Past Paper Questions by Topic › Level 2 FM Algebra › Manipulating Formulae/ Changing the Subject
Manipulating Formulae/ Changing the Subject Topic Factorising (28) Expanding Brackets (27) Comparing Coefficients (10) Drawing & Using Graphs (29) Solving Basic Equations (6) Algebraic Fractions (24) Binomial Expansion (6) Linear Graphs (40) Indices (41) Solving Quadratics (27) Inequalities (18) Polynomials (10) Circles in Coordinate Geometry (17) Manipulating Formulae/ Changing the Subject (11) Completing the Square (13) Differentiation (52) Simultaneous Equations (18) Functions (26) Algebraic Proof (7) Sequences (20) Course Support Current PowerPoint version
More questions on this topic: GCSE Algebra: Manipulating Formulae/ Changing the Subject (91) IGCSE Algebra: Manipulating Formulae/ Changing the Subject (83)
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13 Rearrange \(\quad x = \dfrac{3k^2}{5k^2 - 2} \quad\) to make \(k\) the subject. [4 marks]
Mark scheme
Mark scheme Answer Mark Comments Alternative method 1: eliminates fraction \(x(5k^2 - 2) = 3k^2\) M1 oe elimination of fraction \(5xk^2 - 3k^2 = 2x\) M1dep oe expansion of bracket and collection of terms in \(k^2\) \(k^2(5x - 3) = 2x\) M1dep oe full factorisation eg \(-2x = k^2(3 - 5x)\) \(k = \sqrt{\dfrac{2x}{5x - 3}}\) A1 oe with \(k\) the subject Alternative method 2: takes reciprocal of both sides \(\dfrac{1}{x} = \dfrac{5k^2 - 2}{3k^2}\) M1 oe reciprocal of both sides \(\dfrac{2}{3k^2} = \dfrac{5x - 3}{3x}\) M1dep oe collection of term in \(\dfrac{1}{k^2}\) \(\dfrac{3k^2}{2} = \dfrac{3x}{5x - 3}\) M1dep oe reciprocal of both sides \(k = \sqrt{\dfrac{2x}{5x - 3}}\) A1 oe with \(k\) the subject
Additional guidance M1 may be awarded for correct work with no answer or incorrect answer, even if this is seen amongst multiple attempts Condone one transcription error for A1 after correct answer seen No miscopies for M marks Condone eg \(k = \sqrt{\dfrac{2x}{5x - 3}}\) in working with \(\sqrt{\dfrac{2x}{5x - 3}}\) on answer line M3A1 For A1 accept eg \(k = +\sqrt{\dfrac{2x}{5x - 3}}\) or \(k = \pm\dfrac{\sqrt{2x}}{\sqrt{5x - 3}}\) Correct answer followed by incorrect further work M3A0 For M1 the terms may all be on one side of an equation but must be recovered to award further marks M1 Choose the alt that favours the student
18 Rearrange \(\quad m = \dfrac{2k^3 - 7}{5m - 3k^3} \quad\) to make \(k\) the subject. [4 marks]
Mark scheme
Mark scheme Answer Mark Comments \(m(5m - 3k^3) = 2k^3 - 7\) M1 oe elimination of fraction \(5m^2 + 7 = 2k^3 + 3mk^3\) M1dep oe expansion of brackets and collection of terms in \(k^3\) \(5m^2 + 7 = k^3(2 + 3m)\) M1dep oe factorisation \(k = \sqrt[3]{\dfrac{5m^2 + 7}{2 + 3m}}\) A1 oe with \(k\) the subject
Additional guidance Up to M1 may be awarded for correct work with no answer or incorrect answer, even if this is seen amongst multiple attempts Condone one transcription error for A1 after fully correct answer seen Condone eg \(k = \sqrt[3]{\dfrac{5m^2 + 7}{2 + 3m}}\) in working lines with \(\sqrt[3]{\dfrac{5m^2 + 7}{2 + 3m}}\) on answer line M3A1 Correct answer followed by incorrect further work M3A0
9 Rearrange \(\quad w = \dfrac{y^2 + 5}{y^2 - 2} \quad\) to make \(y\) the subject. [4 marks]
Mark scheme
