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Question
- solve the inequality. use the inequality -\frac{3}{2}y+\frac{2}{3}\leq-\frac{1}{3}(y + 12) to complete questions 8 - 10. 9. write the solution in set - builder notation. 10. graph the solution set.
Step1: Expand the right - hand side
First, expand $-\frac{1}{3}(y + 12)$ using the distributive property $a(b + c)=ab+ac$. Here $a =-\frac{1}{3}$, $b = y$, $c = 12$. So $-\frac{1}{3}(y + 12)=-\frac{1}{3}y-4$. The inequality becomes $-\frac{3}{2}y+\frac{2}{3}\leq-\frac{1}{3}y - 4$.
Step2: Add $\frac{1}{3}y$ to both sides
To get all the $y$ - terms on one side, add $\frac{1}{3}y$ to both sides of the inequality: $-\frac{3}{2}y+\frac{1}{3}y+\frac{2}{3}\leq-\frac{1}{3}y+\frac{1}{3}y - 4$. Combine like - terms: $(-\frac{3}{2}+\frac{1}{3})y+\frac{2}{3}\leq - 4$. Calculate $-\frac{3}{2}+\frac{1}{3}=-\frac{9}{6}+\frac{2}{6}=-\frac{7}{6}$. So the inequality is $-\frac{7}{6}y+\frac{2}{3}\leq - 4$.
Step3: Subtract $\frac{2}{3}$ from both sides
Subtract $\frac{2}{3}$ from both sides: $-\frac{7}{6}y+\frac{2}{3}-\frac{2}{3}\leq - 4-\frac{2}{3}$. Calculate $-4-\frac{2}{3}=-\frac{12}{3}-\frac{2}{3}=-\frac{14}{3}$. So $-\frac{7}{6}y\leq-\frac{14}{3}$.
Step4: Multiply both sides by $-\frac{6}{7}$ and reverse the inequality sign
When multiplying or dividing an inequality by a negative number, the direction of the inequality sign changes. Multiply both sides of $-\frac{7}{6}y\leq-\frac{14}{3}$ by $-\frac{6}{7}$: $y\geq(-\frac{14}{3})\times(-\frac{6}{7})$. Calculate $(-\frac{14}{3})\times(-\frac{6}{7}) = 4$. So $y\geq4$.
Step5: Write the solution in set - builder notation
The set - builder notation for the solution is $\{y|y\geq4,y\in R\}$.
Step6: Graph the solution set
On a number line, draw a closed circle at $y = 4$ (because $y$ can equal 4, indicated by the $\geq$ sign) and shade to the right of 4 to represent all values of $y$ that are greater than or equal to 4.
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The solution of the inequality is $y\geq4$, the set - builder notation is $\{y|y\geq4,y\in R\}$, and on the number line, there is a closed circle at 4 and shading to the right.