Problem 36 - Intermediate Algebra

Question

Solve each proportion. $$ \frac{-3 c}{c-2}=\frac{c}{c+2} $$

Step-by-Step Solution

Answer

The solution to the proportion is \(c = 0\).

Step 1: Set up the Equation

To solve the proportion \( \frac{-3c}{c-2} = \frac{c}{c+2} \), we start by recognizing that cross-multiplication is needed to eliminate the fractions.

Step 2: Cross-multiply

Cross-multiply to get rid of the fractions: \[ (-3c) \cdot (c + 2) = c \cdot (c - 2) \] Distribute both sides.

Step 3: Distribute Terms

Distribute the terms on both sides of the equation:\[ -3c^2 - 6c = c^2 - 2c \]

Step 4: Rearrange and Simplify

Rearrange the equation to bring all terms to one side:\[ -3c^2 - 6c - c^2 + 2c = 0 \]Combine like terms:\[ -4c^2 - 4c = 0 \]

Step 5: Factor the Equation

Factor out the greatest common factor, which is \(-4c\):\[ -4c(c + 1) = 0 \]

Step 6: Solve for c

Set each factor equal to zero to solve for \(c\):1. \(-4c = 0\) gives \(c = 0\)2. \(c + 1 = 0\) gives \(c = -1\)These are the potential solutions, but we must check them in the original equation.

Step 7: Check for Extraneous Solutions

Substitute \(c = 0\) into the original proportion:\[ \frac{-3(0)}{0-2} = \frac{0}{0+2} \] Both sides are 0, so \(c=0\) holds true.Substitute \(c = -1\):\[ \frac{-3(-1)}{-1-2} = \frac{-1}{-1+2} \]\[ \frac{3}{-3} = 1 \] and \(1eq -1\), hence \(c = -1\) is not a valid solution.

Key Concepts

Cross-Multiplication

Cross-multiplication is a technique often used to solve proportions. A proportion is simply an equation that states two ratios are equivalent. In this problem, we have the equation \( \frac{-3c}{c-2} = \frac{c}{c+2} \). Cross-multiplication helps us to eliminate the fractions and simplify the equation. When performing cross-multiplication, we multiply the numerator of each fraction by the denominator of the opposite fraction. For this equation, we cross-multiply to get

• \((-3c) \cdot (c+2) = c \cdot (c-2)\)

This technique is crucial because it simplifies the equation into a polynomial form, removing the denominators. This step sets the groundwork for solving or factoring the equation further. Always remember, cross-multiplication is applicable only in equations where two fractions are set equal to each other.

Factoring Equations

Factoring is a vital skill in solving polynomial equations, especially after cross-multiplying a proportion. Once the cross-multiplication is carried out in our current problem, we distributed:

• \(-3c^2 - 6c = c^2 - 2c\).

We then brought all expressions to one side to form a polynomial:

• \(-4c^2 - 4c = 0\).

The process of factoring involves breaking down the polynomial into simpler expressions, which can be multiplied to give the original polynomial. In this example, we factor out the greatest common factor, which is \(-4c\):

• \(-4c(c + 1) = 0\)

Solving the factored equation involves setting each factor equal to zero to find the roots \(c = 0\) and \(c = -1\). Factoring allows us to find potential solutions quickly, but these solutions need to be verified to make sure they are not extraneous solutions.

Checking Extraneous Solutions

Not every solution obtained from factoring equations is valid due to potential extraneous solutions. These are solutions that arise while solving the problem but do not satisfy the original equation. It's crucial to substitute the solutions back into the original proportion to verify them.Let's check the solutions from our problem:

• Substitute \(c = 0\): \(\frac{-3(0)}{0-2} = \frac{0}{0+2}\). This simplifies to \(0 = 0\), confirming that \(c = 0\) is valid.
• Substitute \(c = -1\): \(\frac{-3(-1)}{-1-2} = \frac{-1}{-1+2}\). This simplifies to \(\frac{3}{-3} = 1\), which is not true. Therefore, \(c = -1\) is an extraneous solution.

Checking for extraneous solutions ensures the integrity of your results by confirming they actually work within the context of the original equation. It’s a crucial step in the problem-solving process, as it helps avoid incorrect answers when dealing with complex equations.