Absolute Value Equations and Inequalities: Free Response

The expression inside the bars changes sign where $2x-6=0$, that is $x=3$. Apply the definition $\lvert X \rvert = X$ when $X \ge 0$ and $\lvert X \rvert = -X$ when $X < 0$, with $X = 2x-6$:

$\lvert 2x-6 \rvert = \begin{cases} 2x-6 & x \ge 3, \\ 6-2x & x < 3. \end{cases}$

Both formulas agree at the boundary $x=3$, where each gives $0$.

Compare each input with the boundary $x=3$ found in part A.

At $x=0$, since $0<3$, use $6-2x$:

$6-2(0)=6.$

At $x=4$, since $4\ge3$, use $2x-6$:

$2(4)-6=2.$

A single linear formula, say $2x-6$ itself, agrees with $\lvert 2x-6 \rvert$ only where $2x-6\ge0$; the moment $2x-6$ turns negative, its true absolute value flips sign relative to that formula. The definition of absolute value

$\lvert X \rvert = X \ (X\ge0), \qquad \lvert X \rvert = -X \ (X<0)$

builds exactly this fork in for every $X$, so $\lvert 2x-6 \rvert$ needs both branches, one for each side of its boundary $x=3$; no single linear formula can cover both sides at once.

With $X=x+4$ and $Y=2x-7$, solve $X=Y$:

$x+4 = 2x-7 \quad\Longrightarrow\quad 11 = x.$

Solve $X=-Y$:

$x+4 = -(2x-7) = -2x+7 \quad\Longrightarrow\quad 3x = 3, \qquad x=1.$

Both are candidates only, not yet confirmed.

Check $Y=2x-7$ at each candidate.

$x=11: \ Y=2(11)-7=15\ge0 \quad\checkmark \qquad x=1: \ Y=2(1)-7=-5<0$

Only $x=11$ passes, so the solution set is $\{11\}$; $x=1$ is extraneous, since the original equation would demand an absolute value equal to a negative number.

The left side of $\lvert X \rvert = Y$ is an absolute value, so it can never be negative, which forces $Y\ge0$ as a hidden requirement of the original equation:

$\lvert X \rvert = Y \quad\Longrightarrow\quad Y \ge 0.$

The bare split into $X=Y$ or $X=-Y$ only tracks whether the two sides could match in size; it carries no memory of that sign requirement, so a value of $x$ making $Y$ negative can still satisfy one of the two split equations without ever satisfying the original.

With $X=3x-2$ and $Y=x+8$, solve $X=Y$:

$3x-2=x+8 \quad\Longrightarrow\quad 2x=10, \qquad x=5.$

Solve $X=-Y$:

$3x-2=-(x+8)=-x-8 \quad\Longrightarrow\quad 4x=-6, \qquad x=-\tfrac32.$

Substitute each value into both sides separately.

$x=5: \ \lvert 3(5)-2 \rvert=13, \ \lvert 5+8 \rvert=13 \quad\checkmark \qquad x=-\tfrac32: \ \left\lvert 3\left(-\tfrac32\right)-2 \right\rvert=6.5, \ \left\lvert -\tfrac32+8 \right\rvert=6.5 \quad\checkmark$

Both values check, so neither is discarded.

In $\lvert X \rvert = \lvert Y \rvert$, both sides are absolute values, hence both automatically nonnegative:

$\lvert X \rvert \ge 0, \qquad \lvert Y \rvert \ge 0 \quad \text{always.}$

There is no hidden demand left to forget, so splitting into $X=Y$ or $X=-Y$ loses nothing. In $\lvert X \rvert = Y$, only the left side is guaranteed nonnegative; the right side $Y$ is an ordinary expression that can go negative, so the equation secretly demands $Y\ge0$, a demand the bare split does not track. The shared root cause is the same fact in both cases, an absolute value can never be negative; it is automatically satisfied on both sides of $\lvert X \rvert = \lvert Y \rvert$, but only on one side of $\lvert X \rvert = Y$.

Reading $\lvert x-5 \rvert < 8$ as a distance, $x$ lies within $8$ of $5$:

$-8 < x-5 < 8.$

Add $5$ across every part of the compound inequality:

$-3 < x < 13.$

Reading $\lvert x+6 \rvert > 3$ as a distance, $x$ lies farther than $3$ from $-6$, splitting into two rays:

$x+6 < -3 \quad \text{or} \quad x+6 > 3.$

Solve each branch:

$x < -9 \quad \text{or} \quad x > -3.$

A distance is always $\ge0$, so it can never be at or below a negative number such as $-5$. The inequality $\lvert 2x-1 \rvert \le -5$ therefore asks the impossible, and no algebra is needed to see its solution set is empty:

$\lvert 2x-1 \rvert \ge 0 > -5 \quad \text{for every real } x.$

The same reasoning settles any $\lvert X \rvert \le c$ or $\lvert X \rvert \ge c$ the moment the sign of $c$ makes the comparison impossible or automatic, without ever splitting into a case.

The two absolute values change sign at $x=2$ and $x=9$, the breakpoints. On the interval between them, $2\le x\le9$, the first term is $x-2$ (nonnegative there) and the second is $9-x$ (nonnegative there), so

$(x-2)+(9-x)=7$

for every $x$ in $[2,9]$, a constant equal to the gap between the two breakpoints.

Since the target $7$ exactly matches the constant value found in part A, every point of the interval $[2,9]$ satisfies the equation.

Outside that interval, each unit traveled beyond a breakpoint adds $1$ to each of the two distances, hence $2$ to their sum, so the sum only grows past $7$ for $x<2$ or $x>9$ and never returns to it:

$\text{sum} > 7 \quad \text{for every } x \notin [2,9].$

The complete solution set is therefore $[2,9]$, the segment between the two breakpoints.

Between the two breakpoints the sum of the two distances is fixed at the gap $g$, and it grows by $2$ for every unit traveled beyond either breakpoint. Comparing the target $d$ with $g$ therefore predicts the outcome without any further algebra:

$d<g: \ \varnothing, \qquad d=g: \ [p,q], \qquad d>g: \ \text{two points, each } (d-g)/2 \text{ past its breakpoint}.$

In part B, $d=7$ and the gap is $g=\lvert 2-9 \rvert=7$, so $d=g$ and the equation lands in the entire-segment case, matching the solution set $[2,9]$ found there.