Mathematics • Year 8 • Unit 2 • Lesson 11
y = mx + c — Mixed Challenge
Pull together everything from Lesson 11: reading m and c, writing equations, substituting, rearranging into gradient-intercept form. Six mixed problems, one "find the mistake", one open-ended challenge.
1. Mixed problems — choose the right move
Each question uses a different combination of ideas from Lesson 11. Decide which move applies before you start writing. 3 marks each
1.1 State the gradient and y-intercept of y = 7 − 2x. (Careful: x-term is second; what is m really?)
1.2 A line has gradient −½ and crosses the y-axis at (0, 6). Write its equation in y = mx + c form.
1.3 For the line y = 3x − 4, find y when x = 5, then find the value of x when y = 2.
1.4 Rearrange 4x − y = 7 into y = mx + c form, then state m and c.
1.5 Two lines have equations y = 2x + 1 and y = 2x − 5. State one thing they have in common, and one thing that's different.
1.6 Show whether the point (3, 5) lies on the line y = 2x − 1. Substitute and explain.
2. Find the mistake
Another student has tried to rearrange 2x + y = 5 into y = mx + c form. Their working is shown below. Exactly one line contains a mistake. Spot it, explain why it's wrong, then re-do the working correctly. 3 marks
Student's working — rearrange 2x + y = 5:
Line 1: 2x + y = 5
Line 2: y = 5 + 2x (subtract 2x from both sides)
Line 3: y = 2x + 5
Line 4: So m = 2 and c = 5.
(a) Which line contains the mistake?
(b) Explain in one or two sentences why that line is wrong.
(c) Write out the corrected working in full, including the corrected m and c.
Stuck? Subtracting 2x from the left turns +2x into 0, but on the right it turns +0 into −2x — not +2x.3. Open-ended challenge — design three lines
This question has more than one valid answer. 4 marks
3.1 Design three different straight lines in y = mx + c form that all pass through the point (1, 4).
For each line you find:
(i) Choose a gradient m (different for each line — at least one positive, at least one negative).
(ii) Work out what c must be so that y = 4 when x = 1.
(iii) Write out the final equation.
Bonus: One of your lines must be horizontal (m = 0). What does its equation look like?
How did this worksheet feel?
What I'll revisit before next class:
1.1 — y = 7 − 2x
Rewrite with the x-term first: y = −2x + 7. So m = −2, c = 7. (The minus belongs to the 2x.)
1.2 — gradient −½, through (0, 6)
m = −½, c = 6, so y = −½x + 6 (or equivalently y = 6 − x/2).
1.3 — y = 3x − 4
At x = 5: y = 3(5) − 4 = 11.
When y = 2: 2 = 3x − 4 → 6 = 3x → x = 2.
1.4 — 4x − y = 7
Subtract 4x: −y = 7 − 4x. Multiply both sides by −1: y = 4x − 7. So m = 4, c = −7.
1.5 — y = 2x + 1 and y = 2x − 5
Same: both have gradient m = 2 (so they are parallel — same steepness, same direction).
Different: their y-intercepts (c = 1 vs c = −5) — they cross the y-axis at different heights.
1.6 — Does (3, 5) lie on y = 2x − 1?
Substitute x = 3: y = 2(3) − 1 = 5. The line gives y = 5 when x = 3, which matches the point. Yes, (3, 5) lies on the line.
2 — Find the mistake
(a) The mistake is on Line 2 (and the wrong sign is then carried into Line 3).
(b) Subtracting 2x from both sides should give y = 5 − 2x, NOT y = 5 + 2x. The student kept the sign of 2x instead of flipping it.
(c) Corrected working:
2x + y = 5
y = 5 − 2x (subtract 2x from both sides — sign flips)
y = −2x + 5
So m = −2, c = 5. ✓
3 — Open-ended challenge (sample solution)
For any gradient m, we need 4 = m(1) + c, so c = 4 − m.
Line 1 (positive gradient): pick m = 2 → c = 2 → y = 2x + 2. Check: 2(1) + 2 = 4 ✓.
Line 2 (negative gradient): pick m = −3 → c = 7 → y = −3x + 7. Check: −3(1) + 7 = 4 ✓.
Line 3 (horizontal, m = 0): c = 4 → y = 4. Every point on this line has y = 4, including (1, 4) ✓.
Many other valid answers exist — any pair (m, 4 − m) works.
Marking: 1 mark per valid distinct line (3 marks total). 1 bonus mark for the horizontal line y = 4 and explaining that m = 0 collapses the equation to y = c.