Math101 · Independent Ontario learning library
Math101learn.math101.caAbsolute Value Equations
An absolute value equation describes a distance condition. If $|u|=k$ with $k>0$, then $u=k$ or $u=-k$ because two points lie distance $k$ from zero.
Distance equations model tolerances, deviations, and symmetric locations. The sign cases teach careful logical branching used in many algebraic equations.
Intuition and core definition
An absolute value equation describes a distance condition. If $|u|=k$ with $k>0$, then $u=k$ or $u=-k$ because two points lie distance $k$ from zero. If $k=0$, only $u=0$; if $k<0$, there is no real solution. These cases must be checked before splitting.
Notation, language, and conditions
For $|x-a|=k$, solutions lie $k$ units from centre $a$: $x=a\pm k$. The bars must be isolated before applying the two-case rule. An equation containing extra terms, such as $2|x-1|+3=11$, is first reduced to $|x-1|=4$.
Why this idea matters
An absolute-value equation translates a fixed distance into two possible directions, producing two branches unless the target distance is zero.
A dependable method
- Isolate the absolute-value expression.
- Inspect the other side: negative means no solution, zero means one central solution, positive means two cases.
- For a positive value, write the inside equal to both the positive and negative target.
- Solve each resulting equation.
- Substitute every candidate into the original, especially when other algebraic operations are present.
Worked example
Representations and interpretation
The graph of $y=|2x-5|$ meets the horizontal line $y=7$ at two points. On a number line, $|2x-5|=7$ expresses distance after a scale and shift; isolating the bars exposes the symmetric pair.
Reasoning about variations
For $|x+4|=-2$, no real point can have negative distance, so splitting into cases would be invalid. For $|x+4|=0$, both proposed cases coincide at $x=-4$.
Common mistakes
How to check your work
- Evaluate the expression inside the bars for every candidate.
- Use symmetry about the centre of $|x-a|$ when applicable.
- Graph both sides and count intersection points.
Practice
- Solve $|x-3|=8$.
- Solve $2|x+1|=0$.
- Solve $|3x|-4=-7$.
Answers and brief solutions
Show answers
- $x=11$ or $x=-5$ $x-3=8$ or $x-3=-8$.
- $x=-1$ $|x+1|=0$ only when its inside is zero.
- No real solution Isolation gives $|3x|=-3$, impossible over the reals.
Synthesis and transfer
Locations exactly five kilometres from a station lie on opposite sides of it; solving the two linear branches captures both positions and substitution rejects any extraneous result.
If the station is coordinate $2$ and the required distance is $5$, the model $|x-2|=5$ splits into $x-2=5$ and $x-2=-5$. The positions $7$ and $-3$ lie on opposite sides but are equally far from the reference point. A negative target distance would produce no solution, while target zero would collapse the two branches into one. This geometry also explains why writing only the positive branch loses information. Substituting each candidate into the original absolute value checks distance directly and is especially important after more complicated algebra creates possible extraneous values.
Related topics
Teaching and accessibility note
Try it yourself
Hints are part of learning. Open one whenever it makes the next step feel possible.
Solve $|x-3|=8$.
- $x-3=8$ or $x-3=-8$.
End of lesson
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