## Thinking Mathematically (6th Edition)

$\left\{ x\le 50 \right\}$.
Let us suppose the number is $x$. Then, the algebraic form of the inequality is: $\frac{3x}{5}\,+4\,\le 34$ And now to find out possible values of $x$ satisfying the inequality the inequality is solved as follows: \begin{align} & \frac{3x}{5}\,+4\,\le 34 \\ & \frac{3x\times 5}{5}\,+4\times 5\,\le 34\times 5 \\ & 3x\,+20\le 170 \end{align} This can be further simplified as: \begin{align} & 3x\le 150 \\ & \frac{3x}{3}\le \frac{150}{3} \\ & x\le \text{ }50 \end{align} Hence all the real numbers less than or equal to 50 will satisfy the condition. The set-builder form of the inequality obtained is: $\left\{ x\,x\le \,50 \right\}$ Therefore, the number can be represented in set builder form as $\left\{ x\le 50 \right\}$.