Moderate -0.8 This is a straightforward algebraic manipulation question requiring students to rearrange a rational expression to make x the subject. It involves standard techniques (multiplying both sides by the denominator, expanding, collecting terms, factorizing) that are well-practiced at this level. The 4 marks suggest multiple steps but no conceptual difficulty or problem-solving insight is required—it's a routine procedural exercise easier than the typical A-level question.
for multiplying to eliminate denominator and for expanding brackets, or for correct division by \(y\) and writing as separate fractions: \(x + 3 = \frac{1}{y} - \frac{2x}{y}\); each mark is for carrying out the operation correctly; ft earlier errors for equivalent steps if error does not simplify problem; some common errors shown in table
\(yx + 2x = 1 - 3y\) oe or ft
M1
for collecting terms; dep on having an \(ax\) term and an \(xy\) term, oe after division by \(y\)
\(x(y + 2) = 1 - 3y\) oe or ft
M1
for taking out \(x\) factor; dep on having an \(ax\) term and an \(xy\) term, oe after division by \(y\)
\(\left[x = \right]\frac{1-3y}{y+2}\) oe or ft as final answer
M1
for division with no wrong work after; dep on dividing by a two-term expression; last M not earned for triple-decker fraction as final answer
Common Errors Table
\(y(x + 3) = 1 - 2x\) → \(yx + 3 = 1 - 2x\)
M0
\(yx + 3y = 1 - 2x\)
M0
\(yx + 5x = 1\)
M1 ft
\(x(y + 5) = 1\)
M1 ft
\(x = \frac{1}{y+5}\)
M1 ft
\(yx + 3 = 1 - 2x\)
M0
\(yx + 2x = -2\)
M1 ft
\(x(y + 2) = -2\)
M1 ft
\(x = \frac{-2}{y+2}\)
M1 ft
$yx + 3y = 1 - 2x$ oe or ft | M1 | for multiplying to eliminate denominator **and** for expanding brackets, or for correct division by $y$ and writing as separate fractions: $x + 3 = \frac{1}{y} - \frac{2x}{y}$; each mark is for carrying out the operation correctly; ft earlier errors for equivalent steps if error does not simplify problem; some common errors shown in table
$yx + 2x = 1 - 3y$ oe or ft | M1 | for collecting terms; dep on having an $ax$ term and an $xy$ term, oe after division by $y$
$x(y + 2) = 1 - 3y$ oe or ft | M1 | for taking out $x$ factor; dep on having an $ax$ term and an $xy$ term, oe after division by $y$
$\left[x = \right]\frac{1-3y}{y+2}$ oe or ft as final answer | M1 | for division with no wrong work after; dep on dividing by a two-term expression; last M not earned for triple-decker fraction as final answer | for M4, must be completely correct
| Common Errors Table |
|---|
| $y(x + 3) = 1 - 2x$ → $yx + 3 = 1 - 2x$ | M0 |
| $yx + 3y = 1 - 2x$ | M0 |
| $yx + 5x = 1$ | M1 ft |
| $x(y + 5) = 1$ | M1 ft |
| $x = \frac{1}{y+5}$ | M1 ft |
| $yx + 3 = 1 - 2x$ | M0 |
| $yx + 2x = -2$ | M1 ft |
| $x(y + 2) = -2$ | M1 ft |
| $x = \frac{-2}{y+2}$ | M1 ft |