This collection presents 20 practice problems on absolute value inequalities, arranged in progressive order of difficulty. Each exercise is solved step by step, with particular attention to the meaning of the equivalences used and to the correct notation for the solution set.
Recall that, for \(k>0\), the following fundamental equivalences hold:
\[ |A(x)|<k \iff -k<A(x)<k \]
\[ |A(x)|\le k \iff -k\le A(x)\le k \]
\[ |A(x)|>k \iff A(x)<-k \quad \text{or} \quad A(x)>k \]
\[ |A(x)|\ge k \iff A(x)\le -k \quad \text{or} \quad A(x)\ge k \]
Exercise 1 — level ★☆☆☆☆
Solve the inequality:
\[ |x-2|<5 \]
Answer
\[ S=(-3,7) \]
Solution
The inequality has the form:
\[ |A(x)|<k \]
with:
\[ A(x)=x-2, \qquad k=5 \]
Since \(5>0\), we can rewrite the absolute value inequality as the compound inequality:
\[ -5<x-2<5 \]
Adding \(2\) to all three parts:
\[ -5+2<x-2+2<5+2 \]
that is:
\[ -3<x<7 \]
Therefore the solution set is:
\[ S=(-3,7) \]
Exercise 2 — level ★☆☆☆☆
Solve the inequality:
\[ |x+4|\le 3 \]
Answer
\[ S=[-7,-1] \]
Solution
The inequality has the form:
\[ |A(x)|\le k \]
with:
\[ A(x)=x+4, \qquad k=3 \]
Since \(3>0\), we can write:
\[ -3\le x+4\le 3 \]
Subtracting \(4\) from all three parts:
\[ -3-4\le x+4-4\le 3-4 \]
We obtain:
\[ -7\le x\le -1 \]
Since the original inequality uses \(\le\), the endpoints are included.
Therefore:
\[ S=[-7,-1] \]
Exercise 3 — level ★☆☆☆☆
Solve the inequality:
\[ |2x-1|>7 \]
Answer
\[ S=(-\infty,-3)\cup(4,+\infty) \]
Solution
The inequality has the form:
\[ |A(x)|>k \]
with:
\[ A(x)=2x-1, \qquad k=7 \]
Since \(7>0\), the absolute value exceeds \(7\) precisely when the expression inside is less than \(-7\) or greater than \(7\). Hence:
\[ 2x-1<-7 \quad \text{or} \quad 2x-1>7 \]
Solving the first inequality:
\[ 2x-1<-7 \]
Adding \(1\):
\[ 2x<-6 \]
Dividing by \(2\):
\[ x<-3 \]
Solving the second inequality:
\[ 2x-1>7 \]
Adding \(1\):
\[ 2x>8 \]
Dividing by \(2\):
\[ x>4 \]
Taking the union of the two solution sets:
\[ S=(-\infty,-3)\cup(4,+\infty) \]
Exercise 4 — level ★☆☆☆☆
Solve the inequality:
\[ |3x+2|\ge 4 \]
Answer
\[ S=\left(-\infty,-2\right]\cup\left[\frac{2}{3},+\infty\right) \]
Solution
The inequality has the form:
\[ |A(x)|\ge k \]
with:
\[ A(x)=3x+2, \qquad k=4 \]
Since \(4>0\), we can write:
\[ 3x+2\le -4 \quad \text{or} \quad 3x+2\ge 4 \]
Solving the first inequality:
\[ 3x+2\le -4 \]
Subtracting \(2\):
\[ 3x\le -6 \]
Dividing by \(3\):
\[ x\le -2 \]
Solving the second inequality:
\[ 3x+2\ge 4 \]
Subtracting \(2\):
\[ 3x\ge 2 \]
Dividing by \(3\):
\[ x\ge \frac{2}{3} \]
Since the original inequality uses \(\ge\), the endpoints are included.