Mark scheme Answer Mark Comments \(w(y^2 - 2) = y^2 + 5\) M1 \(wy^2 - 2w = y^2 + 5\) M1dep \(y^2(w - 1) = 2w + 5\) M1dep rearranges and factorises \(y = (\pm)\sqrt{\dfrac{(2w + 5)}{(w - 1)}}\) A1
Additional guidance Root must cover the whole function for final mark (or separate roots over both numerator and denominator) Must be \(y =\) Penalise incorrect further working
14 Rearrange \(\quad ef = \dfrac{5e + 4}{3} \quad\) to make \(e\) the subject. [3 marks]
Mark scheme
Mark scheme Answer Mark Comments Alternative method 1 \(3ef = 5e + 4\) or \(ef - \dfrac{5e}{3} = \dfrac{4}{3}\) M1 \(e(3f - 5) = 4\) or \(e\left(f - \dfrac{5}{3}\right) = \dfrac{4}{3}\) M1dep oe where they are one step away from answer \(e = \dfrac{4}{3f - 5}\) A1 oe eg \(e = \dfrac{\frac{4}{3}}{\left(f - \frac{5}{3}\right)}\) or \(e = \dfrac{-4}{5 - 3f}\) Alternative method 2 \(3f = 5 + \dfrac{4}{e}\) M1 \(\dfrac{4}{e} = 3f - 5\) M1dep oe where they are one step away from answer \(e = \dfrac{4}{3f - 5}\) A1 oe eg \(e = \dfrac{\frac{4}{3}}{\left(f - \frac{5}{3}\right)}\) or \(e = \dfrac{-4}{5 - 3f}\)
Additional guidance Must have \(e =\) on the answer line for full marks
6 Rearrange \(\quad y = \sqrt{\dfrac{x + 2w}{3}} \quad\) to make \(w\) the subject. [3 marks]
Mark scheme
Mark scheme Answer Mark Comments Alternative method 1 \(y^2 = \dfrac{x + 2w}{3}\) M1 \(3y^2 - x = 2w\) M1dep \(w = \dfrac{3y^2 - x}{2}\) A1 Alternative method 2 \(y^2 = \dfrac{x}{3} + \dfrac{2w}{3}\) M1 \(y^2 - \dfrac{x}{3} = \dfrac{2w}{3}\) M1dep \(w = \dfrac{3}{2}\left(y^2 - \dfrac{x}{3}\right)\) A1
Additional guidance Condone eg \(w = \dfrac{3y^2 - x}{2}\) seen in working with \(\dfrac{3y^2 - x}{2}\) on answer line M2A1 \(w = \dfrac{3}{2}y^2 - \dfrac{1}{2}x\) etc M2A1
Specimen 2020 Paper 1 Q13 13 Rearrange \(\quad t = \dfrac{3w^3 + a}{w^3 - 2} \quad\) to make \(w\) the subject. [5 marks]
Mark scheme
Mark scheme Answer Mark Comments \(t(w^3 - 2) = 3w^3 + a\) M1 \(tw^3 - 2t = 3w^3 + a\) M1dep \(tw^3 - 3w^3 = a + 2t\) M1dep \(w^3(t - 3) = a + 2t\) or \(w^3 = \dfrac{a + 2t}{t - 3}\) M1dep \(w = \sqrt[3]{\dfrac{a + 2t}{t - 3}}\) A1
15 Rearrange \(\quad m = \dfrac{2p + 1}{p} + \dfrac{p + 5}{3p} \quad\) to make \(p\) the subject. [4 marks]
Mark scheme
Mark scheme Answer Mark Comments Alternative method 1 \(3mp = 3(2p + 1) + p + 5\) M1 oe fractions eliminated \(3mp = 6p + 3 + p + 5\) M1dep oe brackets expanded and fractions eliminated \(3mp - 7p = 8\) M1dep oe terms collected \(p = \dfrac{8}{3m - 7}\) or \(p = \dfrac{-8}{7 - 3m}\) A1 oe eg \(\dfrac{8}{3m - 7} = p\) Alternative method 2 \((m =)\ \dfrac{3(2p + 1)}{3p} + \dfrac{p + 5}{3p}\) M1 oe common denominator \(m = \dfrac{7p + 8}{3p}\) M1dep simplifies numerator and isolates term in \(p\) \(\dfrac{3m - 7}{3} = \dfrac{8}{3p}\) M1dep converts \(m - \dfrac{7}{3}\) to a single fraction \(p = \dfrac{8}{3m - 7}\) or \(p = \dfrac{-8}{7 - 3m}\) A1 oe eg \(\dfrac{8}{3m - 7} = p\)
Additional guidance \(p = \dfrac{8}{3m - 7}\) in working but \(\dfrac{8}{3m - 7}\) on answer line M3A1 Allow recovery of missing brackets \(p = \dfrac{8}{3m - 7}\) followed by incorrect further work M3A0 Allow equivalences for A1 eg \(p = \dfrac{\frac{8}{3}}{\frac{3m - 7}{3}}\) M3A1 Do not regard eg \(3m(p) = 7p + 8\) as having unexpanded brackets M1M1dep
