Chapter 7.6, Problem 52SR

To determine

ToDetermineThe sum $6\frac{4}{5}+\left(-1\frac{3}{4}\right)$ and write the answer in simplest form.

Answer to Problem 52SR

  $\frac{34}{5}+\left(-\frac{7}{4}\right)=5\frac{1}{20}$

Explanation of Solution

Given:

The sum $6\frac{4}{5}+\left(-1\frac{3}{4}\right)$

Concept Used:

The like fractions are those fractions whose denominators are equal.

The unlike fractions are those fractions whose denominators are unequal.

In addition of two or more like fractions, only numerators of all the fractions are to be added and the denominator will remain the same.

To add two or more unlike fractions, first convert all the fractions into like fractions, then do the addition of like fractions.

Calculation:

For the sum $6\frac{4}{5}+\left(-1\frac{3}{4}\right)$

Converting the mixed fractions into improper fractions

i.e., $6\frac{4}{5}=\frac{34}{5}$

and $-1\frac{3}{4}=-\frac{7}{4}$

The sum is $6\frac{4}{5}+\left(-1\frac{3}{4}\right)=\frac{34}{5}+\left(-\frac{7}{4}\right)$

As the denominators are different, so they are unlike fractions.

Converting them into like fractions.

Consider

  $\begin{array}{c}\frac{34}{5}=\frac{34\times 4}{5\times 4}\\ =\frac{136}{20}\end{array}$

And

  $\begin{array}{c}\frac{7}{4}=\frac{7\times 5}{4\times 5}\\ =\frac{35}{20}\end{array}$

Substituting the equivalent fractions

  $\begin{array}{c}\frac{34}{5}+\left(-\frac{7}{4}\right)=\frac{34}{5}-\frac{7}{4}\\ =\frac{136}{20}-\frac{35}{20}\\ =\frac{136-35}{20}\\ =\frac{101}{20}\end{array}$

$=5\frac{1}{20}$

Therefore, $\frac{34}{5}+\left(-\frac{7}{4}\right)=5\frac{1}{20}$

Conclusion:

  $\frac{34}{5}+\left(-\frac{7}{4}\right)=5\frac{1}{20}$