Problem 50 - Algebra 2

Question

Solve each equation or formula for the specified variable. $$ \frac{a(b-2)}{c-3}=x, \text { for } b $$

Step-by-Step Solution

Answer

\( b = \frac{x(c-3) + 2a}{a} \)

Step 1: Isolate the term with 'b'

Begin by isolating the term that contains the variable 'b' on one side of the equation. Multiply both sides by \( c-3 \) to get rid of the fraction: \( a(b-2) = x(c-3) \).

Step 2: Distribute 'a'

Distribute the 'a' through the expression \( (b-2) \) on the left side to get \( ab - 2a = x(c-3) \).

Step 3: Isolate 'b'

To isolate 'b', you want to get all terms with 'b' on one side. Add \( 2a \) to both sides: \( ab = x(c-3) + 2a \).

Step 4: Solve for 'b'

Finally, divide both sides by 'a' to solve for 'b': \( b = \frac{x(c-3) + 2a}{a} \).

Key Concepts

Variables

In the world of algebra, variables are symbols, often letters, that represent numbers or values that can change. In equations, we use variables to express relationships and solve for unknown quantities.
Commonly used variables include letters like \( x \), \( y \), and \( b \). These are placeholders for numbers we either don't know yet or want to generalize.
In our specific problem, the variable we are interested in is \( b \). Our goal is to express \( b \) in terms of the other variables and constants given in the equation.

Isolation Method

The isolation method in algebra is a technique used to solve for one variable in an equation. By isolating the variable, we aim to get it alone on one side of the equation.

In our exercise, the first step to solving for \( b \) involves isolating the term \( a(b-2) \). This is done by multiplying both sides by \( c-3 \) to eliminate the fraction, resulting in \( a(b-2) = x(c-3) \).
The isolation method helps us focus on the specific variable we want to solve for, making the equation simpler and easier to understand.

Distribution Property

The distribution property is an essential principle in algebra. It allows us to multiply a single term by each term in a parenthesis. This property is helpful for simplifying expressions.

In the equation from our exercise, after isolating the term with \( b \), we use the distribution property with \( a(b-2) \). This means \( a \) multiplies both terms inside the parenthesis, transforming it into \( ab - 2a \).
This step is crucial because it allows us to rewrite the equation clearly and sets the stage for further manipulation.

Algebraic Manipulation

Algebraic manipulation involves rearranging and simplifying equations to solve for a variable. This process often includes combining like terms, factoring, and performing operations like addition, subtraction, multiplication, and division on both sides of the equation.

In solving for \( b \), we first added \( 2a \) to both sides to move all constants away from \( ab \), resulting in \( ab = x(c-3) + 2a \).
This rearrangement is essential for keeping the equation balanced while isolating the desired variable. Finally, dividing both sides by \( a \) gives us the final solution \( b = \frac{x(c-3) + 2a}{a} \), effectively solving for \( b \).