080e. Solving Equations Involving Fractions

Learning Intentions

• To understand that to solve an equation with a fraction, it is helpful to multiply both sides by its denominator
• solve equations involving fractions

Pre-requisite Summary

• Understand that an equation states that two expressions are equal
• Know that equivalent equations have the same solution
• Understand that the same operation can be applied to both sides of an equation
• Know that multiplying by a denominator can remove a fraction
• Recall multiplication facts and simple fraction notation
• Be able to solve one-step and two-step equations with whole numbers
• Be able to simplify expressions after multiplying brackets where needed
• Understand that solutions should be checked by substitution

Worked Examples

Worked Example 1

a) Explain why multiplying both sides of ${\displaystyle \frac{x}{3}}=5$ by $3$ is helpful.
b) Solve ${\displaystyle \frac{x}{3}}=5$.
c) Check the solution.

Worked Example 2

Solve each equation involving a fraction:
a) ${\displaystyle \frac{x}{4}}=6$
b) ${\displaystyle \frac{x}{5}}=9$
c) ${\displaystyle \frac{x}{8}}=7$

Worked Example 3

Solve each equation involving a fraction:
a) ${\displaystyle \frac{x}{3}}+2=7$
b) ${\displaystyle \frac{x}{4}}-1=5$
c) ${\displaystyle \frac{x}{6}}+3=8$

Worked Example 4

Solve each equation involving a fraction:
a) ${\displaystyle \frac{x+2}{5}}=3$
b) ${\displaystyle \frac{x-4}{3}}=6$
c) ${\displaystyle \frac{2x}{7}}=4$

Worked Example 5

Solve each equation involving a fraction:
a) ${\displaystyle \frac{x}{2}}+{\displaystyle \frac{1}{2}}=4$
b) ${\displaystyle \frac{x}{5}}-{\displaystyle \frac{2}{5}}=3$
c) ${\displaystyle \frac{x+1}{4}}=2$

Worked Example 6

For each equation:
a) multiply both sides by the denominator
b) solve the simpler equation
c) check the solution by substitution
For ${\displaystyle \frac{x-3}{6}}=2$

Problems

Problem 1

a) Explain why multiplying both sides of ${\displaystyle \frac{x}{7}}=4$ is helpful.
b) Solve ${\displaystyle \frac{x}{7}}=4$.
c) Check the solution.

Problem 2

Solve each equation involving a fraction:
a) ${\displaystyle \frac{x}{3}}=8$
b) ${\displaystyle \frac{x}{6}}=5$
c) ${\displaystyle \frac{x}{9}}=4$

Problem 3

Solve each equation involving a fraction:
a) ${\displaystyle \frac{x}{5}}+1=6$
b) ${\displaystyle \frac{x}{2}}-4=3$
c) ${\displaystyle \frac{x}{7}}+2=6$

Problem 4

Solve each equation involving a fraction:
a) ${\displaystyle \frac{x+3}{4}}=5$
b) ${\displaystyle \frac{x-2}{6}}=3$
c) ${\displaystyle \frac{3x}{8}}=6$

Problem 5

Solve each equation involving a fraction:
a) ${\displaystyle \frac{x}{3}}+{\displaystyle \frac{2}{3}}=5$
b) ${\displaystyle \frac{x}{4}}-{\displaystyle \frac{1}{4}}=2$
c) ${\displaystyle \frac{x-5}{2}}=7$

Problem 6

For each equation:
a) multiply both sides by the denominator
b) solve the simpler equation
c) check the solution by substitution
For ${\displaystyle \frac{x+4}{5}}=3$

Exercises

Understanding and Fluency

1. State the denominator in each fraction:
   a) ${\displaystyle \frac{x}{3}}$
   b) ${\displaystyle \frac{x+1}{5}}$
   c) ${\displaystyle \frac{2x}{7}}$

2. State what you would multiply both sides by to remove the fraction:
   a) ${\displaystyle \frac{x}{4}}=9$
   b) ${\displaystyle \frac{x+3}{6}}=2$
   c) ${\displaystyle \frac{5x}{8}}=7$

3. Solve each equation:
   a) ${\displaystyle \frac{x}{2}}=6$
   b) ${\displaystyle \frac{x}{3}}=9$
   c) ${\displaystyle \frac{x}{5}}=4$

4. Solve each equation:
   a) ${\displaystyle \frac{x}{4}}+2=7$
   b) ${\displaystyle \frac{x}{6}}-1=3$
   c) ${\displaystyle \frac{x}{8}}+5=9$

5. Solve each equation:
   a) ${\displaystyle \frac{x+1}{3}}=4$
   b) ${\displaystyle \frac{x-2}{5}}=6$
   c) ${\displaystyle \frac{x+4}{7}}=3$

6. Solve each equation:
   a) ${\displaystyle \frac{2x}{3}}=8$
   b) ${\displaystyle \frac{3x}{4}}=9$
   c) ${\displaystyle \frac{5x}{6}}=10$

7. Solve each equation:
   a) ${\displaystyle \frac{x}{2}}+{\displaystyle \frac{1}{2}}=5$
   b) ${\displaystyle \frac{x}{3}}-{\displaystyle \frac{2}{3}}=4$
   c) ${\displaystyle \frac{x+2}{4}}=3$

8. Solve each equation and check by substitution:
   a) ${\displaystyle \frac{x-1}{5}}=2$
   b) ${\displaystyle \frac{x+6}{3}}=7$
   c) ${\displaystyle \frac{2x}{5}}=6$

Reasoning

9. Explain why multiplying both sides of an equation by the denominator does not change the solution.

10. A student solves ${\displaystyle \frac{x}{4}}=6$ by adding $4$ to both sides. Explain the mistake.

11. Explain why ${\displaystyle \frac{x+3}{5}}=2$ becomes $x+3=10$ after multiplying both sides by $5$.

12. A student solves ${\displaystyle \frac{2x}{3}}=8$ by multiplying only the left side by $3$. Explain why this is incorrect.

13. Explain why checking by substitution is useful after solving an equation with a fraction.

14. A student says ${\displaystyle \frac{x}{6}}+2=5$ can be solved by multiplying by $6$ first or subtracting $2$ first. Explain which approach is clearer and why.

Problem-solving

15. A recipe equation is ${\displaystyle \frac{x}{4}}=3$, where $x$ is the number of tablespoons of flour. Solve for $x$.

16. A ticket problem is modelled by ${\displaystyle \frac{x}{5}}+2=8$, where $x$ is the ticket cost in dollars before the fixed charge. Solve for $x$.

17. A container problem is modelled by ${\displaystyle \frac{x-3}{6}}=4$. Solve for $x$.

18. A phone data problem is modelled by ${\displaystyle \frac{2x}{7}}=6$, where $x$ is the number of gigabytes used. Solve for $x$.

19. A length problem is modelled by ${\displaystyle \frac{x+5}{3}}=9$. Solve for $x$.

20. A saving problem is modelled by ${\displaystyle \frac{x}{8}}-1=4$, where $x$ is the number of dollars saved. Solve for $x$.

Potential Misunderstandings

• Students may multiply only one side of the equation by the denominator
• Students may confuse the denominator with the numerator
• Students may try to add or subtract the denominator instead of multiplying by it
• Students may forget that the entire numerator is divided by the denominator
• Students may fail to use brackets correctly when multiplying both sides
• Students may solve the simplified equation incorrectly after removing the fraction
• Students may not check whether the final value makes the original equation true
• Students may think multiplying by the denominator changes the solution rather than creating an equivalent equation

Next: 081e. Expanding Brackets and Simplifying Expressions