Math Core

Real inequalities rarely come in one or two neat steps. They have parentheses, like terms scattered around, variables on both sides, and sometimes fractions. The good news: you already have every tool you need from solving multi-step equations. This lesson puts them together, adds the one rule about negatives, and shows two surprising outcomes that equations can have too.

The plan

Solving a multi-step inequality follows the same outline as a multi-step equation.

Steps for solving a multi-step inequality

Simplify each side. Use the distributive property and combine like terms.
Collect the variable terms on one side by adding or subtracting.
Collect the constants on the other side.
Divide by the coefficient of the variable. If that coefficient is negative, reverse the symbol.
Check the boundary and one test point.

Only step 4 is different from an equation, and only when the coefficient is negative.

Worked example: Distributing first

Solve $4(x - 2) + 3 \ge 2x + 9$ .

\begin{aligned} 4(x - 2) + 3 &\ge 2x + 9 \\ 4x - 8 + 3 &\ge 2x + 9 && \text{distribute} \\ 4x - 5 &\ge 2x + 9 && \text{combine like terms} \\ 2x - 5 &\ge 9 && \text{subtract } 2x \\ 2x &\ge 14 && \text{add } 5 \\ x &\ge 7 && \text{divide by } 2 \end{aligned}

Check. Boundary $x = 7$ : the left side is $4(5) + 3 = 23$ and the right side is $2(7) + 9 = 23$ . Equal, as expected. Test point $x = 8$ : $4(6) + 3 = 27$ and $2(8) + 9 = 25$ , and $27 \ge 25$ is true.

x ≥ 7

Choosing where to collect the variable

When variables appear on both sides, you get to choose which side keeps them. Either choice gives the same answer, but one of them may let you skip the sign flip.

Worked example: Two ways to finish

Solve $5 - 3(x + 1) < x + 10$ .

First simplify the left side: $5 - 3x - 3 = 2 - 3x$ . The inequality is now $2 - 3x < x + 10$ .

Method 1: variables on the left. Subtract $x$ from both sides to get $2 - 4x < 10$ . Subtract $2$ : $-4x < 8$ . Divide by $-4$ and reverse: $x > -2$ .

Method 2: variables on the right. Add $3x$ to both sides to get $2 < 4x + 10$ . Subtract $10$ : $-8 < 4x$ . Divide by positive $4$ : $-2 < x$ .

Both methods say the same thing: $x > -2$ . Method 2 never divided by a negative, so there was no flip to remember.

Check. Test point $x = 0$ : $5 - 3(1) = 2$ and $0 + 10 = 10$ , and $2 < 10$ is true. A point outside, $x = -3$ : $5 - 3(-2) = 11$ and $-3 + 10 = 7$ , and $11 < 7$ is false.

Tip

Move the variable terms to whichever side has the larger coefficient of $x$ . The variable's coefficient then stays positive, and you never have to reverse the symbol. If you'd rather keep the variable on the left, that's fine too; just remember the flip.

Common mistake

When you rewrite an answer like $-2 < x$ as $x > -2$ , the symbol turns around because the sides switched, not because of a negative. The meaning doesn't change. Don't confuse this with the flip that happens when you divide by a negative, which is a real change you must make to keep the statement true.

Clearing fractions

If an inequality has fractions, multiply both sides by the least common denominator. The LCD is positive, so the symbol stays the same, and the fractions disappear.

Worked example: An inequality with fractions

Solve $\dfrac{x}{2} - 1 \le \dfrac{x + 4}{3}$ .

The denominators are $2$ and $3$ , so multiply every term by $6$ :

\begin{aligned} 6 \cdot \frac{x}{2} - 6 \cdot 1 &\le 6 \cdot \frac{x + 4}{3} \\ 3x - 6 &\le 2(x + 4) \\ 3x - 6 &\le 2x + 8 \\ x - 6 &\le 8 && \text{subtract } 2x \\ x &\le 14 && \text{add } 6 \end{aligned}

Check. At $x = 14$ : $\dfrac{14}{2} - 1 = 6$ and $\dfrac{18}{3} = 6$ . Equal. At $x = 0$ : $-1 \le \dfrac{4}{3}$ , true.

Remember to multiply every term by the LCD, including whole numbers like the $1$ above. Forgetting one term is the most common error when clearing fractions.

All real numbers or no solution

Sometimes the variable terms cancel completely, leaving a statement with no variable at all. That statement is either always true or always false.

Worked example: When the variable disappears

Solve each inequality.

$2(x + 3) > 2x + 1$
$3x - 4 < 3(x - 2)$

Solutions.

Distribute: $2x + 6 > 2x + 1$ . Subtract $2x$ : $6 > 1$ . That is always true, no matter what $x$ is. Every real number is a solution. The graph shades the entire number line.
Distribute: $3x - 4 < 3x - 6$ . Subtract $3x$ : $-4 < -6$ . That is always false. No value of $x$ can make it true, so the inequality has no solution. The graph is an empty number line.

It helps to see why. In part 1, the left side $2x + 6$ is always exactly $5$ more than the right side $2x + 1$ , so it is always greater. In part 2, the left side is always $2$ more than the right, so it can never be less.

Multi-step word problems

Worked example: Comparing two plans

A climbing gym offers two plans. Plan A costs a $40 monthly fee plus $5 per visit. Plan B costs $15 per visit with no fee. For how many visits per month is Plan A cheaper?

Let $v$ be the number of visits. Plan A is cheaper when its cost is less than Plan B's:

\begin{aligned} 40 + 5v &< 15v \\ 40 &< 10v && \text{subtract } 5v \\ 4 &< v && \text{divide by } 10 \end{aligned}

Plan A is cheaper for more than $4$ visits, which means $5$ or more visits a month, since visits are whole numbers. At exactly $4$ visits the plans tie: $40 + 20 = 60$ and $15 \cdot 4 = 60$ .