Question

for each inequality, choose the statement that describes its solution. if applicable, give the solution. (a) $-3(w + 6)+26\geq3(4 - w)$ \bigcirc no solution \bigcirc $w\leq\square$ \bigcirc $w\geq\square$ \bigcirc all real numbers are solutions (b) $5(4v + 6)<18v + 16$ \bigcirc no solution \bigcirc $v<\square$ \bigcirc $v>\square$ \bigcirc all real numbers are solutions

Explanation:

Step1: Expand both sides of (a)

$-3w - 18 + 26 \geq 12 - 3w$

Step2: Simplify both sides of (a)

$-3w + 8 \geq 12 - 3w$

Step3: Isolate like terms in (a)

$-3w + 3w \geq 12 - 8$
$0 \geq 4$

Step4: Expand both sides of (b)

$20v + 30 < 18v + 16$

Step5: Isolate v terms in (b)

$20v - 18v < 16 - 30$

Step6: Simplify to solve for v

$2v < -14$
$v < -7$

Answer:

(a) All real numbers are solutions
(b) $v < -7$