The problem asks us to solve two equations. The first equation is $-5(a-3) = -2a$ for $a$. The second equation is $\frac{3b}{4} + \frac{b}{2} = 10$ for $b$.

1. Problem Description

The problem asks us to solve two equations. The first equation is $-5(a-3) = -2a$ for $a$. The second equation is $\frac{3b}{4} + \frac{b}{2} = 10$ for $b$.

2. Solution Steps

First, let's solve the equation $-5(a-3) = -2a$.

Step 1: Distribute the $-5$ to the terms inside the parentheses.

$-5a + 15 = -2a$

Step 2: Add $5a$ to both sides of the equation.

$15 = -2a + 5a$

$15 = 3a$

Step 3: Divide both sides by

3. $a = \frac{15}{3}$

$a = 5$

Next, let's solve the equation $\frac{3b}{4} + \frac{b}{2} = 10$.

Step 1: Find a common denominator for the fractions. The least common denominator for 4 and 2 is

4. Rewrite the equation with the common denominator.

$\frac{3b}{4} + \frac{2b}{4} = 10$

Step 2: Combine the fractions.

$\frac{3b + 2b}{4} = 10$

$\frac{5b}{4} = 10$

Step 3: Multiply both sides by

4. $5b = 40$

Step 4: Divide both sides by

5. $b = \frac{40}{5}$

$b = 8$

3. Final Answer

The solutions are:

$a = 5$

$b = 8$