Therefore:
\[ S=\left(-\infty,-2\right]\cup\left[\frac{2}{3},+\infty\right) \]
Exercise 5 — level ★☆☆☆☆
Solve the inequality:
\[ |5-x|<2 \]
Answer
\[ S=(3,7) \]
Solution
The inequality is:
\[ |5-x|<2 \]
Since \(2>0\), we can rewrite it as:
\[ -2<5-x<2 \]
To isolate \(x\), subtract \(5\) from all three parts:
\[ -2-5<5-x-5<2-5 \]
that is:
\[ -7<-x<-3 \]
Multiplying all three parts by \(-1\). Since we are multiplying by a negative number, the inequality signs reverse:
\[ 7>x>3 \]
Writing the interval in increasing order along the real line:
\[ 3<x<7 \]
Therefore:
\[ S=(3,7) \]
Exercise 6 — level ★★☆☆☆
Solve the inequality:
\[ |4-2x|\le 6 \]
Answer
\[ S=[-1,5] \]
Solution
Since \(6>0\), we rewrite the inequality as:
\[ -6\le 4-2x\le 6 \]
Subtracting \(4\) from all three parts:
\[ -10\le -2x\le 2 \]
Dividing by \(-2\). Since we are dividing by a negative number, the inequality signs reverse:
\[ 5\ge x\ge -1 \]
Rewriting the compound inequality in increasing order:
\[ -1\le x\le 5 \]
Therefore:
\[ S=[-1,5] \]
Exercise 7 — level ★★☆☆☆
Solve the inequality:
\[ |2x+3|<|x-1| \]
Answer
\[ S=\left(-4,-\frac{2}{3}\right) \]
Solution
Both sides of this inequality are non-negative. We may therefore square both sides without changing the solution set:
\[ |2x+3|<|x-1| \iff (2x+3)^2<(x-1)^2 \]
Expanding the squares:
\[ 4x^2+12x+9<x^2-2x+1 \]
Bringing all terms to the left-hand side:
\[ 4x^2+12x+9-x^2+2x-1<0 \]
that is:
\[ 3x^2+14x+8<0 \]
Factoring the trinomial:
\[ 3x^2+14x+8=(3x+2)(x+4) \]
The inequality becomes:
\[ (3x+2)(x+4)<0 \]
The zeros of the factors are:
\[ 3x+2=0 \iff x=-\frac{2}{3}, \qquad x+4=0 \iff x=-4 \]
We can verify the factorization:
\[ (3x+2)(x+4)=3x^2+14x+8 \]
which is correct. We now analyse the sign of the product:
\[ (3x+2)(x+4)<0 \]
The product is negative between the two zeros:
\[ -4<x<-\frac{2}{3} \]
As a check, note that \(|2x+3|<|x-1|\) means that \(2x+3\) is closer to \(0\) than \(x-1\) is. Squaring both sides yields:
\[ (2x+3)^2<(x-1)^2 \]
which confirms our approach. Therefore:
\[ S=\left(-4,-\frac{2}{3}\right) \]
Exercise 8 — level ★★☆☆☆
Solve the inequality:
\[ |x-3|>0 \]
Answer
\[ S=(-\infty,3)\cup(3,+\infty) \]
Solution
The absolute value of any expression is always greater than or equal to zero:
\[ |x-3|\ge 0 \]
The inequality requires the absolute value to be strictly greater than zero:
\[ |x-3|>0 \]
An absolute value equals zero if and only if its argument equals zero:
\[ |x-3|=0 \iff x-3=0 \iff x=3 \]
Hence the inequality holds for every real number except \(x=3\).
Therefore:
\[ S=\mathbb{R}\setminus\{3\} \]
or, written as a union of intervals:
\[ S=(-\infty,3)\cup(3,+\infty) \]
Exercise 9 — level ★★☆☆☆
Solve the inequality:
\[ |2x-5|\le 0 \]
Answer
\[ S=\left\{\frac{5}{2}\right\} \]
Solution
The absolute value is always non-negative:
\[ |2x-5|\ge 0 \]
The inequality:
\[ |2x-5|\le 0 \]
can therefore be satisfied only when the absolute value equals zero:
\[ |2x-5|=0 \]
An absolute value is zero if and only if its argument is zero:
\[ 2x-5=0 \]
Solving:
\[ 2x=5 \]
hence:
\[ x=\frac{5}{2} \]
Therefore:
\[ S=\left\{\frac{5}{2}\right\} \]
Exercise 10 — level ★★☆☆☆
Solve the inequality:
\[ |x+2|<-1 \]
Answer
\[ S=\varnothing \]
Solution
The absolute value of any real expression is always greater than or equal to zero:
\[ |x+2|\ge 0 \]
The given inequality requires:
\[ |x+2|<-1 \]
that is, it asks for a non-negative quantity to be less than a negative number. This is impossible.
