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Chapter 2: Problem 31

Solve each equation, and check the solution. $$ -6 x=-72 $$

Short Answer

Expert verified
x = 12

Step by step solution

01

Isolate the Variable

To solve for \(x\), divide both sides of the equation by -6 to isolate \(x\).
02

Perform the Division

Divide \(-72\) by \(-6\): \(x = \frac{-72}{-6}\).
03

Simplify the Division

Divide the numbers: \(x = 12\).
04

Check the Solution

Substitute \(x = 12\) back into the original equation to verify: \(-6 \cdot 12 = -72\).

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Key Concepts

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

Isolating Variables
Solving algebraic equations often starts with isolating the variable. Isolation means getting the variable you are solving for, usually denoted as \( x \), alone on one side of the equation. This is crucial because it simplifies the equation and makes it solvable.
For example, in \( -6x = -72 \), the variable \( x \) is multiplied by \( -6 \). To isolate \( x \), we need to undo this multiplication. This involves performing the opposite operation, which, in this case, means division. By dividing both sides of the equation by \( -6 \), we ensure that we preserve the equality.
Division in Algebra
After isolating the variable, the next step is to perform the division. Division in algebra helps to simplify expressions and solve for the unknown.
In the equation \( -6x = -72 \), dividing both sides by \( -6 \) looks like this:
  • Divide the left side: \( \frac{-6x}{-6} = x \)
  • Divide the right side: \( \frac{-72}{-6} = 12 \)

The equation transforms to \( x = 12 \). Division in algebra not only simplifies the equation but ensures that both sides remain equal.
Checking Solutions
After finding a solution, it鈥檚 essential to check if it鈥檚 correct. Substituting the solution back into the original equation helps verify its accuracy.
Using our solution \( x = 12 \), we substitute \( 12 \) back into \( -6x = -72 \):
  • \( -6 \times 12 = -72 \)

This confirms that our solution is correct. Checking solutions helps catch errors and ensures the answer satisfies the original equation. It鈥檚 a fundamental practice in algebra to ensure precision and understanding.

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