Problem 24 - Algebra for College Students

Question

Solve each equation. $$ \frac{x+1}{2 x^{2}+7 x-4}-\frac{x}{2 x^{2}-7 x+3}=\frac{1}{x^{2}+x-12} $$

Step-by-Step Solution

Answer

The solution is \( x = -\frac{1}{4} \).

Step 1: Simplify the Denominators

First, factor the denominators of each fraction. For the first fraction, the denominator is \(2x^2 + 7x - 4\). Factoring gives \((2x - 1)(x + 4)\). For the second fraction, \(2x^2 - 7x + 3\) factors to \((2x - 1)(x - 3)\). The third fraction denominator, \(x^2 + x - 12\), factors to \((x - 3)(x + 4)\).

Step 2: Identify the Least Common Denominator (LCD)

The least common denominator for the fractions is the product of all different linear factors: \((2x - 1)(x + 4)(x - 3)\).

Step 3: Rewrite Fractions with the LCD

Rewrite each fraction so they all have the LCD:- First fraction: \(\frac{x+1}{(2x - 1)(x + 4)}\) becomes \(\frac{(x + 1)(x - 3)}{LCD}\).- Second fraction: \(\frac{x}{(2x - 1)(x - 3)}\) becomes \(\frac{x(x + 4)}{LCD}\).- Third fraction: Already \(\frac{1}{(x - 3)(x + 4)}\) becomes \(\frac{2x - 1}{LCD}\).

Step 4: Set Up the New Equation

With a common denominator, the equation becomes:\(\frac{(x + 1)(x - 3) - x(x + 4) = 2x - 1}{LCD}\).

Step 5: Solve the Numerator Equation

Ignore the common denominator and solve for the numerator:\((x^2 - 3x + x - 3) - (x^2 + 4x) = 2x - 1\)Simplify to:\(-6x - 3 = 2x - 1\).

Step 6: Solve for x

Add \(6x\) to both sides: \(-3 = 8x - 1\)Add 1 to both sides:\(-2 = 8x\). Divide by 8:\(x = -\frac{1}{4}\).

Step 7: Check for Extraneous Solutions

Plug \(x = -\frac{1}{4}\) back into the original denominators to ensure there are no zero values:- None of the expressions \((2x - 1), (x + 4), (x - 3)\) yield zero when \(x = -\frac{1}{4}\), thus it's a valid solution.

Key Concepts

Factoring Quadratic Expressions

Factoring quadratic expressions is a crucial step when solving rational equations. It simplifies complex expressions, making them easier to manage. In our exercise, we begin by factoring each denominator separately.

- For the expression \(2x^2 + 7x - 4\), we factor it into \((2x - 1)(x + 4)\).- The next expression, \(2x^2 - 7x + 3\), is factored into \((2x - 1)(x - 3)\).- Finally, \(x^2 + x - 12\) becomes \((x - 3)(x + 4)\).These factored forms reveal the roots of the quadratic expressions. Factoring breaks down the complex equations into products of simpler expressions, helping us find the least common denominator.

Least Common Denominator

Finding the least common denominator (LCD) in rational equations is pivotal as it allows us to combine fractions. The LCD is derived from the distinct factors of the denominators.

For our problem:- The factors gathered are \((2x - 1)\), \((x + 4)\), and \((x - 3)\).The LCD is therefore the product of these factors: \((2x - 1)(x + 4)(x - 3)\).

Using the LCD, we can rewrite each fraction to have the same denominator, simplifying the subtraction and ultimately solving the equation more efficiently.

Extraneous Solutions

Extraneous solutions are results that solve the manipulated equation but not the original equation. They often appear when cross-multiplying or squaring both sides of an equation.

In this task, it’s essential to check if the found solution of \(x = -\frac{1}{4}\) remains valid when substituted back into the original denominators:- None of these factors \((2x - 1)\), \((x + 4)\), \((x - 3)\) become zero at this point.If substituting back yields zero in any denominator, that solution would be invalid as it makes the original equation undefined. Checking for extraneous solutions ensures the correctness and validity of the solution.

Substitution Method

After solving the simplified equations, checking your solution using substitution ensures accuracy. It involves plugging the solution back into the original equation.In our case, substitute \(x = -\frac{1}{4}\) into the original fractions to verify the outcome:- None of the expressions in the denominators yield zero, confirming that \(x = -\frac{1}{4}\) doesn't make any part undefined.Substitution not only checks for extraneous solutions but also boosts confidence in the accuracy of the outcome. It acts as a verification step, which is good practice in solving mathematical equations.