Hence the inequality has no solution:
\[ S=\varnothing \]
Exercise 11 — level ★★☆☆☆
Solve the inequality:
\[ |3x-6|>-2 \]
Answer
\[ S=\mathbb{R} \]
Solution
The absolute value is always greater than or equal to zero:
\[ |3x-6|\ge 0 \]
The inequality requires:
\[ |3x-6|>-2 \]
Since every absolute value is non-negative, it is certainly greater than \(-2\).
The inequality is therefore satisfied for every real number:
\[ S=\mathbb{R} \]
Exercise 12 — level ★★★☆☆
Solve the inequality:
\[ |x-1|+2<5 \]
Answer
\[ S=(-2,4) \]
Solution
Before applying the rules for absolute values, we isolate the absolute value term.
Starting from:
\[ |x-1|+2<5 \]
Subtracting \(2\) from both sides:
\[ |x-1|<3 \]
Since \(3>0\), we can write:
\[ -3<x-1<3 \]
Adding \(1\) to all three parts:
\[ -2<x<4 \]
Therefore:
\[ S=(-2,4) \]
Exercise 13 — level ★★★☆☆
Solve the inequality:
\[ 2|x+3|-1\ge 7 \]
Answer
\[ S=(-\infty,-7]\cup[1,+\infty) \]
Solution
We begin by isolating the absolute value.
Starting from:
\[ 2|x+3|-1\ge 7 \]
Adding \(1\) to both sides:
\[ 2|x+3|\ge 8 \]
Dividing by \(2\):
\[ |x+3|\ge 4 \]
Since \(4>0\), the inequality is equivalent to:
\[ x+3\le -4 \quad \text{or} \quad x+3\ge 4 \]
Solving the first inequality:
\[ x\le -7 \]
Solving the second:
\[ x\ge 1 \]
Therefore:
\[ S=(-\infty,-7]\cup[1,+\infty) \]
Exercise 14 — level ★★★☆☆
Solve the inequality:
\[ 3-2|x-4|>1 \]
Answer
\[ S=(3,5) \]
Solution
In this exercise the absolute value is multiplied by a negative coefficient. We proceed carefully.
Starting from:
\[ 3-2|x-4|>1 \]
Subtracting \(3\) from both sides:
\[ -2|x-4|>-2 \]
Dividing by \(-2\). Since we are dividing by a negative number, the inequality sign reverses:
\[ |x-4|<1 \]
Since \(1>0\), we can write:
\[ -1<x-4<1 \]
Adding \(4\) to all three parts:
\[ 3<x<5 \]
Therefore:
\[ S=(3,5) \]
Exercise 15 — level ★★★☆☆
Solve the inequality:
\[ |x^2-4|\le 5 \]
Answer
\[ S=[-3,3] \]
Solution
The inequality has the form:
\[ |A(x)|\le k \]
with:
\[ A(x)=x^2-4, \qquad k=5 \]
Since \(5>0\), we can rewrite it as the compound inequality:
\[ -5\le x^2-4\le 5 \]
Adding \(4\) to all three parts:
\[ -1\le x^2\le 9 \]
Since \(x^2\ge 0\) for every \(x\in\mathbb{R}\), the condition:
\[ -1\le x^2 \]
is always satisfied.
It remains to impose:
\[ x^2\le 9 \]
This inequality is equivalent to:
\[ -3\le x\le 3 \]
Therefore:
\[ S=[-3,3] \]
Exercise 16 — level ★★★☆☆
Solve the inequality:
\[ |x^2-1|>3 \]
Answer
\[ S=(-\infty,-2)\cup(2,+\infty) \]
Solution
The inequality has the form:
\[ |A(x)|>k \]
with:
\[ A(x)=x^2-1, \qquad k=3 \]
Since \(3>0\), we must solve the union of two inequalities:
\[ x^2-1<-3 \quad \text{or} \quad x^2-1>3 \]
Considering the first:
\[ x^2-1<-3 \]
Adding \(1\):
\[ x^2<-2 \]
This inequality has no real solutions, since the square of any real number is always greater than or equal to zero.
Considering the second:
\[ x^2-1>3 \]
Adding \(1\):
\[ x^2>4 \]
The inequality \(x^2>4\) holds when \(x\) lies outside the interval \([-2,2]\):
\[ x<-2 \quad \text{or} \quad x>2 \]
Therefore:
\[ S=(-\infty,-2)\cup(2,+\infty) \]
Exercise 17 — level ★★★★☆
Solve the inequality:
\[ |x-1|+|x+2|\le 5 \]
Answer
\[ S=[-3,2] \]
Solution
This inequality involves two absolute values. We therefore use the definition directly, partitioning the real line into the intervals determined by the zeros of the expressions inside the absolute values.
