Question

use the inequality $-\frac{1}{2}y+\frac{1}{3}leq-\frac{1}{3}(y + 12)$ to complete questions 8 - 10.

1. solve the inequality.
2. write the solution in set - builder notation.
3. graph the solution set.

Explanation:

Step1: Expand the right - hand side

First, expand $-\frac{1}{3}(y + 12)$ using the distributive property $a(b + c)=ab+ac$. So $-\frac{1}{3}(y + 12)=-\frac{1}{3}y-4$. The inequality becomes $-\frac{2}{3}y+\frac{1}{3}\leq-\frac{1}{3}y - 4$.

Step2: Add $\frac{2}{3}y$ to both sides

$-\frac{2}{3}y+\frac{2}{3}y+\frac{1}{3}\leq-\frac{1}{3}y+\frac{2}{3}y - 4$. This simplifies to $\frac{1}{3}\leq\frac{1}{3}y - 4$.

Step3: Add 4 to both sides

$\frac{1}{3}+4\leq\frac{1}{3}y-4 + 4$. Since $\frac{1}{3}+4=\frac{1 + 12}{3}=\frac{13}{3}$, the inequality is $\frac{13}{3}\leq\frac{1}{3}y$.

Step4: Multiply both sides by 3

$3\times\frac{13}{3}\leq3\times\frac{1}{3}y$. This gives $13\leq y$, or $y\geq13$.

Step5: Set - builder notation

The set - builder notation for the solution $y\geq13$ is $\{y|y\geq13,y\in\mathbb{R}\}$.

Step6: Graph the solution set

On the number line, draw a closed circle at 13 (because $y$ can equal 13, indicated by $\geq$) and draw an arrow to the right to represent all numbers greater than 13.

Answer:

1. $y\geq13$
2. $\{y|y\geq13,y\in\mathbb{R}\}$
3. Draw a closed circle at 13 on the number - line and an arrow to the right.