Answer
False. Change to
$\displaystyle \frac{3}{4}+(-\frac{3}{5})=\frac{3}{20}$
Work Step by Step
Writing the fractions with common denominator of 20,
$\displaystyle \frac{3}{4}=\frac{3\times 5}{4\times 5}=\frac{15}{20},\qquad \frac{3}{5}=\frac{3\times 4}{5\times 4}=\frac{12}{20}$
$\displaystyle \frac{3}{4}+(-\frac{3}{5})=\frac{15}{20}+(-\frac{12}{20})$
... adding numbers with different signs, the sum has the sign of the number with the greater absolute value.
Here, it is positive, because $\displaystyle \frac{15}{20}$ has greater value.
So, the statement is false.
The true statement is obtained by:
$\displaystyle \frac{15}{20}+(-\frac{12}{20})$= $)=\displaystyle \frac{15}{20}+\frac{-12}{20}=\frac{15-12}{20}=\frac{3}{20}$
That is,
$\displaystyle \frac{3}{4}+(-\frac{3}{5})=\frac{3}{20}$
