Solving Basic Equations
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More questions on this topic: GCSE Algebra: Solving Simple Equations (184) IGCSE Algebra: Solving Simple Equations (250)
1 Solve \(\quad \dfrac{8d - 3}{3d - 7} = \dfrac{5}{2}\) [3 marks]
Mark scheme| Answer | Mark | Comments |
|---|
\(2(8d - 3)\) or \(5(3d - 7)\)
or \(\dfrac{5}{2}(3d - 7)\) or \(\dfrac{2}{5}(8d - 3)\) | M1 | oe eg \(16d - 6\) or \(15d - 35\)
may be seen in an equation |
\(16d - 6 = 15d - 35\)
or \(8d - 3 = \dfrac{15}{2}d - \dfrac{35}{2}\)
or \(\dfrac{16}{5}d - \dfrac{6}{5} = 3d - 7\) | M1dep | oe equation with brackets expanded |
| \(-29\) | A1 | |
Additional guidance
| Up to M2 may be awarded for correct work with no answer or incorrect answer, even if this is seen amongst multiple attempts | |
| Missing brackets must be recovered | |
| Embedded answer | M2A0 |
| \(2\left(\dfrac{8d - 3}{3d - 7}\right) = 5\) with no further correct work | M0 |
2 Solve \(\quad \sqrt[3]{7x - 13} = 2\) [2 marks]
Mark scheme| Answer | Mark | Comments |
|---|
| \(7x - 13 = 2^3\) or \(7x - 13 = 8\) | M1 | oe |
| 3 | A1 | |
16 The curve \(\quad y = 2\sqrt{x - a} + 5 \quad\) passes through the point \((1,\ 8)\)
Work out the value of \(a\). [3 marks]
Mark scheme| Answer | Mark | Comments |
|---|
\(\dfrac{8 - 5}{2} = \sqrt{1 - a}\) or \(\dfrac{3}{2} = \sqrt{1 - a}\)
or
\(3^2 = 2^2(1 - a)\) or \(9 = 4(1 - a)\) | M1 | |
\(1 - a = \left(\dfrac{3}{2}\right)^2\)
or \(1 - a = \dfrac{9}{4}\)
or \(9 = 4 - 4a\)
or \(\dfrac{4 - 9}{4}\) | M1dep | oe equation or calculation
eg \(1 - a = \left(\dfrac{8 - 5}{2}\right)^2\)
or \(1 - a = 2.25\)
or \(\dfrac{9 - 4}{-4}\)
implies M2 |
| \(-\dfrac{5}{4}\) or \(-1.25\) or \(-1\dfrac{1}{4}\) | A1 | |
Additional guidance
| \(3 = 2\sqrt{1 - a}\) | M0 |
| Allow recovery of missing brackets | |
5 Solve \(\quad \sqrt[3]{\left(2\sqrt{x} - 10\right)} = 2\) [3 marks]
Mark scheme| Answer | Mark | Comments |
|---|
\(2\sqrt{x} - 10 = 2^3\) or \(2\sqrt{x} - 10 = 8\)
or \(2\sqrt{x} = 18\) | M1 | |
\(\sqrt{x} = \dfrac{2^3+10}{2}\) or \(\sqrt{x} = \dfrac{8+10}{2}\)
or \(\sqrt{x} = 9\) or \(4x = 18^2\) or \(x = 9^2\) | M1dep | |
| \(x = 81\) | A1 | \(\pm 81\) scores A0 |
15
(a) Solve \(\quad \dfrac{2}{5}\sqrt{x} = 1\) [2 marks]
(b) Solve \(\quad x^3 = 5x^2\) [2 marks]
Mark scheme (a)| Answer | Mark | Comments |
|---|
\(\sqrt{x} = \dfrac{5}{2}\) or \(\sqrt{x} = 2.5\) or \(\sqrt{x} = \dfrac{1}{0.4}\)
or \(\dfrac{4}{25}x = 1\) or \(4x = 25\) | M1 | oe
Must have \(\sqrt{x} = \ldots\) or have eliminated \(\sqrt{\phantom{x}}\) with no errors |
| \(\dfrac{25}{4}\) or 6.25 | A1 | oe |
Additional guidance
| \(2\sqrt{x} = 5\) (no further correct work) | M0 |
| Allow unprocessed values for M1 eg \(\dfrac{4}{25}x = 1^2\) or \(\left(\dfrac{2}{5}\right)^2x = 1\) | M1 |
| \(\sqrt{x}\) may be seen as \(x^{\frac{1}{2}}\) or \(x^{0.5}\) | |
Mark scheme (b)| Answer | Mark | Comments |
|---|
\(x^2(x - 5)\ (= 0)\) or \(x^2(5 - x)\ (= 0)\)
or \((x =)\ 0\) or \((x =)\ 5\) | M1 | oe factorisation
eg1 \((x^2 - 0)(x - 5)\)
eg2 \(x(x^2 - 5x)\) |
| 0 and 5 with no other solutions | A1 | |
Additional guidance
For A1, the word ‘and’ is not needed
If their answer has both 0 and 5 with no other solutions award M1 A1
eg 0, 5 with no other solutions | M1 A1 |
| 0, 5, \(-5\) | M1 A0 |
| Either or both solutions seen embedded | M1 A0 |
8 Solve \(\qquad (3 - \sqrt{x})^{\frac{1}{3}} = -2\) [3 marks]
Mark scheme| Answer | Mark | Comments |
|---|
| \((-2)^3\) or \(-8\) seen | B1 | |
\(-\sqrt{x} =\) (their \(-8\)) \(- 3\) or \(-\sqrt{x} = -11\)
or \(\sqrt{x} = 11\) | M1 | |
| 121 | A1 | |
Additional guidance
\(-2^3\) (no brackets) is B0 unless \(-8\) seen
For M1 it must say \(\sqrt{x} = \ldots\ldots\) or \(-\sqrt{x} = \ldots\ldots\) Note: ... (their \(-8\)) cannot be \(-2\)
... and it must be correct manipulation from their \(-8\)
eg \(\quad 3 - \sqrt{x} = (-2)^3\) or \(3 - \sqrt{x} = -8 \qquad\) B1
\(\sqrt{x} = -11 \qquad\) M0 (error in manipulating terms)
\(x = 121 \qquad\) A0 (correct answer from wrong working)
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