/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 42 Add or subtract as indicated. ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Add or subtract as indicated. $$\frac{8}{x-2}+\frac{2}{x-3}$$

Short Answer

Expert verified
The simplified form of the given expression is \(\frac{5x-14}{(x-2)(x-3)}\).

Step by step solution

01

Find the Least Common Denominator (LCD)

Firstly, the Least Common Denominator (LCD) needs to be found. In this case, since the denominators \(x-2\) and \(x-3\) are both different and there are no common factors, the LCD will be the product of the two denominators. Therefore, the LCD is \((x-2)(x-3)\).
02

Rewrite the fractions with the LCD

Rewrite the fractions using the LCD. This is done by multiplying the numerator and denominator of the first fraction by \(x-3\) and the numerator and denominator of the second fraction by \(x-2\). The expression becomes: \(\frac{8(x-3)}{(x-2)(x-3)} + \frac{2(x-2)}{(x-2)(x-3)}\).
03

Simplify the fractions

Now, simply add the two fractions together (since they have the same denominator, they can be added directly). The resulting fraction is: \(\frac{8(x-3)+2(x-2)}{(x-2)(x-3)}\). Distribute the numerators to get \(\frac{8x-24+2x-4}{(x-2)(x-3)}\). This simplifies further to: \(\frac{10x-28}{(x-2)(x-3)}\) or further simplified to \(\frac{5x-14}{(x-2)(x-3)}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Study anywhere. Anytime. Across all devices.