Question

question 5
20 pts
solve $\frac{3}{6}x<\frac{1}{6}x + 10$.
$x < 60$
$x < 10$
$x < 30$
$x < 6$

Explanation:

Step1: Subtract $\frac{1}{6}x$ from both sides

$\frac{3}{6}x-\frac{1}{6}x<\frac{1}{6}x + 10-\frac{1}{6}x$
$\frac{3 - 1}{6}x<10$
$\frac{2}{6}x<10$
$\frac{1}{3}x<10$

Step2: Multiply both sides by 3

$3\times\frac{1}{3}x<10\times3$
$x < 30$

Answer:

$x < 30$