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.

\(\frac{2}{9}\) – \(\frac{1}{6}\) _____________

Answer:

\(\frac{4}{18}\) – \(\frac{3}{18}\) = \(\frac{1}{18}\)

Question 2.

\(\frac{1}{2}\) – \(\frac{1}{3}\) _______________

Answer:

\(\frac{3}{6}\)–\(\frac{2}{6}\) = \(\frac{1}{6}\).

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.

\(\frac{3}{8}\) – \(\frac{1}{3}\) _______________

Answer:

\(\frac{3}{6}\)–\(\frac{2}{6}\) = \(\frac{1}{6}\).

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.

\(\frac{3}{4}\) – \(\frac{1}{5}\) _______________

Answer:

\(\frac{15}{20}\)–\(\frac{4}{20}\) = \(\frac{11}{20}\).

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.

Answer:

\(\frac{7}{14}\)–\(\frac{6}{14}\) = \(\frac{1}{14}\).

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.

Answer:

\(\frac{8}{12}\)–\(\frac{3}{12}\) = \(\frac{5}{12}\).

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.

Answer:

\(\frac{11}{99}\)–\(\frac{9}{99}\) = \(\frac{2}{99}\).

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.

Answer:

\(\frac{7}{42}\)–\(\frac{6}{42}\) = \(\frac{1}{42}\).

Explanation:

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 \(\frac{7-6}{42}\)

= \(\frac{1}{42}\).