# McGraw Hill Math Grade 5 Chapter 6 Lesson 7 Answer Key Subtracting Fractions with Unlike Denominators

All the solutions provided in McGraw Hill Math Grade 5 Answer Key PDF Chapter 6 Lesson 7 Subtracting Fractions with Unlike Denominators are as per the latest syllabus guidelines.

## McGraw-Hill Math Grade 5 Answer Key Chapter 6 Lesson 7 Subtracting Fractions with Unlike Denominators

Solve

Find equivalent fractions and subtract. Write the difference in simplest form.

Question 1.
$$\frac{2}{9}$$ – $$\frac{1}{6}$$ _____________
$$\frac{4}{18}$$ – $$\frac{3}{18}$$ = $$\frac{1}{18}$$

Question 2.
$$\frac{1}{2}$$ – $$\frac{1}{3}$$ _______________
$$\frac{3}{6}$$–$$\frac{2}{6}$$ = $$\frac{1}{6}$$.

Explanation:
The equivalent fraction of $$\frac{1}{2}$$ – $$\frac{1}{3}$$ which is $$\frac{3}{6}$$ and $$\frac{2}{6}$$. So the addition of $$\frac{3}{6}$$–$$\frac{2}{6}$$ is $$\frac{3-2}{6}$$
= $$\frac{1}{6}$$.

Question 3.
$$\frac{3}{8}$$ – $$\frac{1}{3}$$ _______________
$$\frac{3}{6}$$–$$\frac{2}{6}$$ = $$\frac{1}{6}$$.

Explanation:
The equivalent fraction of $$\frac{1}{2}$$ – $$\frac{1}{3}$$ which is $$\frac{3}{6}$$ and $$\frac{2}{6}$$. So the addition of $$\frac{3}{6}$$–$$\frac{2}{6}$$ is $$\frac{3-2}{6}$$
= $$\frac{1}{6}$$.

Question 4.
$$\frac{3}{4}$$ – $$\frac{1}{5}$$ _______________
$$\frac{15}{20}$$–$$\frac{4}{20}$$ = $$\frac{11}{20}$$.

Explanation:
The equivalent fraction of $$\frac{3}{4}$$ – $$\frac{1}{5}$$ which is $$\frac{15}{20}$$ and $$\frac{4}{20}$$. So the addition of $$\frac{15}{20}$$–$$\frac{4}{20}$$ is $$\frac{15-4}{20}$$
= $$\frac{11}{20}$$.

Question 5. $$\frac{7}{14}$$–$$\frac{6}{14}$$ = $$\frac{1}{14}$$.

Explanation:
The equivalent fraction of $$\frac{1}{2}$$ – $$\frac{3}{7}$$ which is $$\frac{7}{14}$$ and $$\frac{6}{14}$$. So the addition of $$\frac{7}{14}$$–$$\frac{6}{14}$$ is $$\frac{7-6}{14}$$
= $$\frac{1}{14}$$.

Question 6. $$\frac{8}{12}$$–$$\frac{3}{12}$$ = $$\frac{5}{12}$$.

Explanation:
The equivalent fraction of $$\frac{2}{3}$$ – $$\frac{1}{4}$$ which is $$\frac{8}{12}$$ and $$\frac{3}{12}$$. So the addition of $$\frac{8}{12}$$–$$\frac{3}{12}$$ is $$\frac{8-3}{12}$$
= $$\frac{5}{12}$$.

Question 7. $$\frac{11}{99}$$–$$\frac{9}{99}$$ = $$\frac{2}{99}$$.

Explanation:
The equivalent fraction of $$\frac{1}{9}$$ – $$\frac{1}{11}$$ which is $$\frac{11}{99}$$ and $$\frac{9}{99}$$. So the addition of $$\frac{11}{99}$$–$$\frac{9}{99}$$ is $$\frac{11-9}{99}$$
= $$\frac{2}{99}$$.

Question 8. $$\frac{7}{42}$$–$$\frac{6}{42}$$ = $$\frac{1}{42}$$.
The equivalent fraction of $$\frac{1}{6}$$ – $$\frac{1}{7}$$ which is $$\frac{7}{42}$$ and $$\frac{6}{42}$$. So the addition of $$\frac{7}{42}$$–$$\frac{6}{42}$$ is $$\frac{7-6}{42}$$
= $$\frac{1}{42}$$.