Problem 23 - Algebra 1

Question

Solve the proportion. Check for extraneous solutions. $$\frac{6}{3}=\frac{x+8}{-1}$$

Step-by-Step Solution

Answer

It appears that the solution 'x = -6' does not satisfy the original proportion \(\frac{6}{3}=\frac{x+8}{-1}\), therefore, there are no valid solutions to this proportion, making 'x = -6' an extraneous solution.

Step 1: Setting up the proportion

The proportion is given as \(\frac{6}{3}=\frac{x+8}{-1}\). This can be rewritten as \(\frac{6}{3}=\frac{-(x+8)}{1}\) for simplification purposes.

Step 2: Solving for 'x'

To solve for 'x', cross-multiply the terms which gives us the equation: \(6*(-1) = 3*(-(x+8))\). Simplifying this further yields: \(-6 = -3x - 24\). Adding 24 to both sides, we get: \(-6+24 = -3x\), this simplifies to \(18 = -3x\). Dividing by -3 to isolate 'x' gives us the solution: \(x = -6\).

Step 3: Check for extraneous solutions

To make sure 'x = -6' is a valid solution, substitute it back into the original equation and see if both sides are equal. Substituting 'x' into the equation yields: \(\frac{6}{3}=\frac{-6+8}{-1}\), which simplifies to \(2 = -2\). This means that 'x = -6' is an extraneous solution and does not satisfy the original proportion.

Key Concepts

Extraneous Solutions

When solving equations, especially proportions, you may encounter solutions that do not actually satisfy the original equation. These are known as extraneous solutions. The reason they arise is often due to the manipulation of the equation during the solving process. In this exercise, we solved the proportion \( \frac{6}{3} = \frac{x+8}{-1} \) and found \( x = -6 \) as a solution. However, substituting \( x = -6 \) back into the original equation resulted in \( 2 = -2 \), which is false.

Extraneous solutions often appear when:

• We multiply or divide both sides of an equation by a variable expression.
• We perform operations that might change the nature of the equation, such as squaring or taking the square root.

After finding a solution, always substitute back into the original equation to verify it is true. If it isn’t, the proposed solution is extraneous and should be discarded.

Cross-Multiplication

Cross-multiplication is a technique commonly used to solve proportions. In a proportion, two ratios or fractions are set equal to each other, like \( \frac{a}{b} = \frac{c}{d} \). Cross-multiplication allows us to eliminate the fractions by multiplying the numerator of each fraction by the denominator of the other.

Here's how it works in this problem:

• Original proportion: \( \frac{6}{3} = \frac{x+8}{-1} \)
• Cross-multiply: \( 6 \times (-1) = 3 \times (-(x+8)) \)
• This gives us: \(-6 = -3(x+8)\)

Cross-multiplication simplifies solving because it converts the equation into a linear equation, which is generally easier to solve. However, remember that cross-multiplying alters the equation slightly, so verifying the solution is crucial.

Solving Equations

Solving equations is a fundamental aspect of algebra. It involves finding the value of the variable that makes the equation true. In this exercise, we started with a proportion \( \frac{6}{3} = \frac{x+8}{-1} \) and solved it using cross-multiplication.

After setting up the cross-multiplication, we had the equation: \(-6 = -3x - 24\). The next steps involved:

• Adding 24 to both sides: \(-6 + 24 = -3x\)
• Simplifying: \(18 = -3x\)
• Dividing by \(-3\) to solve for \(x\): \(x = -6\)

Understanding each manipulation and why these steps are necessary helps build a deeper comprehension of equation solving. It's important to perform operations methodically to maintain the balance of the equation and ensure an accurate solution. Always remember to check potential solutions in the original equation to confirm their validity.