Inequalities with brackets and fractions

An inequality with brackets or fractions is changed into the form \(ax+b<c\) (or with \(\le\), \(>\), \(\ge\)) before it is solved. Expand the brackets, and clear the fractions by multiplying both sides by the lowest common multiple (LCM) of the denominators.

Why it is allowed: expanding brackets does not change any value, and the LCM is a positive number, so multiplying by it keeps the inequality sign. After that, the steps are the ones you already know: add or subtract, then divide, reversing the sign only for a negative divisor.

Method

1. Expand any brackets, taking care with a negative number in front of a bracket.
2. If there are fractions, multiply every term on both sides by the LCM of the denominators.
3. Collect the terms in the unknown on one side and the numbers on the other.
4. Divide by the coefficient of the unknown, reversing the sign if it is negative.
5. Represent the solution on a number line, and answer any integer question asked.

Example

Solve \(2(3-x)\ge-4\) and represent the solution on a number line.

\(\begin{aligned} 2(3-x)&\ge-4 \\ 6-2x&\ge-4 \\ -2x&\ge-10 \\ x&\le5 \end{aligned}\)

Second example: solve \(\dfrac{x+2}{3}<\dfrac{1}{2}\). The LCM of \(3\) and \(2\) is \(6\).

\(\begin{aligned} 6\times\dfrac{x+2}{3}&<6\times\dfrac{1}{2} \\ 2(x+2)&<3 \\ 2x+4&<3 \\ 2x&<-1 \\ x&<-\tfrac{1}{2} \end{aligned}\)

Check: multiply every term, including terms without a fraction. In \(\dfrac{x}{2}-1>3\), multiplying by \(2\) gives \(x-2>6\), not \(x-1>6\). With \(-3(x-2)\), both terms change sign: \(-3x+6\).