Problem 39 - Algebra for College Students

Question

Find the solution set to each equation. $$\frac{x-2}{x-3}=\frac{x+5}{x+2}$$

Step-by-Step Solution

Answer

The solution is \(x = \frac{11}{2}\).

Step 1 - Understand the equation

The given equation is \(\frac{x-2}{x-3} = \frac{x+5}{x+2}\). Our goal is to find the value of \(x\) that makes this equation true.

Step 2 - Cross-multiply the fractions

To eliminate the fractions, cross-multiply both sides of the equation: \[(x-2)(x+2) = (x-3)(x+5)\]

Step 3 - Expand both sides

Expand both sides of the equation: \[x^2 - 4 = x^2 + 2x - 15\]

Step 4 - Simplify the equation

Subtract \(x^2\) from both sides to simplify: \[-4 = 2x - 15\]

Step 5 - Solve for \(x\)

Add 15 to both sides of the equation: \[11 = 2x\]. Then, divide by 2: \[x = \frac{11}{2}\]

Step 6 - Check for extraneous solutions

Ensure that \(x = \frac{11}{2}\) does not make the denominators zero. Since \(x - 3 = \frac{11}{2} - 3 = \frac{5}{2} eq 0\) and \(x + 2 = \frac{11}{2} + 2 = \frac{15}{2} eq 0\), there are no issues.

Key Concepts

Cross-Multiplication

Cross-multiplication is a reliable method for solving equations involving fractions. In this example, we start with \(\frac{x-2}{x-3}=\frac{x+5}{x+2}\). This equation has fractions on both sides, which complicates solving for \(x\). To eliminate the fractions, we use cross-multiplication.

Cross-multiplication involves multiplying the numerator of one fraction by the denominator of the other. Therefore, we get: \[ (x-2)(x+2) = (x-3)(x+5) \]
This means we're now looking at a polynomial equation instead of a fraction, which can be easier to solve.

Expanding Expressions

After cross-multiplying the fractions, the next step is to expand the expressions. Expanding means applying the distributive property to remove parentheses and clarify the equation.

So we have: \[ (x-2)(x+2) = x^2 - 4 \] and \[ (x-3)(x+5) = x^2 + 2x - 15 \]
Now the equation looks like this: \[ x^2 - 4 = x^2 + 2x - 15 \]
Expanding expressions simplifies the equation and makes it easier to move to the next step.

Simplifying Equations

Simplifying equations is a crucial step to isolate the variable. Once the expressions are expanded, we can simplify the equation further. In this case, after expanding both sides, we have: \[ x^2 - 4 = x^2 + 2x - 15 \]
First, noticing that \(x^2\) appears on both sides of the equation, we subtract \(x^2\) from each side, leaving: \[ -4 = 2x - 15 \]
Then, solve for \(x\) by isolating it to one side. Add 15 to each side: \[ -4 + 15 = 2x \] which simplifies to \[ 11 = 2x \]
Divide by 2: \[ x = \frac{11}{2} \]
The equation is now simplified, giving us the solution.

Extraneous Solutions

When solving rational equations, it's essential to check for extraneous solutions. These are solutions that the math might suggest but don't satisfy the original equation when plugged back in.

In this case, we found \(x = \frac{11}{2}\). We need to ensure this doesn't make any denominator zero, because division by zero is undefined.

• Check \(x-3\): \( \frac{11}{2}-3 = \frac{5}{2} eq 0\)
• Check \(x+2\): \( \frac{11}{2} + 2 = \frac{15}{2} eq 0\)

Our solution does not cause any denominator to be zero, so it's valid. Always double-checking for extraneous solutions ensures the integrity of your answer.