Equations & Inequalities

An equation says that two things are equal, joined by an equals sign, like $x + 3 = 7$. Think of the $=$ sign as the middle of a balance scale. Whatever you do to one side you must do to the other, or the scale tips and the equation is no longer true.

To solve an equation means to find the value of the variable that makes it true. We do this by using inverse operations (opposite operations) to undo what has been done to the variable, one step at a time, until the variable is alone.

The variable has one operation attached to it. Undo that operation on both sides.

When two operations act on the variable, undo them in the reverse of the order of operations: first undo addition/subtraction, then undo multiplication/division.

Before you isolate the variable, clean up each side: use the distributive property to remove parentheses, then combine like terms. Then solve as usual.

Move all variable terms to one side and all number terms to the other by adding or subtracting. A good habit: move the smaller variable term so the coefficient stays positive.

Multiply every term on both sides by the least common denominator (LCD). This turns a fraction equation into a whole-number equation. For decimals, multiply every term by 10, 100, $\dots$ to clear the decimal points.

A literal equation has several letters. To “solve for” one letter means to get that letter alone, treating all the other letters as if they were numbers. Use the same inverse operations as always.

Sometimes, while solving, all the variable terms cancel out. Look at what is left:

• A false statement (like $3 = 7$) means no solution. No value of $x$ works.
• A true statement (like $5 = 5$) means infinitely many solutions. Every value of $x$ works; the two sides are really the same expression.

Inequalities use $<, >, \le, \ge$ instead of $=$. Solve them the same way you solve equations, with one special rule.

Graph the solution of a one-variable inequality on a number line:

• Use an open circle $\circ$ for $<$ or $>$ (the endpoint is not included).
• Use a closed (filled) circle $\bullet$ for $\le$ or $\ge$ (the endpoint is included).
• Shade an arrow to the right for “greater than” and to the left for “less than.”

A compound inequality joins two inequalities.

• And (written $a < x < b$): $x$ must satisfy both. The graph is the overlap between the two endpoints.
• Or: $x$ satisfies either one. The graph shades outward in two directions.

For an “and” inequality, do the same operation to all three parts.

$|x|$ means the distance of $x$ from $0$, which is never negative. So $|x| = 5$ has two answers: $x = 5$ or $x = -5$. Get the absolute value alone first, then split into two cases.

A parameter is a letter that stands for a fixed but unknown number (often $a$, $b$, $k$, $m$). When a linear equation contains a parameter, the number of solutions can change depending on the parameter's value. Every linear equation can be rearranged into the form

Everything then depends on whether the coefficient $A$ is zero:

• If $A \ne 0$: exactly one solution, $x = \dfrac{B}{A}$.
• If $A = 0$ and $B = 0$ (you get $0 = 0$): infinitely many solutions.
• If $A = 0$ and $B \ne 0$ (you get $0 = B$, a contradiction): no solution.

The whole strategy is: collect all $x$-terms on one side, factor $x$ out, then study its coefficient.

An expression like $\bigl||x-3| - 2\bigr|$ has one absolute value inside another. Treat the whole inner absolute value as a single block and undo the outer bars first, exactly as you would solve $|\,\text{block}\,| = c$. Each branch then leaves you with an ordinary absolute-value equation to finish. Always discard any branch that asks an absolute value to equal a negative number.

Read $|x - a|$ as “the distance between $x$ and $a$ on the number line.” Then $|x-a| + |x-b|$ is the total distance from $x$ to the two fixed points $a$ and $b$. If $a$ and $b$ are $d$ units apart, that total is smallest exactly when $x$ sits between them, where it equals $d$ and can go no lower. So:

• $|x-a| + |x-b| = c$ has no solution if $c < d$, and infinitely many (a whole segment) if $c = d$.
• If $c > d$, there are exactly two solutions, one on each side of the pair.

To solve algebraically, split the line at the critical points $x = a$ and $x = b$ and handle each region with casework.

When you solve a formula for a letter that appears in more than one term, you cannot simply divide — you must first collect every term containing that letter on one side, then factor the letter out, and only then divide by the leftover expression. This is the same “$Ax = B$” idea used for parameters, applied to rearranging formulas.

Some problems ask not for the solution set but for how many integers lie in it. First solve the inequality to get a range, then count the whole numbers inside, watching the endpoints:

• A strict endpoint ($<$ or $>$) does not include that boundary value.
• A non-strict endpoint ($\le$ or $\ge$) does include it.

For a run of consecutive integers from $m$ to $n$ inclusive, the count is $n - m + 1$.