Problem 15 - Algebra and Trigonometry

Question

1–54 ? Find all real solutions of the equation. $$ \frac{1}{x-1}+\frac{1}{x+2}=\frac{5}{4} $$

Step-by-Step Solution

Answer

The solutions are \( x = 2 \) and \( x = -1.4 \).

Step 1: Identify Common Denominator

To solve the equation, the first step is to find a common denominator for the fractions on the left side of the equation: \( \frac{1}{x-1} \) and \( \frac{1}{x+2} \). The common denominator is \((x-1)(x+2)\).

Step 2: Rewrite Fractions with Common Denominator

Express each fraction with the common denominator. Rewrite the equation as:\[\frac{x+2}{(x-1)(x+2)} + \frac{x-1}{(x-1)(x+2)} = \frac{5}{4}\] Combine the fractions on the left:\[\frac{(x+2) + (x-1)}{(x-1)(x+2)} = \frac{5}{4}\] Simplify the numerator:\[\frac{2x+1}{(x-1)(x+2)} = \frac{5}{4}\]

Step 3: Cross-Multiply to Eliminate Fractions

To eliminate the fractions, cross-multiply:\[4(2x+1) = 5(x-1)(x+2)\] Expand both sides:\[8x + 4 = 5(x^2 + x - 2)\]

Step 4: Simplify and Collect Like Terms

Expand the right side and simplify:\[8x + 4 = 5x^2 + 5x - 10\] Rearrange all terms to one side to form a quadratic equation:\[5x^2 - 3x - 14 = 0\]

Step 5: Use the Quadratic Formula

Identify the coefficients from the quadratic equation \(5x^2 - 3x - 14 = 0\) as \(a = 5\), \(b = -3\), \(c = -14\). Use the quadratic formula:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] Substitute the values:\[x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 \times 5 \times (-14)}}{2 \times 5}\]Simplify to find:\[x = \frac{3 \pm \sqrt{9 + 280}}{10} = \frac{3 \pm \sqrt{289}}{10}\] Since \( \sqrt{289} = 17 \), we have:\[x = \frac{3 \pm 17}{10}\]

Step 6: Solve for x

Calculate the two possible solutions:\[x = \frac{3 + 17}{10} = 2\]\[x = \frac{3 - 17}{10} = -1.4\] Both values seem valid, but we need to verify them against the original possible restrictions in the denominators.

Step 7: Verify Solutions

Check whether these solutions make any denominator zero in the original equation:1. For \( x = 2 \), neither \(x-1\) nor \(x+2\) is zero.2. For \( x = -1.4 \), \(x-1\) is not zero and \(x+2\) is not zero.Thus, both values are valid solutions.

Key Concepts

Cross-Multiplication

To solve equations involving fractions, a handy technique is cross-multiplication. This process is similar to balancing a seesaw. You eliminate the denominators by swapping and multiplying.

In our equation, \[\frac{2x+1}{(x-1)(x+2)} = \frac{5}{4}\]cross-multiplying eliminates fractions. You multiply each numerator by the opposite side's denominator:- Multiply 4 by \(2x + 1\)- Multiply 5 by \((x-1)(x+2)\)

This gives us the proportion:\[4(2x+1) = 5(x^2 + x - 2)\]Now, it's a simple polynomial equation. This method is simpler than dealing with fractions directly, making calculations easier and keeps equations balanced.

Common Denominator

Fractions can be tricky, especially when solving equations. Finding a common denominator simplifies the problem by creating equivalent fractions with a shared base.

For example, in \[\frac{1}{x-1} + \frac{1}{x+2}\]we identify the common denominator as \((x-1)(x+2)\). Then, we rewrite each fraction using that base:

• The first fraction becomes \(\frac{x+2}{(x-1)(x+2)}\)
• The second becomes \(\frac{x-1}{(x-1)(x+2)}\)

This step is crucial. It transforms different fractions into terms you can easily compare or add. Simplifying the expression then looks like:\[\frac{(x+2)+(x-1)}{(x-1)(x+2)}\]resulting in a single fraction. This foundational skill in algebra sets up the equation for cross-multiplication or further simplification.

Quadratic Formula

Quadratic equations like \(5x^2 - 3x - 14 = 0\)are common in algebra. The quadratic formula helps solve them when factoring seems tough or impossible.

The formula is:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Identifying coefficients \(a = 5\), \(b = -3\), and \(c = -14\) is the start. Plug these into the formula.

Use:\[x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 \times 5 \times (-14)}}{2 \times 5}\]Calculating within the square root is key: \(9 + 280 = 289\), leading to \(\sqrt{289} = 17\).
Two solutions result from\(3 \pm 17\):

• \(x = \frac{3 + 17}{10} = 2\)
• \(x = \frac{3 - 17}{10} = -1.4\)

Using the quadratic formula ensures each step is clear, reduces errors, and helps find real solutions efficiently.

Rational Equations

Rational equations involve fractions with variables in denominators. Understanding and solving these requires special techniques.

The original problem, \[\frac{1}{x-1} + \frac{1}{x+2} = \frac{5}{4}\]forms a rational equation. \(x-1\) and \(x+2\) are crucial since they affect the domain. Important steps include:

• Finding a common denominator for fractions (like \((x-1)(x+2)\)).
• Cross-multiplying to handle equations without fractional terms.
• Using the quadratic formula to solve resulting quadratic equations.

Each decision impacts the solution's validity. Sometimes, solutions (like with \(x = 2\)and \(-1.4\))must be checked back in the original equation due to potential undefined values. Rational equations prepare you for more complex algebra, emphasizing procedure and precision.