The expressions inside the absolute values are:
\[ x-1, \qquad x+2 \]
Their zeros are:
\[ x-1=0 \iff x=1 \]
\[ x+2=0 \iff x=-2 \]
The critical points are therefore:
\[ -2, \qquad 1 \]
These divide the real line into three intervals:
\[ (-\infty,-2), \qquad [-2,1), \qquad [1,+\infty) \]
Case 1: \(x<-2\)
If \(x<-2\), then:
\[ x-1<0, \qquad x+2<0 \]
So:
\[ |x-1|=-(x-1)=-x+1 \]
\[ |x+2|=-(x+2)=-x-2 \]
The inequality becomes:
\[ -x+1-x-2\le 5 \]
that is:
\[ -2x-1\le 5 \]
Adding \(1\):
\[ -2x\le 6 \]
Dividing by \(-2\), remembering to reverse the inequality sign:
\[ x\ge -3 \]
Intersecting with the condition \(x<-2\), we obtain:
\[ -3\le x<-2 \]
Case 2: \(-2\le x<1\)
If \(-2\le x<1\), then:
\[ x-1<0, \qquad x+2\ge 0 \]
So:
\[ |x-1|=-(x-1)=-x+1 \]
\[ |x+2|=x+2 \]
The inequality becomes:
\[ -x+1+x+2\le 5 \]
that is:
\[ 3\le 5 \]
This is always true, so the entire interval is part of the solution:
\[ -2\le x<1 \]
Case 3: \(x\ge 1\)
If \(x\ge 1\), then:
\[ x-1\ge 0, \qquad x+2>0 \]
So:
\[ |x-1|=x-1 \]
\[ |x+2|=x+2 \]
The inequality becomes:
\[ x-1+x+2\le 5 \]
that is:
\[ 2x+1\le 5 \]
Subtracting \(1\):
\[ 2x\le 4 \]
Dividing by \(2\):
\[ x\le 2 \]
Intersecting with the condition \(x\ge 1\), we obtain:
\[ 1\le x\le 2 \]
Taking the union of the solutions from all three cases:
\[ S= [-3,-2) \cup [-2,1) \cup [1,2] \]
Since these intervals are adjacent, we can write more simply:
\[ S=[-3,2] \]
Exercise 18 — level ★★★★☆
Solve the inequality:
\[ |x+1|-|x-3|>2 \]
Answer
\[ S=(2,+\infty) \]
Solution
The expressions inside the absolute values are:
\[ x+1, \qquad x-3 \]
Their zeros are:
\[ x+1=0 \iff x=-1, \qquad x-3=0 \iff x=3 \]
The critical points are:
\[ -1, \qquad 3 \]
These divide the real line into three intervals:
\[ (-\infty,-1), \qquad [-1,3), \qquad [3,+\infty) \]
Case 1: \(x<-1\)
If \(x<-1\), then:
\[ x+1<0, \qquad x-3<0 \]
So:
\[ |x+1|=-x-1, \qquad |x-3|=-x+3 \]
The inequality becomes:
\[ (-x-1)-(-x+3)>2 \]
that is:
\[ -x-1+x-3>2 \]
hence:
\[ -4>2 \]
This is false, so there are no solutions in this interval.
