A progressive collection of step-by-step solved exercises to learn how to correctly transform absolute value equations, distinguish the necessary cases, and verify the solutions obtained.
Exercise 1 — level ★☆☆☆☆
\[ |x|=7 \]
Answer
\[ x=-7 \quad \text{or} \quad x=7 \]
Solution
If the absolute value of a number equals \(7\), then the number can be either \(7\) or \(-7\).
\[ |x|=7 \iff x=7 \quad \text{or} \quad x=-7 \]
Therefore: \[ \boxed{x=-7 \quad \text{or} \quad x=7} \]
Exercise 2 — level ★☆☆☆☆
\[ |x-3|=5 \]
Answer
\[ x=-2 \quad \text{or} \quad x=8 \]
Solution
We use the property: \[ |A|=k \iff A=k \quad \text{or} \quad A=-k \] with \(k\ge 0\).
We then have: \[ x-3=5 \quad \text{or} \quad x-3=-5 \]
We solve the first equation: \[ x-3=5 \Rightarrow x=8 \]
We solve the second: \[ x-3=-5 \Rightarrow x=-2 \]
Therefore: \[ \boxed{x=-2 \quad \text{or} \quad x=8} \]
Exercise 3 — level ★☆☆☆☆
\[ |2x-1|=9 \]
Answer
\[ x=-4 \quad \text{or} \quad x=5 \]
Solution
The absolute value equals \(9\), so the inner expression can be either \(9\) or \(-9\).
\[ 2x-1=9 \quad \text{or} \quad 2x-1=-9 \]
First case: \[ 2x-1=9 \Rightarrow 2x=10 \Rightarrow x=5 \]
Second case: \[ 2x-1=-9 \Rightarrow 2x=-8 \Rightarrow x=-4 \]
Therefore: \[ \boxed{x=-4 \quad \text{or} \quad x=5} \]
Exercise 4 — level ★★☆☆☆
\[ |x+2|=0 \]
Answer
\[ x=-2 \]
Solution
An absolute value equals zero only when the expression inside it equals zero.
\[ |x+2|=0 \iff x+2=0 \]
We solve: \[ x+2=0 \Rightarrow x=-2 \]
Therefore: \[ \boxed{x=-2} \]
Exercise 5 — level ★★☆☆☆
\[ |3x+6|=12 \]
Answer
\[ x=-6 \quad \text{or} \quad x=2 \]
Solution
The expression inside the absolute value can equal either \(12\) or \(-12\).
\[ 3x+6=12 \quad \text{or} \quad 3x+6=-12 \]
First case: \[ 3x+6=12 \Rightarrow 3x=6 \Rightarrow x=2 \]
Second case: \[ 3x+6=-12 \Rightarrow 3x=-18 \Rightarrow x=-6 \]
Therefore: \[ \boxed{x=-6 \quad \text{or} \quad x=2} \]
Exercise 6 — level ★★☆☆☆
\[ |x-4|=-3 \]
Answer
\[ \varnothing \]
Solution
The absolute value of any expression is always greater than or equal to zero: \[ |x-4|\ge 0 \]
For this reason, it can never equal a negative number.
Since: \[ -3<0 \] the equation has no solution.
Therefore: \[ \boxed{\varnothing} \]
Exercise 7 — level ★★☆☆☆
\[ |x-1|=x+3 \]
Answer
\[ x=-1 \]
Solution
The right-hand side must be non-negative: \[ x+3\ge 0 \Rightarrow x\ge -3 \]
We solve by splitting into the two cases of the absolute value.
First case: \[ x-1=x+3 \] \[ -1=3 \] impossible.
Second case: \[ -(x-1)=x+3 \] \[ -x+1=x+3 \] \[ -2x=2 \] \[ x=-1 \]
The solution we found satisfies the condition \(x\ge -3\), so it is acceptable.
Therefore: \[ \boxed{x=-1} \]
Exercise 8 — level ★★☆☆☆
\[ |2x-5|=x+1 \]
Answer
\[ x=\frac{4}{3} \quad \text{or} \quad x=6 \]
Solution
We first require the right-hand side to be non-negative: \[ x+1\ge 0 \Rightarrow x\ge -1 \]
Then we solve the two cases.
First case: \[ 2x-5=x+1 \] \[ x=6 \]
Second case: \[ -(2x-5)=x+1 \] \[ -2x+5=x+1 \] \[ -3x=-4 \] \[ x=\frac{4}{3} \]
Both solutions satisfy \(x\ge -1\).
