Moderate -0.3 This is a straightforward rational equation requiring finding a common denominator, clearing fractions, and solving a resulting quadratic. It's slightly easier than average as it's a standard technique with no conceptual surprises, though it requires careful algebraic manipulation across multiple steps.
M1*: Multiplying throughout by \((2x + 1)(x + 1)\) or combining fractions and multiplying up or equivalent (e.g. can retain denominator throughout). Condone a single numerical error, sign error or slip provided that there is no conceptual error in the process involved. Do not condone omission of brackets unless it is clear from subsequent work that they were assumed.
- \(5x(x + 1) - 3(2x + 1) = (2x + 1)\) gets M1, just, for slip in omission of \((x + 1)\)
M1dep*: Multiplying out, collecting like terms and forming quadratic (\(= 0\)). Follow through from their equation provided the algebra is not significantly eased and it is a quadratic. Condone a further sign or numerical error or a minor slip when rearranging or equivalent (www, not fortuitously obtained – check for double errors) provided \(b^2 - 4ac \geq 0\).
A1: Or equivalent www (not fortuitously obtained – check for double errors).
M1: Solving their three term quadratic (\(= 0\)). Use of correct quadratic equation formula (if formula is quoted correctly then only one sign slip is permitted; if the formula is quoted incorrectly M0; if not quoted at all substitution must be completely correct to earn the M1) or factorising (giving their \(x^2\) term and one other term when factors multiplied out) or completing the square (must get to the square root stage).
A1: Correct answers for both obtained www (condone \(-0.667\) or better). If no factorisation (or equivalent) seen, B1 for each answer stated following correct quadratic.
# Question 1
**Mark scheme:**
$5x(x + 1) - 3(2x + 1) = (2x + 1)(x + 1)$ | M1*
$3x^2 - 4x - 4 = 0$ | M1dep*
$(3x + 2)(x - 2) = 0$ | A1
$x = -\frac{2}{3}$ or $2$ | M1, A1
| [5]
**Guidance:**
M1*: Multiplying throughout by $(2x + 1)(x + 1)$ or combining fractions and multiplying up or equivalent (e.g. can retain denominator throughout). Condone a single numerical error, sign error or slip provided that there is no conceptual error in the process involved. Do not condone omission of brackets unless it is clear from subsequent work that they were assumed.
Examples:
- $5x(x + 1) - 3(2x + 1) = (2x + 1)(x - 1)$ gets M1
- $5x(x + 1) - 3(2x + 1) = 1$ gets M0
- $5x(x + 1)(2x + 1) - 3(2x + 1)(x + 1) = (x + 1)(2x + 1)$ gets M0
- $5x(x + 1) - 3(2x + 1) = (2x + 1)$ gets M1, just, for slip in omission of $(x + 1)$
M1dep*: Multiplying out, collecting like terms and forming quadratic ($= 0$). Follow through from their equation provided the algebra is not significantly eased and it is a quadratic. Condone a further sign or numerical error or a minor slip when rearranging or equivalent (www, not fortuitously obtained – check for double errors) provided $b^2 - 4ac \geq 0$.
A1: Or equivalent www (not fortuitously obtained – check for double errors).
M1: Solving their three term quadratic ($= 0$). Use of correct quadratic equation formula (if formula is quoted correctly then only one sign slip is permitted; if the formula is quoted incorrectly M0; if not quoted at all substitution must be completely correct to earn the M1) or factorising (giving their $x^2$ term and one other term when factors multiplied out) or completing the square (must get to the square root stage).
A1: Correct answers for both obtained www (condone $-0.667$ or better). If no factorisation (or equivalent) seen, B1 for each answer stated following correct quadratic.