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)

$-3(w+6)+26 = -3w -18 +26 = -3w +8$
$3(4-w) = 12 -3w$
Inequality becomes: $-3w +8 \geq 12 -3w$

Step2: Simplify inequality (a)

Add $3w$ to both sides:
$-3w +3w +8 \geq 12 -3w +3w$
$8 \geq 12$

Step3: Expand both sides of (b)

$5(4v+6) = 20v +30$
Inequality becomes: $20v +30 < 18v +16$

Step4: Isolate variable term (b)

Subtract $18v$ from both sides:
$20v -18v +30 < 18v -18v +16$
$2v +30 < 16$

Step5: Isolate constant terms (b)

Subtract 30 from both sides:
$2v +30 -30 < 16 -30$
$2v < -14$

Step6: Solve for v (b)

Divide by 2:
$\frac{2v}{2} < \frac{-14}{2}$
$v < -7$

Answer:

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