Question

what are all values of x for which the inequality $5x + \frac{5}{3} \leq -2x - \frac{2}{3}$ is true?
a. $x \leq -\frac{7}{9}$
b. $x \leq -\frac{1}{3}$
c. $x \geq 0$
d. $x \geq \frac{7}{3}$

Explanation:

Step1: Add \(2x\) to both sides

To isolate the \(x\) terms on one side, we add \(2x\) to both sides of the inequality \(5x+\frac{5}{3}\leq - 2x-\frac{2}{3}\). This gives us \(5x + 2x+\frac{5}{3}\leq-2x + 2x-\frac{2}{3}\), which simplifies to \(7x+\frac{5}{3}\leq-\frac{2}{3}\).

Step2: Subtract \(\frac{5}{3}\) from both sides

Next, we subtract \(\frac{5}{3}\) from both sides to get the \(x\) term alone on the left. So, \(7x+\frac{5}{3}-\frac{5}{3}\leq-\frac{2}{3}-\frac{5}{3}\). Simplifying the right - hand side: \(-\frac{2 + 5}{3}=-\frac{7}{3}\), and the left - hand side becomes \(7x\). So we have \(7x\leq-\frac{7}{3}\).

Step3: Divide both sides by 7

Now, we divide both sides of the inequality \(7x\leq-\frac{7}{3}\) by 7. When we divide both sides of an inequality by a positive number, the direction of the inequality sign remains the same. So \(x\leq\frac{-\frac{7}{3}}{7}\). Using the rule \(\frac{a/b}{c}=\frac{a}{b\times c}\), we get \(x\leq-\frac{7}{3\times7}=-\frac{1}{3}\).

Answer:

B. \(x\leq-\frac{1}{3}\)