Exercises — Absolute value equations and inequalities

Distance from zero equal to $9$ has two answers, one on each side. $x=9\text{or}x=-9$ Answer: $x\in \{-9,9\}$.

The center is where the inside is zero: $x-4=0\Rightarrow x=4$. We step $6$ units each way.

• Case 1: $x-4=6\Rightarrow x=10$
• Case 2: $x-4=-6\Rightarrow x=-2$

Check: $|10-4|=6$ ✓ and $|-2-4|=|-6|=6$ ✓. Answer: $x\in \{-2,10\}$.

An absolute value is a distance, so it is never negative. There is nothing on the number line whose distance from zero is $-2$. Answer: no solution, $x\in \varnothing$.

• Case 1: $2x-1=7\Rightarrow 2x=8\Rightarrow x=4$
• Case 2: $2x-1=-7\Rightarrow 2x=-6\Rightarrow x=-3$

Check: $|2(4)-1|=|7|=7$ ✓, $|2(-3)-1|=|-7|=7$ ✓. Answer: $x\in \{-3,4\}$.

"Less than" means between (an AND). Distance from $x$ to $2$ is at most $5$: $-5\le x-2\le 5$ Add $2$ to every part: $-3\le x\le 7$ Answer: $x\in [-3,7]$. (Center $2$, reach $5$ each way: $2-5=-3$, $2+5=7$.)

"Greater than" means outside (an OR) — the split we justified at the top of the page.

• Case 1: $3x+2>4\Rightarrow 3x>2$. Now divide both sides by $+3$; since $3$ is positive the inequality direction is unchanged: $x>\frac{2}{3}$.
• Case 2: $3x+2<-4\Rightarrow 3x<-6$. Divide by $+3$ (positive, direction kept): $x<-2$.

Answer: $x\in (-\infty ,-2)\cup \left(\frac{2}{3},\infty \right)$.

The figure marks the two open circles at $-2$ and $\frac{2}{3}$ (excluded, because "$>$" is strict) and shades the two coral rays that make up the answer.

This is the standard "$\ge$" case — the closed-end cousin of Exercise 2.3. By the OR-split (with $\ge$ keeping the endpoints):

• Case 1: $x-1\ge 3\Rightarrow x\ge 4$
• Case 2: $x-1\le -3\Rightarrow x\le -2$

Because it is "$\ge$" (not strict), the endpoints are included, so we use square brackets at $-2$ and $4$. Answer: $x\in (-\infty ,-2]\cup [4,\infty )$. (Center $1$, reach $3$: $1-3=-2$, $1+3=4$; keep everything at or beyond those.)

The right side must be $\ge 0$ (a distance can't equal a negative): $2x+3\ge 0\Rightarrow x\ge -\frac{3}{2}$.

• Case 1: $x-1=2x+3\Rightarrow -1-3=2x-x\Rightarrow x=-4$
  • Check RHS sign: $x=-4<-\frac{3}{2}$ ✗ → reject.
• Case 2: $x-1=-(2x+3)=-2x-3\Rightarrow 3x=-2\Rightarrow x=-\frac{2}{3}$
  • Check RHS sign: $-\frac{2}{3}\ge -\frac{3}{2}$ ✓. Verify: $|-\frac{2}{3}-1|=\frac{5}{3}$ and $2(-\frac{2}{3})+3=\frac{5}{3}$ ✓.

Answer: $x=-\frac{2}{3}$ only.

RHS non-negative: $x+5\ge 0\Rightarrow x\ge -5$.

• Case 1: $2x+1=x+5\Rightarrow x=4$. Check: $4\ge -5$ ✓; $|9|=9=4+5$ ✓.
• Case 2: $2x+1=-(x+5)=-x-5\Rightarrow 3x=-6\Rightarrow x=-2$. Check: $-2\ge -5$ ✓; $|2(-2)+1|=|-3|=3=-2+5$ ✓.

Answer: $x\in \{-2,4\}$.

RHS must be positive for any solution: $2x-1>0\Rightarrow x>\frac{1}{2}$. "Less than" (with RHS treated as $b$) unfolds to $-(2x-1)<x+4<2x-1$, valid only where $2x-1>0$.

• Left: $-(2x-1)<x+4\Rightarrow -2x+1<x+4\Rightarrow -3<3x$. Now divide both sides by $+3$ — $3$ is positive, so the "$<$" stays a "$<$": $-1<x$, i.e. $x>-1$.
• Right: $x+4<2x-1\Rightarrow 5<x\Rightarrow x>5$.

