4 + 1 = 5. RHS = 5. LHS = RHS ( ✓ ).
Step 2: LHS = 2(4) − 1 = 8 − 1 = 7. RHS = 7. LHS = RHS ( ✓ ).
Step 3: Pair satisfies both equations, so it is the solution.
3.1 — (2, 3)
Eq 1: 2 + 3 = 5 ✓. Eq 2: y = 2x − 1 → 3 = 2(2) − 1 = 3 ✓. Both pass → YES, (2, 3) is a solution.
3.2 — (5, 0)
Eq 1: 5 + 0 = 5 ✓. Eq 2: 5 − 0 = 5, but RHS = 4. 5 ≠ 4 ✗. NOT a solution — Eq 2 fails.
3.3 — Intersection at (−1, 4)
The solution is read straight off the intersection: x = −1, y = 4.
3.4 — y = 2x + 1 and y = 2x − 3
No solutions. Same gradient (m = 2) but different y-intercepts → the lines are parallel and never cross.
3.5 — y = x + 2 and 2y = 2x + 4
Infinite solutions. Divide the second equation by 2: y = x + 2 — the SAME line as the first. Every point on the line satisfies both equations.
3.6 — y = 3x − 2 and y = −x + 6
Exactly one solution. Different gradients (3 vs −1) mean the lines must cross at exactly one point.
3.7 — Canteen prices
Eq 1: s + 2d = 9. Eq 2: s + d = 7.
3.8 — Sum 14, difference 4
Eq 1: x + y = 14. Eq 2: x − y = 4.
Check (9, 5): Eq 1: 9 + 5 = 14 ✓. Eq 2: 9 − 5 = 4 ✓. (9, 5) IS the solution.