Question

solve the inequality. show all of your work. give your answer as an inequality or in set - builder no
$4n-\frac{2}{3}leq\frac{7}{6}-2n$

1. three inequalities are given below. decide whether each has limited solutions, no solution, or infinite solutions.

inequality 1: $5(3x - 2)geq7x + 18$
inequality 2: $8-3(2n + 4)lt - 6n-10$
inequality 3: $17p - 8leq3(4p - 2)+5p$
solve the compound inequality. show all of your work. graph your solution on the number line $x + 4geq6$ or $-2x + 10gt4$

Explanation:

Step1: Add 2n to both sides

$4n + 2n-\frac{2}{3}\leq\frac{7}{6}-2n + 2n$
$6n-\frac{2}{3}\leq\frac{7}{6}$

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

$6n-\frac{2}{3}+\frac{2}{3}\leq\frac{7}{6}+\frac{2}{3}$
$6n\leq\frac{7 + 4}{6}$
$6n\leq\frac{11}{6}$

Step3: Divide both sides by 6

$n\leq\frac{11}{6}\div6$
$n\leq\frac{11}{6}\times\frac{1}{6}$
$n\leq\frac{11}{36}$

for Inequality 1:

Step1: Expand left - hand side

$5(3x - 2)=15x-10$, so the inequality becomes $15x-10\geq7x + 18$

Step2: Subtract 7x from both sides

$15x-7x-10\geq7x-7x + 18$
$8x-10\geq18$

Step3: Add 10 to both sides

$8x-10 + 10\geq18+10$
$8x\geq28$

Step4: Divide both sides by 8

$x\geq\frac{28}{8}=\frac{7}{2}$
It has infinite solutions.

for Inequality 2:

Step1: Expand left - hand side

$8-3(2n + 4)=8-6n-12=-6n - 4$, so the inequality becomes $-6n-4<-6n - 10$

Step2: Add 6n to both sides

$-6n+6n-4<-6n+6n - 10$
$-4<-10$ which is false. So it has no solution.

for Inequality 3:

Step1: Expand right - hand side

$3(4p - 2)+5p=12p-6 + 5p=17p-6$, so the inequality becomes $17p-8\leq17p-6$

Step2: Subtract 17p from both sides

$17p-17p-8\leq17p-17p-6$
$-8\leq-6$ which is always true. So it has infinite solutions.

Answer:

$n\leq\frac{11}{36}$