Solving ax + b < c in two steps

An inequality such as \(4x-5\ge11\) or \(7-2x>1\) has the form \(ax+b<c\) (or with \(\le\), \(>\), \(\ge\)). It is solved in two steps, in the same order as the equation \(ax+b=c\): first remove the constant \(b\), then divide by the coefficient \(a\).

Why this order: adding or subtracting never changes the sign, so doing it first leaves a one-step inequality. Only the division in step 2 can reverse the sign, and you can see clearly whether its divisor is positive or negative.

Method

1. Add or subtract the same number on both sides so that only the \(x\) term is left on its side.
2. Divide both sides by the coefficient of \(x\). If the coefficient is negative, reverse the sign in the same step.
3. Write the boundary as a fraction or mixed number if it is not whole, and draw the solution on a number line.
4. Check one value from the ray in the original inequality.

Example

Solve \(9-4x\le17\) and represent the solution on a number line.

\(\begin{aligned} 9-4x&\le17 \\ -4x&\le8 \\ x&\ge-2 \end{aligned}\)

Dividing by \(-4\) reverses \(\le\) to \(\ge\).

Check: \(x=0\) gives \(9-0=9\), and \(9\le17\) is true.

Check: the reversal happens at the division by \(-4\), not at the subtraction of \(9\). A common error is to reverse the sign when subtracting \(9\), which gives \(x\le-2\) instead of \(x\ge-2\).