Unit 2 · Lesson 2 of 6
Equations with fractions and decimals
You'll be able to: turn an equation full of fractions or decimals into one with whole numbers only, then solve it.
Step 1 of 5Warm-up
Two quick questions from earlier lessons. Try them from memory before you look anything up.
Solve \(4x - 7 = 2x + 5\). Type the value of \(x\).
Subtract \(2x\) from both sides: \(2x - 7 = 5\).
Add 7: \(2x = 12\). Divide by 2: \(x = 6\).
Add \(\dfrac{1}{4} + \dfrac{5}{6}\).
The LCD of 4 and 6 is 12.
\(\dfrac{1}{4} = \dfrac{3}{12}\) and \(\dfrac{5}{6} = \dfrac{10}{12}\).
Answer: \(\dfrac{3}{12} + \dfrac{10}{12} = \dfrac{13}{12}\).
Step 2 of 5The idea
Fractions make equations look scary, but you can get rid of them in one move. Multiply both sides by the LCD (least common denominator). Every denominator divides into it, so all the fractions cancel.
What is left is an ordinary equation like the ones in the last lesson.
Fractions: find the LCD of all the denominators. Multiply every term on both sides by it. Then solve as usual.
Decimals: multiply every term by 10 if the most decimal places is one, by 100 if it is two, and so on. Then solve as usual.
A top with two terms, like \(\dfrac{x+1}{4}\), gets parentheses when you clear it: \(12 \cdot \dfrac{x+1}{4} = 3(x+1)\).
Why this shows up on the placement test: test equations often use fractions on purpose, and clearing them first is the fastest way to avoid fraction arithmetic mistakes.
Step 3 of 5Worked examples
Example 1: \(\dfrac{x+1}{4} - \dfrac{x}{6} = 1\)
- The denominators are 4 and 6. The LCD is 12.
- Multiply every term by 12: \(12 \cdot \dfrac{x+1}{4} - 12 \cdot \dfrac{x}{6} = 12 \cdot 1\).
- Cancel: \(3(x+1) - 2x = 12\). 12 ÷ 4 = 3 and 12 ÷ 6 = 2. Keep the parentheses around \(x+1\).
- Distribute and combine: \(3x + 3 - 2x = 12\), so \(x + 3 = 12\), and \(x = 9\).
- Check: \(\dfrac{10}{4} - \dfrac{9}{6} = 2.5 - 1.5 = 1\). It works.
Example 2: \(0.4x - 1.2 = 0.1x + 0.3\)
- The most decimal places in any number is one, so multiply every term by 10.
- \(4x - 12 = x + 3\). Moving each decimal point one place right.
- Subtract \(x\): \(3x - 12 = 3\). Add 12: \(3x = 15\). Divide by 3: \(x = 5\).
- Check: \(0.4(5) - 1.2 = 0.8\) and \(0.1(5) + 0.3 = 0.8\). It works.
Step 4 of 5Practice
Clear the fractions or decimals first, then solve. Type your answer and press Check. If you miss twice, open the worked solution.
Solve \(\dfrac{x}{4} + 2 = 7\). Type the value of \(x\).
Multiply every term by 4: \(x + 8 = 28\).
Subtract 8: \(x = 20\).
Check: \(\dfrac{20}{4} + 2 = 5 + 2 = 7\). Answer: \(20\).
Solve \(\dfrac{2x}{3} - 1 = \dfrac{1}{3}\). Type the value of \(x\).
Multiply every term by 3: \(2x - 3 = 1\).
Add 3: \(2x = 4\). Divide by 2: \(x = 2\).
Check: \(\dfrac{4}{3} - 1 = \dfrac{1}{3}\). Answer: \(2\).
Solve \(\dfrac{x}{2} + \dfrac{x}{3} = 10\). Type the value of \(x\).
The LCD of 2 and 3 is 6. Multiply every term by 6: \(3x + 2x = 60\).
Combine: \(5x = 60\), so \(x = 12\).
Check: \(6 + 4 = 10\). Answer: \(12\).
Solve \(\dfrac{x - 2}{3} = \dfrac{x + 1}{4}\). Type the value of \(x\).
The LCD is 12. Multiply both sides by 12: \(4(x - 2) = 3(x + 1)\).
Distribute: \(4x - 8 = 3x + 3\). Subtract \(3x\): \(x - 8 = 3\), so \(x = 11\).
Check: \(\dfrac{9}{3} = 3\) and \(\dfrac{12}{4} = 3\). Answer: \(11\).
Solve \(\dfrac{3x}{4} - \dfrac{1}{2} = \dfrac{x}{8} + 2\). Type the value of \(x\).
The LCD of 4, 2, and 8 is 8. Multiply every term by 8: \(6x - 4 = x + 16\).
Subtract \(x\): \(5x - 4 = 16\). Add 4: \(5x = 20\), so \(x = 4\).
Check: \(3 - \dfrac{1}{2} = 2.5\) and \(\dfrac{4}{8} + 2 = 2.5\). Answer: \(4\).
Solve \(0.3x + 1.5 = 0.8x - 2\). Type the value of \(x\).
The most decimal places is one, so multiply every term by 10: \(3x + 15 = 8x - 20\).
Subtract \(3x\): \(15 = 5x - 20\). Add 20: \(35 = 5x\), so \(x = 7\).
Check: \(2.1 + 1.5 = 3.6\) and \(5.6 - 2 = 3.6\). Answer: \(7\).
Solve \(0.05x + 0.1(x + 4) = 1.6\). Type the value of \(x\).
The most decimal places is two (in 0.05), so multiply every term by 100: \(5x + 10(x + 4) = 160\).
Distribute: \(5x + 10x + 40 = 160\), so \(15x + 40 = 160\).
Subtract 40: \(15x = 120\), so \(x = 8\).
Check: \(0.4 + 0.1(12) = 0.4 + 1.2 = 1.6\). Answer: \(8\).
You multiply both sides of \(\dfrac{x}{6} - \dfrac{x - 1}{4} = \dfrac{1}{3}\) by 12. Which equation do you get?
- \(2x - 3x - 1 = 4\)
- \(2x - 3(x - 1) = 1\)
- \(2x - (x - 1) = 4\)
- \(2x - 3(x - 1) = 4\)
\(12 \cdot \dfrac{x}{6} = 2x\), \(12 \cdot \dfrac{x-1}{4} = 3(x-1)\), and \(12 \cdot \dfrac{1}{3} = 4\). So the equation is \(2x - 3(x - 1) = 4\). Answer (d).
(a) lost the parentheses, (b) multiplied the right side by 3 instead of 12, and (c) forgot the 3.
To finish: \(2x - 3x + 3 = 4\), so \(-x = 1\) and \(x = -1\).
Step 5 of 5Wrap-up
You can now clear fractions by multiplying every term by the LCD, clear decimals by multiplying by 10 or 100, and then solve the whole-number equation that is left.
Read or watch next: OpenStax Elementary Algebra 2e, 2.5 Solve Equations with Fractions or Decimals (free textbook, with more practice at the end of the section).
Also good: Khan Academy: Pre-algebra: Two-step equations with decimals and fractions (short videos). Quick rules: Equations and inequalities cheat sheet.
Something unclear? Ask your teacher bot a follow-up question, for example: "Why do I put parentheses around x + 1 when I clear the fraction?" It knows this course and can explain things another way.