Two non-parallel straight lines meet once. Why can an equation involving a modulus have more solutions than that?

Because the modulus reflects part of a graph upwards, and the reflected piece can cross the other graph as well. That produces an intersection the unreflected lines never had.

What should you do before solving a modulus equation algebraically?

Sketch both graphs , including the reflected parts, and locate the intersections. The sketch tells you how many solutions there should be, and therefore which algebraic answers to keep.

A LevelMathsCambridge (CIE)Pure 3FlashcardsAlgebra & FunctionsModulus Functions

Modulus Functions (Cambridge (CIE) A Level Maths: Pure 3): Flashcards

Exam code: 9709

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FrontSketching Graphs of Modulus Functions
Define the modulus function.

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Cards in this collection (12)

Define the modulus function.
The modulus of a number is its size with any minus sign removed, so $|x| = x$ when $x \geq 0$ , and $|x| = - x$ when $x < 0$ .
It is also called the absolute value, so $|5| = 5$ and $|- 5| = 5$ .
Starting from the line $y = a x + b$ , how do you sketch $y = |a x + b|$ ?
Sketch the line first, then reflect the part below the $x$ -axis in the $x$ -axis.
Whatever was already above the axis stays exactly where it is, which is why the result has a sharp corner rather than a smooth turn.
Complete the coordinates of the vertex of the graph $y = a |x + p| + q$ :
$(______,______)$
The completed coordinates are:
$(- p, q)$
The sign flips because the vertex sits where the bracket is zero, and $x + p = 0$ gives $x = - p$ .
What decides whether $y = a |x + p| + q$ opens upwards or downwards?
The sign of $a$ decides it: the graph opens upwards into a V shape when $a > 0$ , and is turned upside down when $a < 0$ .
A modulus is never negative, so only a negative $a$ can send the graph downwards.
True or False?
For a function $f$ , the graphs of $y = |f (x)|$ and $y = f (|x|)$ are the same.
False.
Take $f (x) = 2 x - 6$ : the graph of $y = |2 x - 6|$ has its corner at $(3, 0)$ , while $y = 2 |x| - 6$ has its corner at $(0, - 6)$ and dips below the $x$ -axis.
The modulus is applied to the output in one and to the input in the other, and those are different operations.
For $a \neq 0$ , how many roots can the graph of $y = a |x + p| + q$ have, and what decides the number?
It can have zero, one or two roots.
Which of the three you get depends on where the vertex sits relative to the $x$ -axis and which way the graph opens: a vertex on the axis gives exactly one root, and a graph opening away from the axis gives none at all.
True or False?
Whatever the function $f$ , the graph of $y = f (|x|)$ is symmetrical about the $y$ -axis.
True.
Replacing $x$ by $- x$ leaves $|x|$ unchanged, so the graph reaches the same height at $- x$ as it does at $x$ .
That holds for every $f$ , because the reflection happens to the input before $f$ is applied to it.
Two non-parallel straight lines meet once. Why can an equation involving a modulus have more solutions than that?
Because the modulus reflects part of a graph upwards, and the reflected piece can cross the other graph as well.
That produces an intersection the unreflected lines never had.
To solve $|f (x)| = |g (x)|$ , solve both $f (x) = g (x)$ and $f (x) =______$ .
To solve $|f (x)| = |g (x)|$ , solve both $f (x) = g (x)$ and $f (x) = - g (x)$ .
The two cases arise because each side can take either sign once the modulus is removed.
What should you do before solving a modulus equation algebraically?
Sketch both graphs, including the reflected parts, and locate the intersections.
The sketch tells you how many solutions there should be, and therefore which algebraic answers to keep.
True or False?
Every solution of $x - 4 = 2 x - 5$ is also a solution of $|x - 4| = 2 x - 5$ .
False.
$x = 1$ satisfies the first, but $|1 - 4| = 3$ while $2 (1) - 5 = - 3$ , so it fails the second.
How do you solve $|f (x)| = g (x)$ , with a modulus on one side only?
Solve both $f (x) = g (x)$ and $- f (x) = g (x)$ .
Then test each answer in the original equation, because a modulus can never be negative, so any root making $g (x)$ negative has to be thrown out.

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