Problem 63 - Introductory and Intermediate Algebra for College Students 4th

Question

Simplify each series of additions and subtractions. $$1-\frac{2}{3}-\left(-\frac{5}{6}\right)$$

Step-by-Step Solution

Answer

The simplified value of the given expression is \(2 \frac{1}{2}\).

Step 1: Simplify Negatives

As the first step, take care of the negative before the parentheses. This would mean changing the sign of the fraction inside the parentheses, so the expression becomes \(1 - \frac{2}{3} + \frac{5}{6}\) since a negative times a negative is positive.

Step 2: Common Denominator

In order to add or subtract fractions, they must have the same denominator. The fractions here are in thirds and sixths. The least common denominator is 6. Therefore we turn \(\frac{2}{3}\) into \(\frac{4}{6}\), in order to have the same denominator. The equation now looks like this: \(1 + \frac{4}{6} + \frac{5}{6}\)

Step 3: Simplify Fractions

Add the fractions. This will result in \(\frac{9}{6}\) which is the same as \(1 \frac{3}{6} = 1 \frac{1}{2}\)

Step 4: Write Out the Final Answer

Add the integer part of the answer to the simplified fractions. This results in the final answer of \(1 + 1 \frac{1}{2} = 2 \frac{1}{2}\)

Key Concepts

Understanding Common Denominator

When working with fractions, especially in addition or subtraction, having a common denominator is essential. The denominator is the bottom part of a fraction that shows how many equal parts the whole is divided into. To combine fractions effectively, they should have the same denominator.

For example, if you're working with fractions that have different denominators, like \(\frac{2}{3}\) and \(\frac{5}{6}\), you can't directly add or subtract them. You first need to convert each fraction to an equivalent fraction with the same denominator. This process is known as finding the "least common denominator" (LCD). It is the smallest number that both denominators can divide into evenly.

To find the common denominator for fractions with denominators 3 and 6, we should recognize that 6 is the smallest number that both 3 and 6 can divide. We convert \(\frac{2}{3}\) to \(\frac{4}{6}\) since multiplying the numerator and denominator by the same number keeps the fraction equivalent. After finding a common denominator, the fractions \(\frac{4}{6}\) and \(\frac{5}{6}\) can be added or subtracted easily.

Adding Fractions Simplified

Adding fractions might sound tricky at first, but once you have a common denominator, it's straightforward. When fractions share a common denominator, you simply add their numerators (the top parts) together.

For instance, after finding the common denominator of 6, we have \(\frac{4}{6}\) and \(\frac{5}{6}\). To add these fractions, you sum up the numerators: 4 plus 5 equals 9. Then, you keep the denominator the same. So, \(\frac{4}{6} + \frac{5}{6} = \frac{9}{6}\).

After adding, it's often possible to simplify the result. Here, \(\frac{9}{6}\) can be simplified to \(1\frac{1}{2}\) because 9 divided by 6 is 1 with a remainder of 3, which corresponds to \(\frac{3}{6}\) or \(\frac{1}{2}\). Keep in mind to always simplify fractions when possible for a cleaner form. By mastering these steps, you simplify operations on fractions greatly.

Simplifying Negative Numbers

Manipulating negative numbers can sometimes be confusing. It's crucial to remember that a negative sign before parentheses affects all terms inside when they are multiplied. If we have a negative number inside the parentheses, multiplying it by another negative sign outside turns it positive.

For example, in the expression \(1 - \frac{2}{3} - (-\frac{5}{6})\), the negative before \(-\frac{5}{6}\) indicates that we flip the sign of \(-\frac{5}{6}\) to become positive \(\frac{5}{6}\). Multiplying two negatives (\(-\) times \(-\)) results in a positive according to the rules of multiplication for negative numbers.

This understanding simplifies our calculations significantly, turning potentially tricky operations into simple math. Getting a grasp on these rules allows you to manipulate and simplify expressions with ease, without second-guessing the basic rules of arithmetic.