Intersect all three conditions ($x>\frac{1}{2}$, $x>-1$, $x>5$): the binding one is $x>5$. Answer: $x\in (5,\infty )$.

Critical points (where each inside hits zero): $x=2$ and $x=-1$. They cut the line into three regions.

Region A, $x<-1$ (both insides negative): $-(x-2)-(x+1)=7\Rightarrow -2x+1=7\Rightarrow x=-3$. Check $-3<-1$ ✓.

Region B, $-1\le x<2$ (first negative, second positive): $-(x-2)+(x+1)=7\Rightarrow 3=7$ ✗. No solution here.

Region C, $x\ge 2$ (both positive): $(x-2)+(x+1)=7\Rightarrow 2x-1=7\Rightarrow x=4$. Check $4\ge 2$ ✓.

Answer: $x\in \{-3,4\}$.

The figure plots $y=|x-2|+|x+1|$ (lavender) against the line $y=7$ (dashed coral). The two coral dots where they cross are exactly $x=-3$ and $x=4$; the flat valley in the middle (height $3$) is why Region B could never reach $7$.

Let $A=x^2-4$. Then $A=3$ or $A=-3$.

• $x^2-4=3\Rightarrow x^2=7\Rightarrow x=\pm \sqrt{7}$
• $x^2-4=-3\Rightarrow x^2=1\Rightarrow x=\pm 1$

All four are real and valid (an absolute value equalling $+3$ is fine). Answer: $x\in \{-\sqrt{7},-1,1,\sqrt{7}\}$.

First note $2x-6=2(x-3)$, so $|2x-6|=|2|\cdot |x-3|=2|x-3|=|x-3|+|x-3|$ — true for all $x$ (identity), using $|ab|=|a||b|$. Now solve $|2x-6|\le 4$, i.e. $2|x-3|\le 4\Rightarrow |x-3|\le 2$: $-2\le x-3\le 2\Rightarrow 1\le x\le 5$ Answer: $x\in [1,5]$.

$|x-5|=b$ asks for points at distance $b$ from $5$.

• $b>0$: two solutions, $x=5\pm b$ → (a) any $b>0$.
• $b=0$: the two collapse to one, $x=5$ → (b) $b=0$.
• $b<0$: distance can't be negative → (c) any $b<0$ (answer set $\varnothing$).

Answer: (a) $b>0$; (b) $b=0$; (c) $b<0$.

Critical points $x=1$, $x=-2$; three regions.

Region A, $x<-2$: $-(x-1)-[-(x+2)]=(1-x)-(-x-2)=3$. Equation $3=1$ ✗.

Region B, $-2\le x<1$: $-(x-1)-(x+2)=(1-x)-(x+2)=-2x-1$. Set $-2x-1=1\Rightarrow x=-1$. Check $-2\le -1<1$ ✓.

Region C, $x\ge 1$: $(x-1)-(x+2)=-3$. Equation $-3=1$ ✗.

Answer: $x=-1$. (Notice the outer regions give constants: far left the difference is fixed at $+3$, far right at $-3$ — a signature of $|x-a|-|x-b|$ being flat outside $[a,b]$.)

$|x-1|+|x-4|$ is the total distance from $x$ to the two points $1$ and $4$. Its minimum value is the gap between them, $4-1=3$, reached for any $x\in [1,4]$. Since the smallest possible left side is $3$, it can never be $<2$. Answer: no solution, $x\in \varnothing$.

By the same distance reading, the sum equals its minimum $3$ exactly on the whole segment between the two points.

• Region $x<1$: $(1-x)+(4-x)=5-2x=3\Rightarrow x=1$ (boundary, include).
• Region $1\le x\le 4$: $(x-1)+(4-x)=3$ ✓ for all such $x$.
• Region $x>4$: $(x-1)+(x-4)=2x-5=3\Rightarrow x=4$ (boundary).

Answer: $x\in [1,4]$ — an entire interval, not isolated points.

"$<$" (less than) gives ::: one interval, an AND (between $-b$ and $b$) "$>$" or "$\ge$" (greater than) gives ::: two intervals, an OR (outside $\pm b$), joined by $\cup$ Before accepting a root of $|A|=(\text{expression})$, you must check ::: that the right-hand side is $\ge 0$ The minimum of $|x-a|+|x-b|$ equals ::: the distance $|a-b|$, achieved on the whole segment $[a,b]$