Equations, Inequalities & Absolute Value

An equation is a balance scale. Whatever you do to one side, you must do to the other. Your goal is to isolate the variable. A reliable order of moves:

• Clear parentheses by distributing.
• Clear fractions by multiplying every term by the LCD.
• Collect the variable terms on one side, constants on the other.
• Combine like terms, then divide by the coefficient.

Every step should leave the equation equivalent (same solution set).

A literal equation contains several letters. Solving one for a chosen variable uses the exact same inverse-operation moves as a numeric equation — just treat the other letters as constants. Undo operations in reverse order: get the target variable alone.

Solve an inequality just like an equation, with one crucial rule: when you multiply or divide by a negative number, flip the inequality sign. A solution is a whole range of numbers, best shown on a number line and written in interval notation.

A compound inequality joins two inequalities.

• AND (intersection): both must hold. Graph is the overlap — a segment between two values. Written $a < x < b$, interval $(a,b)$.
• OR (union): at least one holds. Graph is two rays pointing outward, interval $(-\infty,a)\cup(b,\infty)$.

$|x|$ is the distance of $x$ from $0$, so it is never negative. To solve $|ax+b| = c$ with $c > 0$, split into two cases:

Isolate the absolute value first, before splitting.

After isolating the absolute value (with $c>0$):

• $|ax+b| < c$ means $-c < ax+b < c$ — an AND (a segment). Less thAND.
• $|ax+b| > c$ means $ax+b < -c$ or $ax+b > c$ — an OR (two rays). GreatOR.

Turn the words into an equation or inequality, solve, then interpret the answer in context.

• “at least” $\to \ge$,   “at most” $\to \le$,   “more than” $\to >$,   “fewer than” $\to <$.
• Tolerance problems use absolute value: $|\text{measured} - \text{target}| \le \text{tolerance}$.

An equation like $\big|\,|x-2|-3\,\big| = 4$ has an absolute value inside another. Treat the outer bars as a single unit and split first, then split again inside each branch. Each split can double the number of equations, so an outer/inner pair can yield up to four candidate solutions. Always solve from the outside in, and check every candidate back in the original — some branches collapse or produce values that fail.

To solve $|x-a| + |x-b| = c$ you cannot square or split into two — instead find the critical points where each expression changes sign ($x=a$ and $x=b$). These break the number line into three regions. On each region every bar opens with a fixed sign, so the equation becomes ordinary linear. Solve on each region and keep only the roots that land inside that region.

Read $|x-a|$ as “the distance from $x$ to $a$.” Then $|x-a| + |x-b|$ is the total distance from $x$ to the two fixed points $a$ and $b$. This gives instant intuition:

• The sum is minimized for any $x$ between $a$ and $b$, where it equals $|a-b|$, the gap itself.
• $|x-a| + |x-b| = c$ has no solution if $c < |a-b|$, infinitely many (the whole segment) if $c = |a-b|$, and exactly two symmetric solutions if $c > |a-b|$.
• $|x-a| \le |x-b|$ means “$x$ is at least as close to $a$” — the solution is the half-line on $a$'s side of the midpoint $\tfrac{a+b}{2}$.

A single absolute-value inequality in $x$ and $y$ carves the plane into a shape; a system intersects several such shapes.

• $|x| \le p$ is the vertical strip $-p \le x \le p$; $|y| \le q$ is a horizontal strip. Together they form a rectangle.
• $|x| + |y| \le r$ is a diamond (square rotated $45^\circ$) with vertices at $(\pm r, 0)$ and $(0, \pm r)$.
• $|y| \ge |x|$ is the pair of “bowtie” wedges opening up and down.

Sketch each boundary, decide inside vs. outside with a test point (the origin is easiest when it is not on a boundary), then shade the overlap.

Sometimes the unknown is a parameter that controls whether — and how many — solutions exist. Isolate the absolute value and compare with the structure rules:

• $|ax+b| = k$ has two solutions if $k>0$, one if $k=0$, none if $k<0$.
• $|ax+b| < k$ has solutions only if $k > 0$ (if $k \le 0$ the solution set is empty).
• $|ax+b| > k$ is all reals when $k < 0$, and all reals except one point when $k = 0$.

Translate the requested behavior into an inequality on the parameter, then solve that.

When an inequality contains a repeated chunk, let a new variable stand for it. Solving $(x^2-1)^2 - 5(x^2-1) + 4 \le 0$ is hopeless head-on, but with $u = x^2 - 1$ it becomes $u^2 - 5u + 4 \le 0$, i.e. $(u-1)(u-4) \le 0$, giving $1 \le u \le 4$. Back-substitute and solve the resulting compound inequality — often itself an absolute-value or interval statement. The same trick tames $|x|^2 - 3|x| \ge 0$ via $u = |x| \ge 0$.