Therefore: \[ \boxed{x=\frac{4}{3} \quad \text{or} \quad x=6} \]
Exercise 9 — level ★★★☆☆
\[ |x+4|=2x-1 \]
Answer
\[ x=5 \]
Solution
The right-hand side must be non-negative: \[ 2x-1\ge 0 \] \[ x\ge \frac{1}{2} \]
First case: \[ x+4=2x-1 \] \[ x=5 \]
Second case: \[ -(x+4)=2x-1 \] \[ -x-4=2x-1 \] \[ -3=3x \] \[ x=-1 \]
The solution \(x=-1\) does not satisfy the condition \(x\ge \frac{1}{2}\), so it is discarded.
What remains is: \[ \boxed{x=5} \]
Exercise 10 — level ★★★☆☆
\[ |x-2|=|x+6| \]
Answer
\[ x=-2 \]
Solution
Two absolute values are equal when the inner expressions are either equal or opposite.
First case: \[ x-2=x+6 \] \[ -2=6 \] impossible.
Second case: \[ x-2=-(x+6) \] \[ x-2=-x-6 \] \[ 2x=-4 \] \[ x=-2 \]
Therefore: \[ \boxed{x=-2} \]
Exercise 11 — level ★★★☆☆
\[ |2x+3|=|x-5| \]
Answer
\[ x=\frac{2}{3} \quad \text{or} \quad x=-8 \]
Solution
We solve by requiring the two expressions to be either equal or opposite.
First case: \[ 2x+3=x-5 \] \[ x=-8 \]
Second case: \[ 2x+3=-(x-5) \] \[ 2x+3=-x+5 \] \[ 3x=2 \] \[ x=\frac{2}{3} \]
Therefore: \[ \boxed{x=\frac{2}{3} \quad \text{or} \quad x=-8} \]
Exercise 12 — level ★★★☆☆
\[ |3x-2|=|x+4| \]
Answer
\[ x=3 \quad \text{or} \quad x=-\frac{1}{2} \]
Solution
In this case as well we use: \[ |A|=|B| \iff A=B \quad \text{or} \quad A=-B \]
First case: \[ 3x-2=x+4 \] \[ 2x=6 \] \[ x=3 \]
Second case: \[ 3x-2=-(x+4) \] \[ 3x-2=-x-4 \] \[ 4x=-2 \] \[ x=-\frac{1}{2} \]
Therefore: \[ \boxed{x=3 \quad \text{or} \quad x=-\frac{1}{2}} \]
Exercise 13 — level ★★★★☆
\[ |x-1|+|x+3|=6 \]
Answer
\[ x=-4 \quad \text{or} \quad x=2 \]
Solution
The critical points are those that make the arguments of the absolute values vanish: \[ x-1=0 \Rightarrow x=1 \] \[ x+3=0 \Rightarrow x=-3 \]
We then study the intervals: \[ x<-3, \quad -3\le x<1, \quad x\ge 1 \]
First interval: \(x<-3\). On this interval, both expressions are negative: \[ |x-1|=-(x-1), \qquad |x+3|=-(x+3) \] So: \[ -x+1-x-3=6 \] \[ -2x-2=6 \] \[ -2x=8 \] \[ x=-4 \] The solution belongs to the interval \(x<-3\), so it is valid.
Second interval: \(-3\le x<1\). On this interval: \[ |x-1|=-(x-1), \qquad |x+3|=x+3 \] So: \[ -x+1+x+3=6 \] \[ 4=6 \] impossible.
Third interval: \(x\ge 1\). On this interval, both expressions are non-negative: \[ |x-1|=x-1, \qquad |x+3|=x+3 \] So: \[ x-1+x+3=6 \] \[ 2x+2=6 \] \[ 2x=4 \] \[ x=2 \] The solution belongs to the interval \(x\ge 1\), so it is valid.
Therefore: \[ \boxed{x=-4 \quad \text{or} \quad x=2} \]
Exercise 14 — level ★★★★☆
\[ |x+2|+|x-4|=10 \]
Answer
\[ x=-4 \quad \text{or} \quad x=6 \]
Solution
The critical points are: \[ x+2=0 \Rightarrow x=-2 \] \[ x-4=0 \Rightarrow x=4 \]
We consider the intervals: \[ x<-2, \quad -2\le x<4, \quad x\ge 4 \]
First interval: \(x<-2\). \[ |x+2|=-(x+2), \qquad |x-4|=-(x-4) \] \[ -x-2-x+4=10 \] \[ -2x+2=10 \] \[ -2x=8 \] \[ x=-4 \] Valid since \(x<-2\).
Second interval: \(-2\le x<4\). \[ |x+2|=x+2, \qquad |x-4|=-(x-4) \] \[ x+2-x+4=10 \] \[ 6=10 \] impossible.