Case 2: \(-1\le x<3\)
If \(-1\le x<3\), then:
\[ x+1\ge 0, \qquad x-3<0 \]
So:
\[ |x+1|=x+1, \qquad |x-3|=-x+3 \]
The inequality becomes:
\[ x+1-(-x+3)>2 \]
that is:
\[ x+1+x-3>2 \]
hence:
\[ 2x-2>2 \]
Adding \(2\):
\[ 2x>4 \]
Dividing by \(2\):
\[ x>2 \]
Intersecting with \(-1\le x<3\), we obtain:
\[ 2<x<3 \]
Case 3: \(x\ge 3\)
If \(x\ge 3\), then:
\[ x+1>0, \qquad x-3\ge 0 \]
So:
\[ |x+1|=x+1, \qquad |x-3|=x-3 \]
The inequality becomes:
\[ x+1-(x-3)>2 \]
that is:
\[ 4>2 \]
This is always true, so the entire interval is part of the solution:
\[ x\ge 3 \]
Taking the union of the results:
\[ S=(2,3)\cup[3,+\infty) \]
hence:
\[ S=(2,+\infty) \]
Exercise 19 — level ★★★★☆
Solve the inequality:
\[ |2x-1|+|x+2|<6 \]
Answer
\[ S=\left(-\frac{7}{3},\frac{5}{3}\right) \]
Solution
The expressions inside the two absolute values are:
\[ 2x-1, \qquad x+2 \]
Their zeros are:
\[ 2x-1=0 \iff x=\frac{1}{2}, \qquad x+2=0 \iff x=-2 \]
The critical points, ordered on the real line, are:
\[ -2, \qquad \frac{1}{2} \]
We study the three intervals:
\[ (-\infty,-2), \qquad \left[-2,\frac{1}{2}\right), \qquad \left[\frac{1}{2},+\infty\right) \]
Case 1: \(x<-2\)
If \(x<-2\), then:
\[ 2x-1<0, \qquad x+2<0 \]
So:
\[ |2x-1|=-(2x-1)=-2x+1, \qquad |x+2|=-(x+2)=-x-2 \]
The inequality becomes:
\[ -2x+1-x-2<6 \]
that is:
\[ -3x-1<6 \]
Adding \(1\):
\[ -3x<7 \]
Dividing by \(-3\), reversing the inequality sign:
\[ x>-\frac{7}{3} \]
Intersecting with \(x<-2\), we obtain:
\[ -\frac{7}{3}<x<-2 \]
Case 2: \(-2\le x<\frac{1}{2}\)
If \(-2\le x<\frac{1}{2}\), then:
\[ 2x-1<0, \qquad x+2\ge 0 \]
So:
\[ |2x-1|=-2x+1, \qquad |x+2|=x+2 \]
The inequality becomes:
\[ -2x+1+x+2<6 \]
that is:
\[ -x+3<6 \]
Subtracting \(3\):
\[ -x<3 \]
Multiplying by \(-1\), reversing the inequality sign:
\[ x>-3 \]
Intersecting with \(-2\le x<\frac{1}{2}\), the entire interval is part of the solution:
\[ -2\le x<\frac{1}{2} \]
Case 3: \(x\ge \frac{1}{2}\)
If \(x\ge \frac{1}{2}\), then:
\[ 2x-1\ge 0, \qquad x+2>0 \]
So:
\[ |2x-1|=2x-1, \qquad |x+2|=x+2 \]
The inequality becomes:
\[ 2x-1+x+2<6 \]
that is:
\[ 3x+1<6 \]
Subtracting \(1\):
\[ 3x<5 \]
Dividing by \(3\):
\[ x<\frac{5}{3} \]
Intersecting with \(x\ge \frac{1}{2}\), we obtain:
\[ \frac{1}{2}\le x<\frac{5}{3} \]
Taking the union of the partial solutions:
\[ S= \left(-\frac{7}{3},-2\right) \cup \left[-2,\frac{1}{2}\right) \cup \left[\frac{1}{2},\frac{5}{3}\right) \]
Since these intervals are adjacent, we obtain:
\[ S=\left(-\frac{7}{3},\frac{5}{3}\right) \]
Exercise 20 — level ★★★★★
Solve the inequality:
\[ |x-2|\ge 2|x+1| \]
Answer
\[ S=[-4,0] \]
Solution
Both sides of the inequality are non-negative. We may therefore square both sides without changing the solution set.
Starting from:
\[ |x-2|\ge 2|x+1| \]
Squaring both sides:
\[ (x-2)^2\ge \left(2|x+1|\right)^2 \]
Since:
\[ \left(2|x+1|\right)^2=4(x+1)^2 \]
we obtain:
\[ (x-2)^2\ge 4(x+1)^2 \]
Expanding:
\[ x^2-4x+4\ge 4(x^2+2x+1) \]
that is:
\[ x^2-4x+4\ge 4x^2+8x+4 \]
Subtracting \(x^2-4x+4\) from both sides:
\[ 0\ge 3x^2+12x \]
that is:
\[ 3x^2+12x\le 0 \]
Factoring out \(3x\):
\[ 3x(x+4)\le 0 \]
Since \(3>0\), the sign depends on the product:
\[ x(x+4)\le 0 \]
The zeros are:
\[ x=0, \qquad x=-4 \]
The product \(x(x+4)\) is less than or equal to zero between the two zeros, endpoints included:
\[ -4\le x\le 0 \]
Therefore:
\[ S=[-4,0] \]