23 For each of these two function machines, when the input is \(a\) the output is \(b\).
\(k \gt 0 \quad\) and \(\quad k \ne 1 \quad\) and \(\quad a \gt 0\)
Work out an expression for \(a\) in terms of \(k\).[6 marks]
Mark scheme
Mark scheme Answer Mark Comments \((a^2 + 3) \times k\) or \(ka^2 + 3k\) M1 oe eg \(b = ka^2 + 3k\) \((ak)^2 + 3\) or \(a^2k^2 + 3\) M1 oe eg \(b = a^2k^2 + 3\) \(ka^2 + 3k = a^2k^2 + 3\) M1dep oe \(a^2(k - k^2) = 3 - 3k\) M1dep oe eg \((a^2 =)\ \dfrac{3 - 3k}{k - k^2}\) or \((a =)\ (\pm)\sqrt{\dfrac{3 - 3k}{k - k^2}}\) \((a^2 =)\ \dfrac{3(1 - k)}{k(1 - k)} \quad\) or \(\quad (a^2 =)\ \dfrac{3}{k}\) M1dep oe eg \((a^2 =)\ \dfrac{3(k - 1)}{k(k - 1)}\) \((a =)\ \sqrt{\dfrac{3}{k}}\) or \((a =)\ \left(\dfrac{3}{k}\right)^{\frac{1}{2}}\) A1 oe eg \((a =)\ \dfrac{\sqrt{3}}{\sqrt{k}}\) or \((a =)\ \left(\dfrac{k}{3}\right)^{-\frac{1}{2}}\)
Additional guidance Only one machine fully correct M1 only Missing brackets must be recovered
10 Rearrange \(\qquad \dfrac{1}{xy} = 4 - \dfrac{3}{y} \qquad\) to make \(x\) the subject. [3 marks]
Mark scheme
Mark scheme Answer Mark Comments A correct first step using algebra M1 Here are some of the possible alternativesand making the LHS an algebraic fraction Further correct algebra which leads to an equation that is one step from the final answer. M1dep Following two of the above alternatives … A correct final answer in any form A1 \(x = \dfrac{1}{4y - 3} \qquad x = \dfrac{-1}{3 - 4y}\)
Additional guidance There are many ways of scoring the first M mark. They do not need to give any reasons but you need to check that what they do is valid. For the M1dep mark you must check that their algebra is correct and will lead to a result that is one step from the final answer. ‘One step from …’ means that when they divide through, they have a correct version where \(x\) is the subject. Some of the final answers are more compact than others, but we didn’t ask for any simplification so we have to accept a correct answer in any form. … and, finally, one to look out for … correct answer from wrong working … 0 marks
15 Rearrange \(\qquad w = \dfrac{8x - y}{y} \qquad\) to make \(y\) the subject. [3 marks]
Mark scheme
Mark scheme Answer Mark Comments Alternative method 1 \(wy = 8x - y\)or M1 \(wy + y = 8x\)or M1dep oe \(y\) term(s) collected \(y = \dfrac{8x}{w + 1}\) or \(y = \dfrac{-8x}{-w - 1}\) A1 oe eg \(y = \dfrac{4x}{0.5w + 0.5}\) Alternative method 2 \(y = \dfrac{8x}{w} - \dfrac{y}{w}\) M1 \(y + \dfrac{y}{w} = \dfrac{8x}{w}\) or \(y\left(1 + \dfrac{1}{w}\right) = \dfrac{8x}{w}\) M1dep oe \(y\) term(s) collected \(y = \dfrac{\frac{8x}{w}}{1 + \frac{1}{w}}\) A1 oe
Additional guidance \(y = \dfrac{8x}{w + 1}\) in working with \(\dfrac{8x}{w + 1}\) on answer line etc M2A1 Allow multiplications signs and 1s throughout \(w = \dfrac{8x}{y} - \dfrac{y}{y}\) with no further simplification M0 Correct answer followed by incorrect further work M2A0
16 Rearrange \(\quad y = \dfrac{8(w - x)}{x} \quad\) to make \(x\) the subject. [4 marks]
Mark scheme
Mark scheme Answer Mark Comments Alternative method 1 \(yx = 8(w - x)\) or \(y = \dfrac{8w - 8x}{x}\) M1 \(yx = 8w - 8x\) M1dep oe eg \(yx - 8w + 8x = 0\) \(yx + 8x = 8w\) or \(x(y + 8) = 8w\) M1dep oe \(x = \dfrac{8w}{y + 8}\) A1 oe eg \(\dfrac{-8w}{-y - 8}\) Alternative method 2 \(y = \dfrac{8w}{x} - 8\) or \(y = \dfrac{8w}{x} - \dfrac{8x}{x}\) M1 \(y + 8 = \dfrac{8w}{x}\) M1dep oe eg \(y + 8 - \dfrac{8w}{x} = 0\) \(yx + 8x = 8w\) or \(x(y + 8) = 8w\) M1dep oe \(x = \dfrac{8w}{y + 8}\) A1 oe eg \(\dfrac{-8w}{-y - 8}\) Alternative method 3 \(yx = 8(w - x)\) M1 \(\dfrac{yx}{8} = w - x\) M1dep oe eg \(\dfrac{yx}{8} - w + x = 0\) \(\dfrac{yx}{8} + x = w\) or \(x\left(\dfrac{y}{8} + 1\right) = w\) M1dep oe \(x = \dfrac{w}{\frac{y}{8} + 1}\) A1 oe eg \(x = \dfrac{-w}{-\frac{y}{8} - 1}\)
Additional guidance \(x = \dfrac{8w}{y + 8}\) in working with \(\dfrac{8w}{y + 8}\) on answer line M3 A1 \(x = \dfrac{8w}{y + 1}\) in working with \(\dfrac{8w}{y + 1}\) on answer line SC2 3rd M1 is for collecting terms in \(x\) (or \(x\) in numerator in Alt 2) Allow multiplications signs and 1s throughout Correct answer followed by incorrect further work M3 A0
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