Problem 34 - Prealgebra

Question

Solve each equation. Check your solution. $$\frac{a-6}{12}=\frac{a-2}{4}$$

Step-by-Step Solution

Answer

The solution is consistent, both sides are equal, but simplification shows no distinct solutions: it's technically solved, verifying only if fractions match initially suggests possibly no unique solution.

Step 1: Cross-Multiply to Eliminate Fractions

To eliminate the fractions, cross-multiply. This means you multiply the numerator of each fraction on one side by the denominator of the other side of the equation.So, multiply the left numerator by the right denominator and the right numerator by the left denominator:\[(a - 6) imes 4 = (a - 2) imes 12\]

Step 2: Distribute and Simplify

Distribute the numbers across the terms in the parentheses:\[4(a - 6) = 12(a - 2)\]This gives:\[4a - 24 = 12a - 24\]

Step 3: Simplify the Equation

Notice that both sides of the equation have \-24\. Add 24 to both sides to cancel them out:\[4a - 24 + 24 = 12a - 24 + 24\]This simplifies to:\[4a = 12a\]

Step 4: Solve for 'a'

To isolate the variable, subtract \4a\ from both sides:\[4a - 4a = 12a - 4a\]This simplifies to:\[0 = 8a\]

Step 5: Check Solution (Divide by Zero Issue)

To solve for \a\, divide both sides by 8:\[0 = 8a\]Dividing gives:\[a = 0\]However, this implies that there is no solution when substituting back into the original equation. Check:For \a = 0:\Left side:\[\frac{0-6}{12} = \frac{-6}{12} = -\frac{1}{2}\]Right side:\[\frac{0-2}{4} = \frac{-2}{4} = -\frac{1}{2}\]Both sides are equal, but this simplifies to \0 = 0\.

Key Concepts

Cross-Multiplication

Cross-multiplication is a useful technique to eliminate fractions from an equation, making it easier to solve. This method involves multiplying the numerator of one fraction by the denominator of the other fraction, and vice versa, essentially swapping and multiplying. This allows us to compare two products instead of dealing with fractions.
When applied to the equation \(\frac{a-6}{12}=\frac{a-2}{4}\), this method helps us make the fractions disappear so we can deal with a simpler equation. Specifically, you'll multiply \(a-6\) by 4 and \(a-2\) by 12. This gives the equation

• \((a-6) \times 4 = (a-2) \times 12\)

By doing so, we get rid of the fractions. Solving for the variable now becomes more intuitive as it resembles a basic linear equation. Cross-multiplication is thus a powerful first step to simplify fraction-based equations.

Distributive Property

The distributive property is a fundamental algebraic principle that allows us to multiply a number by a group of numbers added together, distributing the multiplication across each term inside the parentheses. In our equation, we use this property to open up the parentheses and simplify the equation.
For example, after cross-multiplying, our equation is\(4(a-6)=12(a-2)\). Here, apply the distributive property:

• \(4 \cdot a - 4 \cdot 6 = 12 \cdot a - 12 \cdot 2\)
• This simplifies to \(4a - 24 = 12a - 24\)

This property aids in breaking down complex expressions into manageable pieces, key in solving equations. Notice how applying the distributive property helped transform the equation into a simpler form that is more straightforward to solve.

Solving Linear Equations

Once you have simplified the equation using cross-multiplication and the distributive property, the next step is solving for the variable, which, in our case, is \(a\). With the equation \(4a - 24 = 12a - 24\), we can simplify further by eliminating like terms on both sides.
First, notice that both sides contain \(-24\). Adding 24 to both sides cancels these terms out:

• \(4a - 24 + 24 = 12a - 24 + 24\)
• Which simplifies to \(4a = 12a\)

To isolate \(a\), subtract \(4a\) from both sides:

• \(4a - 4a = 12a - 4a\)
• Reducing to \(0 = 8a\)

Finally, divide both sides by 8 yielding \(a = 0\). However, it’s critical in problem-solving to check your solution by substituting back into the original equation. Here, both sides equate to \(\frac{-1}{2}\), showing the solution is valid for this linear equation, emphasizing thorough checking in algebraic solutions.