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Chapter 1: Problem 36

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

Short Answer

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

Step by step solution

01

Distribute Inside the Brackets

Start by distributing the factor \(-2\) into the brackets \(b - 3(c-x)\). This involves applying the distributive property: \[a - 2(b - 3(c-x)) = a - 2b + 6(c-x)\]
02

Simplify Further

Now distribute the 6 into the expression \(c-x\):\[a - 2b + 6c - 6x = 6\]
03

Isolate the Term with the Variable

Move all terms except the term with \(x\) to the other side of the equation:\[6x = a - 2b + 6c - 6\]
04

Solve for the Variable x

Finally, divide both sides by 6 to solve for the variable \(x\):\[x = \frac{a - 2b + 6c - 6}{6}\]
05

Final Solution

The solution for \(x\) is:\[x = \frac{a - 2b + 6c - 6}{6}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding the Distributive Property
The distributive property is a helpful algebraic tool that allows us to break down expressions and simplify complex equations. Essentially, it involves multiplying a single term by each term inside a bracket. This is written as:
\[ a(b + c) = ab + ac \]
In the given exercise, the distributive property was used to rewrite the expression \(-2(b - 3(c-x))\):
  • We distributed \(-2\) across \(b - 3(c-x)\).
  • This results in: \[a - 2b + 6(c-x)\]
Using the distributive property helps simplify equations, making it easier to manage and eventually solve them. This principle is essential when dealing with variables and constants, especially in linear equations.
  • Always multiply across every term within the brackets.
  • This step creates a simpler expression, setting the stage for further simplification.
The Art of Variable Isolation
Variable isolation is a critical step in solving equations where you aim to "isolate" the variable of interest on one side of the equation. This is a systematic approach that involves moving other terms to the opposite side.
Following the original solution's steps, here's how we isolated the variable \(x\):
  • First, after applying the distributive property, we had \(a - 2b + 6c - 6x = 6\).
  • To isolate \(x\), we rearranged the equation to get \(6x = a - 2b + 6c - 6\).
Methods of isolation include:
  • Moving constants over by addition or subtraction.
  • Dividing or multiplying both sides by numbers to simplify.
Isolating the variable helps us determine its value precisely by simplifying everything else. It's like solving a puzzle where you ensure the piece (variable) stands out individually.
Solving Linear Equations
Linear equations are mathematical statements that present a straight line when graphed. These equations have no variables raised to a power higher than one, making them straightforward to solve.
The standard form of a linear equation is:
  • \(ax + b = c\)
In our problem, we dealt with a more complex version:
  • \(a - 2[b-3(c-x)] = 6\)
To solve linear equations efficiently:
  • First, eliminate parentheses using the distributive property.
  • Combine like terms and simplify the equation.
  • Isolate the variable by balancing the equation.
At the final step in our exercise, once the variable \(x\) was isolated: we divided by the coefficient (6 in this case) to find the variable's precise value. These processes are foundational in algebra and help in determining unknown variables in straightforward and practical ways.

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Most popular questions from this chapter

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