Higher June 2021 Paper 1 Q21
21 Given that \(x = \dfrac{5}{9y + 5}\) and that \(y = \dfrac{5}{5a - 2}\)
find an expression for \(x\) in terms of \(a\).
Give your expression as a single fraction in its simplest form.
(4)
| Scheme | Marks |
|---|---|
| \([x =]\ \dfrac{5}{9\left(\dfrac{5}{5a - 2}\right) + 5}\) oe or \(y = \dfrac{5}{9x} - \dfrac{5}{9}\) oe | M1 |
\([x =]\ \dfrac{5(5a - 2)}{45 + 5(5a - 2)}\) oe or \((5 - 5x)(5a - 2) = 45x\) oe or \(9x = \dfrac{5(45a - 18)}{35 + 25a}\) oe | M1 |
| \([x =]\ \dfrac{25a - 10}{35 + 25a}\) oe or \([x =]\ \dfrac{5(5a - 2)}{5(7 + 5a)}\) | M1 |
Working not required, so correct answer scores full marks (unless from obvious incorrect working) Answer: \(x = \dfrac{5a - 2}{7 + 5a}\) | A1 |
| (4) | |
| (4 marks) |
Notes
M1: A correct substitution for \(y\)
or
writing \(y\) in terms of \(x\)
M1: Multiplying each term in the numerator and denominator by \((5a - 2)\) to eliminate the fraction in the denominator
or
equating \(y\)’s and getting rid of fractions as far as shown on left or single fraction in terms of \(a\)
M1: A correct fraction not in simplest form with all brackets expanded or numerator and denominator factorised with the same common factor taken out
A1: Correctly simplified
\(x =\) needed for the answer, or \(x =\) previously seen in working with correct simplified expression
Do not isw if students have tried to do some incorrect cancelling eg \(x = \dfrac{5a - 2}{7 + 5a} = \dfrac{-2}{7}\) gets M3A0