Third interval: \(x\ge 4\). \[ |x+2|=x+2, \qquad |x-4|=x-4 \] \[ x+2+x-4=10 \] \[ 2x-2=10 \] \[ 2x=12 \] \[ x=6 \] Valid since \(x\ge 4\).
Therefore: \[ \boxed{x=-4 \quad \text{or} \quad x=6} \]
Exercise 15 — level ★★★★☆
\[ |2x-1|+|x+2|=8 \]
Answer
\[ x=-3 \quad \text{or} \quad x=\frac{7}{3} \]
Solution
We find the critical points: \[ 2x-1=0 \Rightarrow x=\frac{1}{2} \] \[ x+2=0 \Rightarrow x=-2 \]
The intervals to study are: \[ x<-2, \quad -2\le x<\frac{1}{2}, \quad x\ge \frac{1}{2} \]
First interval: \(x<-2\). \[ |2x-1|=-(2x-1), \qquad |x+2|=-(x+2) \] \[ -2x+1-x-2=8 \] \[ -3x-1=8 \] \[ -3x=9 \] \[ x=-3 \] Valid since \(x<-2\).
Second interval: \(-2\le x<\frac{1}{2}\). \[ |2x-1|=-(2x-1), \qquad |x+2|=x+2 \] \[ -2x+1+x+2=8 \] \[ -x+3=8 \] \[ x=-5 \] It does not belong to the interval \(-2\le x<\frac{1}{2}\), so it is discarded.
Third interval: \(x\ge \frac{1}{2}\). \[ |2x-1|=2x-1, \qquad |x+2|=x+2 \] \[ 2x-1+x+2=8 \] \[ 3x+1=8 \] \[ 3x=7 \] \[ x=\frac{7}{3} \] Valid since \(x\ge \frac{1}{2}\).
Therefore: \[ \boxed{x=-3 \quad \text{or} \quad x=\frac{7}{3}} \]
Exercise 16 — level ★★★★☆
\[ |x-2|+|x+1|=3 \]
Answer
\[ -1\le x\le 2 \]
Solution
The critical points are: \[ x=-1, \qquad x=2 \]
We study the three intervals: \[ x<-1, \quad -1\le x<2, \quad x\ge 2 \]
First interval: \(x<-1\). \[ |x-2|=-(x-2), \qquad |x+1|=-(x+1) \] \[ -x+2-x-1=3 \] \[ -2x+1=3 \] \[ -2x=2 \] \[ x=-1 \] This solution does not belong to the interval \(x<-1\), so it is not accepted in this case.
Second interval: \(-1\le x<2\). \[ |x-2|=-(x-2), \qquad |x+1|=x+1 \] \[ -x+2+x+1=3 \] \[ 3=3 \] The identity is true for every value in the interval.
All the values below are therefore solutions: \[ -1\le x<2 \]
Third interval: \(x\ge 2\). \[ |x-2|=x-2, \qquad |x+1|=x+1 \] \[ x-2+x+1=3 \] \[ 2x-1=3 \] \[ 2x=4 \] \[ x=2 \] Valid since \(x\ge 2\).
Combining the results: \[ \boxed{-1\le x\le 2} \]
Exercise 17 — level ★★★★★
\[ |x-3|-|x+1|=2 \]
Answer
\[ x=0 \]
Solution
The critical points are: \[ x=-1, \qquad x=3 \]
We study the intervals: \[ x<-1, \quad -1\le x<3, \quad x\ge 3 \]
First interval: \(x<-1\). \[ |x-3|=-(x-3), \qquad |x+1|=-(x+1) \] \[ -x+3-(-x-1)=2 \] \[ -x+3+x+1=2 \] \[ 4=2 \] impossible.
Second interval: \(-1\le x<3\). \[ |x-3|=-(x-3), \qquad |x+1|=x+1 \] \[ -x+3-(x+1)=2 \] \[ -2x+2=2 \] \[ -2x=0 \] \[ x=0 \] Valid since \(0\in[-1,3)\).
Third interval: \(x\ge 3\). \[ |x-3|=x-3, \qquad |x+1|=x+1 \] \[ x-3-(x+1)=2 \] \[ -4=2 \] impossible.
