Problem 34 - Precalculus Mathematics for Calculus

Question

Solve the equation for the indicated variable. $$a-2[b-3(c-x)]=6 ; \text { for } x$$

Step-by-Step Solution

Answer

\(x = \frac{a - 2b + 6c - 6}{6}\)

Step 1: Distribute the -2 inside the brackets

Start by distributing the -2 through the expression inside the brackets. This means multiplying -2 by each term inside the brackets: \[-2[b-3(c-x)]=-2b+6(c-x).\]

Step 2: Expand the brackets further

Next, expand the inner brackets by distributing the 6 through the expression inside them:\[6(c-x)=6c-6x.\]

Step 3: Substitute back into the equation

Replace the expanded terms back into the original equation:\[a - 2b + 6c - 6x = 6.\]

Step 4: Isolate terms involving x

Move all terms involving \(x\) to one side of the equation and the rest to the other side. This gives:\[a - 2b + 6c - 6 = 6x.\]

Step 5: Solve for x

Divide both sides by 6 to solve for \(x\):\[x = \frac{a - 2b + 6c - 6}{6}.\]

Key Concepts

Distributive Property

The distributive property is a fundamental concept in algebra that assists in simplifying expressions and solving equations. This property states that multiplying a number by a group of terms inside brackets is the same as multiplying the number by each term separately and then adding the results. For example, in the equation given, we have an expression inside brackets: \[-2[b-3(c-x)]\]Using the distributive property, you distribute the \(-2\) to both \(b\) and \(-3(c-x)\). This is calculated as:

• \(-2 \times b = -2b\)
• \(-2 \times -3(c-x) = 6(c-x)\)

Breaking it down further involves another use of the distributive property within \(6(c-x)\), distributing \(6\) across the terms \(c\) and \(-x\), leading to:

• \(6 \times c = 6c\)
• \(6 \times -x = -6x\)

After distribution, complex brackets are simplified, which makes the equation more manageable to solve. Understanding this property is crucial for solving equations that involve nested expressions.

Isolation of Variables

When solving for a specific variable, like \(x\), it is often essential to isolate that variable on one side of the equation. This process involves moving other terms to the opposite side, which effectively "clears up" one side containing only the variable of interest. In our step-by-step solution, after expanding and simplifying, the result was:\[a - 2b + 6c - 6x = 6\]To isolate \(x\), we need all terms not involving \(x\) to be on the other side of the equation:

• First, subtract \(a - 2b + 6c - 6\) from both sides, aligning terms with the opposite operation.
• This gives us \(6x = a - 2b + 6c - 6\).

The concept of isolation helps transform complex equations into straightforward ones. By isolating the variable, we significantly simplify the equation's complexity and make it easier to find the solution.

Algebraic Manipulation

Algebraic manipulation involves rearranging and simplifying equations. This includes using operations such as addition, subtraction, multiplication, and division to transform an equation to its simplest form. Crucial during this process is the maintenance of equation balance – whatever operation is performed on one side must be done on the other. Following our example, after distributing and isolating terms involving \(x\):\[6x = a - 2b + 6c - 6\]We use division to simplify further. Dividing both sides by \(6\) provides us:

• \(x = \frac{a - 2b + 6c - 6}{6}\)

This step demonstrates how algebraic manipulation ensures that we can simplify expressions and isolate variables to solve equations. Mastering these techniques allows us to work efficiently with even the most intricate algebraic problems.

Linear Equations

Linear equations are equations of the first degree, meaning they have variables raised only to the power of one. These equations often appear in the form \(ax + b = c\), and their solutions are the values of the variable that make the equation true. In our equation, once we manipulated and simplified it, it represented a linear form concerning \(x\):\[x = \frac{a - 2b + 6c - 6}{6}\]The characteristic feature of linear equations is their straightforward nature. They involve operations such as addition, subtraction, multiplication, and division. Solving such equations typically involves straightforward methods like isolation and algebraic manipulation. Solving a linear equation provides a direct path toward reaching a solution, offering clear steps to ascertain the answer effortlessly. Understanding linear equations forms the foundational basis for progressing to more complex algebraic expressions.