Solution: Step 1: Express the equation in the form b cdot c + b = a. Step 2: Extract the common factor b from the terms b cdot c and b. Step 2.1: Identify b as a common factor in b cdot c to get b(c) + b = a. Step 2.2: Recognize that b is implicitly raised to the power of 1, hence b(c) + b1 = a. Step 2.3: Extract b from b1 to continue factoring, resulting in b(c) + b cdot 1 = a. Step 2.4: Complete the factoring process to obtain b(c + 1) = a. Step 3: Isolate b by dividing the equation b(c + 1) = a by c + 1. Step 3.1: Apply the division to both sides of the equation to get fracb(c + 1)c + 1 = fracac + 1. Step 3.2: Simplify the left-hand side of the equation. Step 3.2.1: Eliminate the common factor (c + 1). Step 3.2.1.1: Cancel out (c + 1) to simplify the expression to fracb cancel(c + 1)cancel(c + 1) = fracac + 1. Step 3.2.1.2: Simplify the left-hand side by dividing b by 1 to finally get b = fracac + 1. Knowledge Notes: The solution involves algebraic manipulation to solve for the variable b in the equation a = bc + b. The key knowledge points include: 1. Rearranging Equations: The ability to rearrange equations and express them in a different form while maintaining equality is fundamental in algebra. 2. Factoring: The process of factoring involves identifying and extracting common factors from terms in an expression. In this case, b is factored out of both bc and b. 3. Simplifying Expressions: After factoring, the expression is simplified by canceling …

Solve for b a=bc+b

The given problem is an algebraic equation where 'a' and 'b' are variables and 'c' is presumably a known quantity or another variable. The goal is to solve for the variable 'b' by isolating it on one side of the equation. This typically involves algebraic manipulation such as combining like terms and isolating the variable of interest.

$a = b c + b$

Answer

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Solution:

Step 1:

Express the equation in the form $b \cdot c + b = a$ .

Step 2:

Extract the common factor $b$ from the terms $b \cdot c$ and $b$ .

Step 2.1:

Identify $b$ as a common factor in $b \cdot c$ to get $b (c) + b = a$ .

Step 2.2:

Recognize that $b$ is implicitly raised to the power of $1$ , hence $b (c) + b^{1} = a$ .

Step 2.3:

Extract $b$ from $b^{1}$ to continue factoring, resulting in $b (c) + b \cdot 1 = a$ .

Step 2.4:

Complete the factoring process to obtain $b (c + 1) = a$ .

Step 3:

Isolate $b$ by dividing the equation $b (c + 1) = a$ by $c + 1$ .

Step 3.1:

Apply the division to both sides of the equation to get $\frac{b (c + 1)}{c + 1} = \frac{a}{c + 1}$ .

Step 3.2:

Simplify the left-hand side of the equation.

Step 3.2.1:

Eliminate the common factor $(c + 1)$ .

Step 3.2.1.1:

Cancel out $(c + 1)$ to simplify the expression to $\frac{b (c + 1)}{(c + 1)} = \frac{a}{c + 1}$ .

Step 3.2.1.2:

Simplify the left-hand side by dividing $b$ by $1$ to finally get $b = \frac{a}{c + 1}$ .

Knowledge Notes:

The solution involves algebraic manipulation to solve for the variable $b$ in the equation $a = b c + b$ . The key knowledge points include:

Rearranging Equations: The ability to rearrange equations and express them in a different form while maintaining equality is fundamental in algebra.
Factoring: The process of factoring involves identifying and extracting common factors from terms in an expression. In this case, $b$ is factored out of both $b c$ and $b$ .
Simplifying Expressions: After factoring, the expression is simplified by canceling out common terms, which is a basic operation in algebra.
Division: The final step to isolate the variable involves dividing both sides of the equation by the same non-zero term, which is a standard technique for solving linear equations.
Latex Formatting: Mathematical expressions are rendered in Latex format to provide clear and professional-looking equations. For example, $\frac{a}{b}$ is the Latex representation of the fraction "a over b".

Understanding these concepts is essential for solving linear equations and manipulating algebraic expressions effectively.