Problem 91 - Algebra for College Students

Question

Perform the indicated operations. $$ \frac{2 x+1}{6 x^{2}-5 x+1}+\frac{2 x-1}{6 x^{2}+x-1} $$

Step-by-Step Solution

Answer

\frac{2(4x^2 + 1)}{(2x - 1)(3x - 1)(2x + 1)}

Step 1 - Factorize the Denominators

First, factorize the denominators of the fractions. For the first denominator, factorize the quadratic expression: \[6x^2 - 5x + 1 = (2x - 1)(3x - 1)\]For the second denominator, factorize the quadratic expression: \[6x^2 + x - 1 = (2x + 1)(3x - 1)\]

Step 2 - Rewrite the Fractions

Rewrite the fractions with the factorized denominators: \[\frac{2x + 1}{(2x - 1)(3x - 1)} + \frac{2x - 1}{(2x + 1)(3x - 1)}\]

Step 3 - Find a Common Denominator

Identify a common denominator for the two fractions, which is the product of the unique factors: \[(2x - 1)(3x - 1)(2x + 1)\]

Step 4 - Rewrite the Fractions with Common Denominator

Adjust the fractions so they have the common denominator: \[\frac{(2x + 1)(2x + 1)}{(2x - 1)(3x - 1)(2x + 1)} + \frac{(2x - 1)(2x - 1)}{(2x + 1)(3x - 1)(2x - 1)}\]

Step 5 - Simplify the Numerators

Simplify the numerators: \[(2x + 1)^2 = 4x^2 + 4x + 1\]and \[(2x - 1)^2 = 4x^2 - 4x + 1\]Rewrite the fractions: \[\frac{4x^2 + 4x + 1}{(2x - 1)(3x - 1)(2x + 1)} + \frac{4x^2 - 4x + 1}{(2x + 1)(3x - 1)(2x - 1)}\]

Step 6 - Combine the Fractions

Combine the fractions over the common denominator: \[\frac{4x^2 + 4x + 1 + 4x^2 - 4x + 1}{(2x - 1)(3x - 1)(2x + 1)} = \frac{8x^2 + 2}{(2x - 1)(3x - 1)(2x + 1)}\]

Step 7 - Factorize the Numerator

Factorize the numerator: \[8x^2 + 2 = 2(4x^2 + 1)\]

Step 8 - Final Simplified Fraction

Rewrite the final form of the fraction: \[\frac{2(4x^2 + 1)}{(2x - 1)(3x - 1)(2x + 1)}\]

Key Concepts

Factoring Quadratics

To start solving algebraic fractions, we often need to factor quadratics. Quadratic equations follow the form \[ax^2 + bx + c = 0\]. Factoring involves rewriting this form into \[(px + q)(rx + s)\]. For example, in our problem, we had two quadratic equations: \[6x^2 - 5x + 1\] and \[6x^2 + x - 1\].
Let's break down each factoring process:

• For \[6x^2 - 5x + 1\], we factor into \[(2x - 1)(3x - 1)\]. This means that when we expand \[(2x - 1)(3x - 1)\], we will get \[6x^2 - 5x + 1\].
• For \[6x^2 + x - 1\], we factor into \[(2x + 1)(3x - 1)\]. This indicates that expanding \[(2x + 1)(3x - 1)\] gives \[6x^2 + x - 1\].

Spotting patterns and practicing factoring different quadratic equations enhance your skills in algebra and make subsequent steps smoother.

Common Denominators

To add or subtract fractions, they need a common denominator. The common denominator is a multiple of each fraction's denominator. For our problem:
The given fractions were: \[\frac{2x+1}{6x^2-5x+1} \]+\[\frac{2x-1}{6x^2+x-1}\]
After factoring the denominators, the fractions became:
\[\frac{2x + 1}{(2x - 1)(3x - 1)}\] and \[\frac{2x - 1}{(2x + 1)(3x - 1)}\]
To find a common denominator, we identified all unique factors present, leading to:
\[(2x - 1)(3x - 1)(2x + 1)\]
We adjusted each fraction to have this common denominator:

• For \[\frac{2x+1}{(2x - 1)(3x - 1)}\], multiply numerator and denominator by \[(2x + 1)\].
• For \[\frac{2x-1}{(2x + 1)(3x - 1)}\], multiply numerator and denominator by \[(2x - 1)\].

This step ensures both fractions are compared on the same baseline allowing valid addition.

Simplifying Fractions

Simplifying fractions makes calculations more manageable and results clearer. Here’s a breakdown of simplifying our combined fraction:
We added our fractions, rewriting with a common denominator:

• First fraction: \[\frac{(2x + 1)^2}{(2x - 1)(3x - 1)(2x + 1)}\]
• Second fraction: \[\frac{(2x - 1)^2}{(2x + 1)(3x - 1)(2x - 1)}\]

We then simplified the numerators:
\[(2x + 1)^2 = 4x^2 + 4x + 1\]
\[(2x - 1)^2 = 4x^2 - 4x + 1\]
Combining these gives:
\[\frac{4x^2 + 4x + 1 + 4x^2 - 4x + 1}{(2x - 1)(3x - 1)(2x + 1)} = \frac{8x^2 + 2}{(2x - 1)(3x - 1)(2x + 1)}\]
Lastly, we factorized the numerator one more time to:
\[\frac{2(4x^2 + 1)}{(2x - 1)(3x - 1)(2x + 1)}\]
This final step underlined why simplifying step-by-step ensures clarity and accuracy in algebra.