Problem 53 - Elementary Algebra

Question

(a) Explain how to do the addition problem \(\frac{3}{x+2}+\frac{5}{x-1} .\) (b) Explain how to solve the equation \(\frac{3}{x+2}+\frac{5}{x-1}=0\).

Step-by-Step Solution

Answer

The sum is \( \frac{8x+7}{(x+2)(x-1)} \) and the solution to the equation is \( x = -\frac{7}{8} \).

Step 1: Find the Common Denominator

To add the fractions \( \frac{3}{x+2} \) and \( \frac{5}{x-1} \), we need to find a common denominator. The denominators are \( x+2 \) and \( x-1 \). The common denominator is the product \( (x+2)(x-1) \).

Step 2: Adjust the Numerators

Write both fractions with the common denominator: \[\frac{3}{x+2} = \frac{3(x-1)}{(x+2)(x-1)} \quad \text{and} \quad \frac{5}{x-1} = \frac{5(x+2)}{(x+2)(x-1)}.\]

Step 3: Combine the Fractions

Add the fractions using the common denominator:\[\frac{3(x-1) + 5(x+2)}{(x+2)(x-1)}.\] Simplify the numerator: \[3(x-1) + 5(x+2) = 3x - 3 + 5x + 10 = 8x + 7.\]Thus, the sum is \( \frac{8x+7}{(x+2)(x-1)} \).

Step 4: Transition to Solving Equation

Now we will solve the equation \( \frac{3}{x+2} + \frac{5}{x-1} = 0 \). Utilize the same common denominator from addition to set up the equation.

Step 4: Set the Equation to Zero

Set up the equation: \[\frac{3(x-1) + 5(x+2)}{(x+2)(x-1)} = 0.\] This implies that the numerator must be zero:\[8x + 7 = 0.\]

Step 5: Solve for x

Solve the equation \( 8x + 7 = 0 \):\[ 8x = -7 \]Divide both sides by 8:\[x = -\frac{7}{8}.\]

Step 6: Verify the Solution

Check that \( x = -\frac{7}{8} \) does not make the denominators zero. Since \( x+2 eq 0 \) and \( x-1 eq 0 \) for \( x = -\frac{7}{8} \), the solution is valid.

Key Concepts

Common Denominator

When adding or subtracting fractions, finding a common denominator is a crucial step. A common denominator is a shared multiple of the denominators in the fractions. For the problem of adding fractions like \( \frac{3}{x+2} \) and \( \frac{5}{x-1} \), the common denominator is obtained by multiplying the two distinct denominators: \((x+2)\) and \((x-1)\). This results in the common denominator \((x+2)(x-1)\).

Using a common denominator allows you to rewrite the fractions so that their denominators are the same. After this step, you can combine the numerators, maintaining the single denominator. Here is a quick bulleted breakdown:

• Identify individual denominators: \(x+2\) and \(x-1\).
• Multiply to find common denominator: \((x+2)(x-1)\).
• Rewrite each fraction using this common denominator before proceeding to addition.

Establishing a common denominator makes the upcoming addition or solving process more straightforward, providing a basis for combining fractions accurately.

Simplifying Fractions

Simplifying fractions means breaking down fractions into their simplest form. It often involves factoring and canceling out terms in the numerator and the denominator. Once you adjust the fractions to have a common denominator, as in \(\frac{3}{x+2}\) and \(\frac{5}{x-1}\) that were rewritten as \(\frac{3(x-1)}{(x+2)(x-1)}\) and \(\frac{5(x+2)}{(x+2)(x-1)}\) respectively, you can combine them.

The next step is to simplify the expression by expanding and combining like terms in the numerator. This is done as follows:

• Distribute to simplify each numerator: \(3(x-1)\) becomes \(3x - 3\), and \(5(x+2)\) becomes \(5x + 10\).
• Add the numerators together: \(3x - 3 + 5x + 10\) results in \(8x + 7\).
• Compose the final fraction: \(\frac{8x + 7}{(x+2)(x-1)}\).

Simplifying the numerator and writing it over the common denominator gives us a neater, more manageable expression. This process is essential for clarity and helps when further solving such expressions or equations.

Rational Equations

Rational equations involve solving equations that contain rational expressions. In the equation \( \frac{3}{x+2} + \frac{5}{x-1} = 0 \), both fractions use the same common denominator found in the addition process: \((x+2)(x-1)\).

When solving rational equations, the equality implies that if \(\frac{a}{b} = 0\), then the numerator \(a\) must be zero, given that the denominator \(b\) is not zero. For this problem:

• Combine numerators over the common denominator: \(\frac{3(x-1) + 5(x+2)}{(x+2)(x-1)} = 0\).
• Since the denominator \((x+2)(x-1)\) doesn't change the equality when the whole expression is zero, set the numerator to zero: \(8x + 7 = 0\).
• Solve for \(x\): Isolate \(x\) to find \(x = -\frac{7}{8}\).
• Verify the solution ensures denominators are not zero: Check that \(x+2 eq 0\) and \(x-1 eq 0\) at \(x = -\frac{7}{8}\).

By turning the equation's numerators to zero and checking each step, solving rational equations becomes intuitive. Verification is always necessary to ensure solutions do not nullify original denominators, maintaining equation validity.