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Chapter 1: Problem 13

Evaluate each expression. $$\left|\frac{27-5}{5-27}\right|$$

Short Answer

Expert verified
The absolute value of the expression is 1.

Step by step solution

01

Simplify the Numerator

First, we need to simplify the numerator of the fraction. It is \(27 - 5\). Calculate that to get \(22\).
02

Simplify the Denominator

Next, simplify the denominator which is \(5 - 27\). Calculate that to get \(-22\).
03

Divide the Numerator by the Denominator

Now divide the simplified numerator by the simplified denominator: \(\frac{22}{-22}\). The result of this division is \(-1\).
04

Take the Absolute Value

The problem asks for the absolute value of the expression. The absolute value of \(-1\) is \(1\). Thus, the final result is \(1\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Numerator and Denominator Simplification
When dealing with fractions, one of the key steps often involves simplifying both the numerator and the denominator. This is essential for making calculations easier and obtaining a clearer understanding of the fraction.
  • Numerator: This is the top part of a fraction. In our exercise, the numerator is the expression \(27 - 5\). By simplifying this, you perform the subtraction, which gives you \(22\).
  • Denominator: This is the bottom part of a fraction. For the problem at hand, the denominator is \(5 - 27\). Simplifying it involves computing the difference, resulting in \(-22\).

Simplifying these parts early makes subsequent steps, like division, much smoother. It sets the stage for handling the fraction in an orderly fashion and reduces potential errors.
Fractions in Mathematics
Fractions represent the division of one quantity by another and are fundamental in many areas of mathematics. Understanding fractions helps you tackle a variety of problems effectively.
  • Each fraction has a numerator (top number) and a denominator (bottom number).
  • The value of a fraction is determined by dividing the numerator by the denominator.
  • Fractions can also be simplified by finding common factors to reduce both the numerator and denominator.

In our example, the fraction was \( \frac{22}{-22} \), which is simplified further by carrying out the division. Fractions give you a way to express precise portions of a whole, making them useful in contexts ranging from basic arithmetic to complex equations.
Negative Numbers Division
Dividing negative numbers is another essential part of handling fractions and expressions involving negative values.
Here are some useful points that might help you understand:
  • When you divide a positive number by a negative number, the result is negative.
  • The division of two negative numbers results in a positive number. However, in our exercise, we have a positive divided by a negative, giving a negative result.
  • In the expression \( \frac{22}{-22} \), dividing gives us \(-1\).

Understanding these rules helps prevent confusion and ensures correct calculations when negative numbers are involved. After obtaining a result from division, you can apply further operations, such as taking the absolute value in this exercise, to complete your solution.

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Most popular questions from this chapter

Find an equation for the line that is described. Write the answer in the two forms \(y=m x+b\) and \(A x+B y+C=0\). Is perpendicular to \(x-y+2=0\) and passes through (3,1).

Graph the equation after determining the \(x\) - and \(y\) -intercepts and whether the graph possesses any of the three types of symmetry described on page 58 $$y=-1 / x$$

(a), three points \(A, B,\) and \(C\) are specified. Determine whether \(A, B\), and C are collinear (lie on the same line) by checking to see whether the slope of \(\overline{A B}\) equals the slope of \(\overline{B C}\). $$A(-8,-2) ; B\left(2, \frac{1}{2}\right) ; C(11,-1)$$

Rewrite each statement using absolute value notation, as in Example 5. The number \(y\) is less than one unit from the number \(t\).

Rewrite each statement using absolute value notation, as in Example 5. The distance between \(x\) and 1 exceeds \(1 / 2\).
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