Chapter 4.4, Problem 65PS

Intermediate Algebra

10th Edition
Jerome E. Kaufmann + 1 other
ISBN: 9781285195728

Chapter
Section

Intermediate Algebra

10th Edition
Jerome E. Kaufmann + 1 other
ISBN: 9781285195728
Textbook Problem

Which of the two techniques presented in the text would use to simplify 1 4 + 1 3 3 4 − 1 6 ? Which technique would you use to simplify 3 8 − 5 7 7 9 + 6 25 ? Explain your choice for each problem.

To determine

To simplify:

The given complex fractions:

14+133416 and 385779+625

Explanation

Approach:

Fraction in a fraction is called complex fraction. Fraction of this kind can resolve to the reduced form in different approaches.

The first method is to find the LCD and multiplied it to the numerator and denominator of the complex fraction.

The second method is to simplify the expression by performing addition in the numerator and denominator and then divide to solve.

Calculation:

Here, the given expression are 14+1334âˆ’16 and 38âˆ’5779+625.

The first method using LCD is more appropriate to solve the expression 14+1334âˆ’16 since in the given complex fraction, LCD can be evaluated simply by taking the common factors in the denominator terms of the numerator and denominator of the fraction.

LCD of the given expression is 24.

So, multiply and divide the expression by 24 as follows:

14+1334âˆ’16=2424(14+1334âˆ’16)=6+83Ã—6âˆ’4=1414=1

Therefore, the complex fraction 14+1334âˆ’16 in reduced form is 1.

The second method is more appropriate to solve the expression 38âˆ’5779+625 since the complex fraction have numerator and denominator terms which give a LCD of 12600. The value of LCD is large which will make the simplification complex. Hence, the second method is preferable

Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started