We verify the solution by substitution: \[ |0-3|-|0+1|=3-1=2 \;\checkmark \]
Therefore: \[ \boxed{x=0} \]
Exercise 18 — level ★★★★★
\[ |2x+1|-|x-4|=3 \]
Answer
\[ x=-8 \quad \text{or} \quad x=2 \]
Solution
The critical points are: \[ 2x+1=0 \Rightarrow x=-\frac{1}{2} \] \[ x-4=0 \Rightarrow x=4 \]
We study the intervals: \[ x<-\frac{1}{2}, \quad -\frac{1}{2}\le x<4, \quad x\ge 4 \]
First interval: \(x<-\frac{1}{2}\). \[ |2x+1|=-(2x+1), \qquad |x-4|=-(x-4) \] \[ -2x-1-(-x+4)=3 \] \[ -2x-1+x-4=3 \] \[ -x-5=3 \] \[ x=-8 \] Valid since \(-8<-\frac{1}{2}\).
Second interval: \(-\frac{1}{2}\le x<4\). \[ |2x+1|=2x+1, \qquad |x-4|=-(x-4) \] \[ 2x+1-(-x+4)=3 \] \[ 2x+1+x-4=3 \] \[ 3x-3=3 \] \[ 3x=6 \] \[ x=2 \] Valid since \(2\in\left[-\frac{1}{2},4\right)\).
Third interval: \(x\ge 4\). \[ |2x+1|=2x+1, \qquad |x-4|=x-4 \] \[ 2x+1-(x-4)=3 \] \[ x+5=3 \] \[ x=-2 \] It does not belong to the interval \(x\ge 4\), so it is discarded.
We verify the solutions by substitution: \[ x=-8: \quad |-15|-|-12|=15-12=3 \;\checkmark \] \[ x=2: \quad |5|-|-2|=5-2=3 \;\checkmark \]
Therefore: \[ \boxed{x=-8 \quad \text{or} \quad x=2} \]
Exercise 19 — level ★★★★★
\[ |x-1|+|2x+4|=9 \]
Answer
\[ x=-4 \quad \text{or} \quad x=2 \]
Solution
The critical points are: \[ x-1=0 \Rightarrow x=1 \] \[ 2x+4=0 \Rightarrow x=-2 \]
We study the intervals: \[ x<-2, \quad -2\le x<1, \quad x\ge 1 \]
First interval: \(x<-2\). \[ |x-1|=-(x-1), \qquad |2x+4|=-(2x+4) \] \[ -x+1-2x-4=9 \] \[ -3x-3=9 \] \[ -3x=12 \] \[ x=-4 \] Valid since \(x<-2\).
Second interval: \(-2\le x<1\). \[ |x-1|=-(x-1), \qquad |2x+4|=2x+4 \] \[ -x+1+2x+4=9 \] \[ x+5=9 \] \[ x=4 \] It does not belong to the interval \(-2\le x<1\), so it is discarded.
Third interval: \(x\ge 1\). \[ |x-1|=x-1, \qquad |2x+4|=2x+4 \] \[ x-1+2x+4=9 \] \[ 3x+3=9 \] \[ 3x=6 \] \[ x=2 \] Valid since \(x\ge 1\).
Therefore: \[ \boxed{x=-4 \quad \text{or} \quad x=2} \]
Exercise 20 — level ★★★★★
\[ |x+2|+|x-1|+|x-4|=9 \]
Answer
\[ x=-2 \quad \text{or} \quad x=4 \]
Solution
The critical points are: \[ x=-2, \qquad x=1, \qquad x=4 \]
We study the intervals: \[ x<-2, \quad -2\le x<1, \quad 1\le x<4, \quad x\ge 4 \]
First interval: \(x<-2\). \[ |x+2|=-(x+2), \quad |x-1|=-(x-1), \quad |x-4|=-(x-4) \] \[ -x-2-x+1-x+4=9 \] \[ -3x+3=9 \] \[ -3x=6 \] \[ x=-2 \] This value does not belong to the interval \(x<-2\), but it will be included in the next interval if it turns out to be valid.
Second interval: \(-2\le x<1\). \[ |x+2|=x+2, \quad |x-1|=-(x-1), \quad |x-4|=-(x-4) \] \[ x+2-x+1-x+4=9 \] \[ -x+7=9 \] \[ x=-2 \] Valid since \(-2\in[-2,1)\).
Third interval: \(1\le x<4\). \[ |x+2|=x+2, \quad |x-1|=x-1, \quad |x-4|=-(x-4) \] \[ x+2+x-1-x+4=9 \] \[ x+5=9 \] \[ x=4 \] This value does not belong to the interval \(1\le x<4\), but it will be included in the next interval if it turns out to be valid.
Fourth interval: \(x\ge 4\). \[ |x+2|=x+2, \quad |x-1|=x-1, \quad |x-4|=x-4 \] \[ x+2+x-1+x-4=9 \] \[ 3x-3=9 \] \[ 3x=12 \] \[ x=4 \] Valid since \(x\ge 4\).
Therefore: \[ \boxed{x=-2 \quad \text{or} \quad x=4} \]