Question

1. solve the inequality.

\\( \frac { 5 } { 3 } ( 2 x + 2 ) - 10 \geq 2 x + 2 \left( \frac { 2 } { 3 } x + 2 \
ight) \\)
\\( \textcircled { a } x \geq - 5 \\)
\\( \textcircled { b } x \leq 6 \\)
\\( \textcircled { c } \\) no solution
\\( \textcircled { d } \\) all real numbers

Explanation:

Step1: Expand both sides of the inequality

$$\begin{align*} \frac{5}{3}(2x + 2)-10&\geq2x+2(\frac{2}{3}x + 2)\\ \frac{10}{3}x+\frac{10}{3}-10&\geq2x+\frac{4}{3}x + 4\\ \frac{10}{3}x-\frac{20}{3}&\geq\frac{6}{3}x+\frac{4}{3}x + 4\\ \frac{10}{3}x-\frac{20}{3}&\geq\frac{10}{3}x + 4 \end{align*}$$

Step2: Subtract $\frac{10}{3}x$ from both sides

$$\begin{align*} \frac{10}{3}x-\frac{10}{3}x-\frac{20}{3}&\geq\frac{10}{3}x-\frac{10}{3}x + 4\\ -\frac{20}{3}&\geq4 \end{align*}$$

Since $-\frac{20}{3}\approx - 6.67$ and $-6.67<4$, this inequality is a contradiction.

Answer:

C. no solution