Question

7 - m2 - ta - lesson 5

1. $-\frac{3}{10}+1\frac{9}{10}$
2. $23.5+(-33.5)$
3. $-1\frac{1}{3}+(-5\frac{1}{3})$
4. $-2\frac{3}{4}+3\frac{1}{4}$
5. $-4\frac{1}{4}+(-2\frac{3}{4})$
6. $4\frac{2}{3}+(-3\frac{1}{3})$

Explanation:

Step1: Convert mixed - numbers to improper fractions (if applicable)

For problem 9: $1\frac{9}{10}=\frac{1\times10 + 9}{10}=\frac{19}{10}$, so $-\frac{3}{10}+1\frac{9}{10}=-\frac{3}{10}+\frac{19}{10}$.
For problem 11: $-1\frac{1}{3}=-\frac{1\times3+1}{3}=-\frac{4}{3}$ and $-5\frac{1}{3}=-\frac{5\times3 + 1}{3}=-\frac{16}{3}$, so $-1\frac{1}{3}+(-5\frac{1}{3})=-\frac{4}{3}-\frac{16}{3}$.
For problem 12: $-2\frac{3}{4}=-\frac{2\times4 + 3}{4}=-\frac{11}{4}$ and $3\frac{1}{4}=\frac{3\times4+1}{4}=\frac{13}{4}$, so $-2\frac{3}{4}+3\frac{1}{4}=-\frac{11}{4}+\frac{13}{4}$.
For problem 13: $-4\frac{1}{4}=-\frac{4\times4+1}{4}=-\frac{17}{4}$ and $-2\frac{3}{4}=-\frac{2\times4 + 3}{4}=-\frac{11}{4}$, so $-4\frac{1}{4}+(-2\frac{3}{4})=-\frac{17}{4}-\frac{11}{4}$.
For problem 14: $4\frac{2}{3}=\frac{4\times3 + 2}{3}=\frac{14}{3}$ and $-3\frac{1}{3}=-\frac{3\times3+1}{3}=-\frac{10}{3}$, so $4\frac{2}{3}+(-3\frac{1}{3})=\frac{14}{3}-\frac{10}{3}$.

Step2: Add or subtract fractions with the same denominator

Problem 9: $-\frac{3}{10}+\frac{19}{10}=\frac{-3 + 19}{10}=\frac{16}{10}=\frac{8}{5}=1\frac{3}{5}$.
Problem 10: $23.5+(-33.5)=23.5 - 33.5=-10$.
Problem 11: $-\frac{4}{3}-\frac{16}{3}=\frac{-4-16}{3}=\frac{-20}{3}=-6\frac{2}{3}$.
Problem 12: $-\frac{11}{4}+\frac{13}{4}=\frac{-11 + 13}{4}=\frac{2}{4}=\frac{1}{2}$.
Problem 13: $-\frac{17}{4}-\frac{11}{4}=\frac{-17-11}{4}=\frac{-28}{4}=-7$.
Problem 14: $\frac{14}{3}-\frac{10}{3}=\frac{14 - 10}{3}=\frac{4}{3}=1\frac{1}{3}$.

Answer:

1. $1\frac{3}{5}$
2. $-10$
3. $-6\frac{2}{3}$
4. $\frac{1}{2}$
5. $-7$
6. $1\frac{1}{3}$