Solve |linear| = linear (no modulus)

Solve equation where modulus of linear expression equals a non-modulus linear expression, e.g. |2x-9| = 5x-3.

16 questions · Moderate -0.8

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Edexcel C3 2013 June Q3
5 marks Moderate -0.8
3. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{0f6fd881-4d4b-4f80-96cc-6da41cc33c60-05_654_967_244_507} \captionsetup{labelformat=empty} \caption{Figure 2}
\end{figure} Figure 2 shows a sketch of the graph with equation \(y = 2 | x | - 5\).
The graph intersects the positive \(x\)-axis at the point \(P\) and the negative \(y\)-axis at the point \(Q\).
  1. State the coordinates of \(P\) and the coordinates of \(Q\).
  2. Solve the equation $$2 | x | - 5 = 3 - x$$
Edexcel C3 Specimen Q1
8 marks Moderate -0.8
  1. The function f is defined by
$$\mathrm { f } : x \mapsto | x - 2 | - 3 , x \in \mathbb { R }$$
  1. Solve the equation \(\mathrm { f } ( x ) = 1\). The function g is defined by $$\mathrm { g } : x \mapsto x ^ { 2 } - 4 x + 11 , x \geq 0$$
  2. Find the range of g .
  3. Find \(g f ( - 1 )\).
OCR MEI C3 2005 June Q1
3 marks Easy -1.2
1 Solve the equation \(| 3 x + 2 | = 1\).
OCR C3 Q8
14 marks Standard +0.3
  1. The functions \(f\) and \(g\) are defined by
$$\begin{aligned} & \mathrm { f } : x \rightarrow | 2 x - 5 | , \quad x \in \mathbb { R } , \\ & \mathrm {~g} : x \rightarrow \ln ( x + 3 ) , \quad x \in \mathbb { R } , \quad x > - 3 \end{aligned}$$
  1. State the range of f .
  2. Evaluate fg(-2).
  3. Solve the equation $$\operatorname { fg } ( x ) = 3$$ giving your answers in exact form.
  4. Show that the equation $$\mathrm { f } ( x ) = \mathrm { g } ( x )$$ has a root, \(\alpha\), in the interval [3,4].
  5. Use the iterative formula $$x _ { n + 1 } = \frac { 1 } { 2 } \left[ 5 + \ln \left( x _ { n } + 3 \right) \right]$$ with \(x _ { 0 } = 3\), to find \(x _ { 1 } , x _ { 2 } , x _ { 3 }\) and \(x _ { 4 }\), giving your answers to 4 significant figures.
  6. Show that your answer for \(x _ { 4 }\) is the value of \(\alpha\) correct to 4 significant figures.
OCR MEI C3 2006 June Q1
3 marks Moderate -0.8
1 Solve the equation \(| 3 x - 2 | = x\).
OCR MEI C3 Q3
6 marks Moderate -0.8
3
  1. Sketch the graph of \(y = | 3 x - 6 |\).
  2. Solve the equation \(| 3 x - 6 | = x + 4\) and illustrate your answer on your graph. \(4 \quad\) Find \(\int x \sin 3 x \mathrm {~d} x\). \(5 \quad\) Make \(x\) the subject of \(t = \ln \sqrt { \frac { 5 } { ( x - 3 ) } }\).
OCR MEI C3 Q11
3 marks Moderate -0.8
11 Solve the equation \(| 3 x - 2 | = x\).
OCR MEI C3 Q12
3 marks Easy -1.2
12 Solve the equation \(| 3 x + 2 | = 1\).
OCR C3 2011 January Q1
3 marks Moderate -0.8
1 Solve the equation \(| 3 x + 4 a | = 5 a\), where \(a\) is a positive constant.
OCR C3 2011 June Q7
8 marks Moderate -0.8
7 The functions \(\mathrm { f } , \mathrm { g }\) and h are defined for all real values of \(x\) by $$\mathrm { f } ( x ) = | x | , \quad \mathrm { g } ( x ) = 3 x + 5 \quad \text { and } \quad \mathrm { h } ( x ) = \mathrm { gg } ( x ) .$$
  1. Solve the equation \(\mathrm { g } ( x + 2 ) = \mathrm { f } ( - 12 )\).
  2. Find \(\mathrm { h } ^ { - 1 } ( x )\).
  3. Determine the values of \(x\) for which $$x + \mathrm { f } ( x ) = 0 .$$
OCR MEI C3 2016 June Q4
4 marks Moderate -0.8
4 By sketching the graphs of \(y = | 2 x + 1 |\) and \(y = - x\) on the same axes, show that the equation \(| 2 x + 1 | = - x\) has two roots. Find these roots.
OCR MEI C3 Q1
3 marks Easy -1.2
1 Solve the equation \(| 3 x + 2 | = 1\).
OCR H240/03 2022 June Q1
3 marks Easy -1.8
1 Solve the equation \(| 2 x - 3 | = 9\).
Edexcel Paper 2 2022 June Q1
4 marks Moderate -0.8
  1. In this question you must show all stages of your working.
Solutions relying entirely on calculator technology are not acceptable. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{824d73c5-525c-4876-ad66-33c8f1664277-02_671_759_383_653} \captionsetup{labelformat=empty} \caption{Figure 1}
\end{figure} Figure 1 shows a sketch of the graph with equation \(y = | 3 - 2 x |\) Solve $$| 3 - 2 x | = 7 + x$$
AQA C3 2007 January Q7
9 marks Moderate -0.3
7
  1. Sketch the graph of \(y = | 2 x |\).
  2. On a separate diagram, sketch the graph of \(y = 4 - | 2 x |\), indicating the coordinates of the points where the graph crosses the coordinate axes.
  3. Solve \(4 - | 2 x | = x\).
  4. Hence, or otherwise, solve the inequality \(4 - | 2 x | > x\).
AQA C3 2015 June Q2
13 marks Moderate -0.3
2
  1. Sketch, on the axes below, the curve with equation \(y = 4 - | 2 x + 1 |\), indicating the coordinates where the curve crosses the axes.
  2. Solve the equation \(x = 4 - | 2 x + 1 |\).
  3. Solve the inequality \(x < 4 - | 2 x + 1 |\).
  4. Describe a sequence of two geometrical transformations that maps the graph of \(y = | 2 x + 1 |\) onto the graph of \(y = 4 - | 2 x + 1 |\).
    [0pt] [4 marks] \section*{Answer space for question 2}
  5. \includegraphics[max width=\textwidth, alt={}, center]{2df59047-3bfe-4b9c-a77f-142bc7506cbc-04_851_1459_1000_319}