Easy -1.8 This is a straightforward algebraic rearrangement requiring only expansion, collection of terms, and factorization—purely mechanical manipulation with no problem-solving element. Significantly easier than average A-level content.
For correct collection of terms, ft. e.g. after M0 for \(5c + 9t = 2ac + t\) allow M1 for \(5c - 2ac = -8t\) oe. For each M, ft previous errors if their equation is of similar difficulty
\(c(5 - 2a) = at - 9t\) oe
M1
For correctly factorising, ft; must be \(c \times\) a two-term factor. May be earned before \(t\) terms collected
\(\left[c =\right] \frac{at - 9t}{5 - 2a}\) or \(\frac{t(a-9)}{5-2a}\) oe as final answer
M1
For correct division, ft their two-term factor. Treat as MR if \(t\) is the subject, with a penalty of 1 mark from those gained, marking similarly
[4]
## Question 8:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $5c + 9t = 2ac + at$ | M1 | For correct expansion of brackets |
| $5c - 2ac = at - 9t$ oe | M1 | For correct collection of terms, ft. e.g. after M0 for $5c + 9t = 2ac + t$ allow M1 for $5c - 2ac = -8t$ oe. For each M, ft previous errors if their equation is of similar difficulty |
| $c(5 - 2a) = at - 9t$ oe | M1 | For correctly factorising, ft; must be $c \times$ a two-term factor. May be earned before $t$ terms collected |
| $\left[c =\right] \frac{at - 9t}{5 - 2a}$ or $\frac{t(a-9)}{5-2a}$ oe as final answer | M1 | For correct division, ft their two-term factor. Treat as MR if $t$ is the subject, with a penalty of 1 mark from those gained, marking similarly |
| **[4]** | | |
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