Question

1. $12b + 18 = 5b + 4$
2. $3d + 8 = 12 + 2d$
3. $5w - 13 = 11 - 3w$
4. $\frac{7}{16}y + 6 = 8 + \frac{3}{8}y$
5. $5 - \frac{1}{8}r = \frac{3}{4}r - 10$

Explanation:

Problem 2: Solve for $b$

Step1: Subtract 18 from both sides

$12b + 18 - 18 = 5b + 4 - 18$
$12b = 5b - 14$

Step2: Subtract $5b$ from both sides

$12b - 5b = 5b - 14 - 5b$
$7b = -14$

Step3: Divide by 7

$\frac{7b}{7} = \frac{-14}{7}$
$b = -2$

Problem 4: Solve for $d$

Step1: Subtract 8 from both sides

$3d + 8 - 8 = 12 + 2d - 8$
$3d = 2d + 4$

Step2: Subtract $2d$ from both sides

$3d - 2d = 2d + 4 - 2d$
$d = 4$

Problem 6: Solve for $w$

Step1: Add 13 to both sides

$5w - 13 + 13 = 11 - 3w + 13$
$5w = 24 - 3w$

Step2: Add $3w$ to both sides

$5w + 3w = 24 - 3w + 3w$
$8w = 24$

Step3: Divide by 8

$\frac{8w}{8} = \frac{24}{8}$
$w = 3$

Problem 8: Solve for $y$

Step1: Subtract 6 from both sides

$\frac{7}{10}y + 6 - 6 = 8 + \frac{3}{5}y - 6$
$\frac{7}{10}y = 2 + \frac{3}{5}y$

Step2: Subtract $\frac{3}{5}y$ (rewrite as $\frac{6}{10}y$)

$\frac{7}{10}y - \frac{6}{10}y = 2 + \frac{6}{10}y - \frac{6}{10}y$
$\frac{1}{10}y = 2$

Step3: Multiply by 10

$\frac{1}{10}y \times 10 = 2 \times 10$
$y = 20$

Problem 10: Solve for $r$

Step1: Add 10 to both sides

$5 - \frac{1}{2}r + 10 = \frac{1}{4}r - 10 + 10$
$15 - \frac{1}{2}r = \frac{1}{4}r$

Step2: Add $\frac{1}{2}r$ (rewrite as $\frac{2}{4}r$)

$15 - \frac{2}{4}r + \frac{2}{4}r = \frac{1}{4}r + \frac{2}{4}r$
$15 = \frac{3}{4}r$

Step3: Multiply by $\frac{4}{3}$

$15 \times \frac{4}{3} = \frac{3}{4}r \times \frac{4}{3}$
$r = 20$

Answer:

1. $b = -2$
2. $d = 4$
3. $w = 3$
4. $y = 20$
5. $r = 20$