Eighth Grade Mathematics: Clearing Fractions and Decimals in Equations

Although you can solve equations containing fractions and decimals by directly applying fractional arithmetic at every step, performing common denominator conversions and multi-digit decimal calculations creates frequent opportunities for arithmetic errors. A powerful algebraic technique known as clearing fractions and decimals transforms complicated rational equations into equivalent integer equations in a single step. In this chapter, you will master using the Least Common Denominator (LCD) to clear fractions and multiplying by powers of ten to clear decimals.

The Principle of Clearing Fractions Using the LCD

The Multiplication Property of Equality allows you to multiply both sides of an entire equation by any non-zero number of your choice without changing the solution. If you choose the Least Common Denominator (LCD) of all fractions appearing in the equation, every denominator divides evenly into that multiplier, eliminating all fractions immediately.

The Fraction Clearing Algorithm:
Step 1: Identify all denominators in the equation.
Step 2: Find the Least Common Multiple (LCM) of those denominators (the LCD).
Step 3: Multiply EVERY single term on BOTH sides of the equation by the LCD.
Step 4: Simplify each term; all denominators cancel out to 1!

Example: Solve (1/4)x + 2/3 = (5/6)x - 1/2

Step 1: Denominators are 4, 3, 6, and 2.
Step 2: The LCM of (4, 3, 6, 2) is 12. So LCD = 12.

Step 3: Multiply every term by 12:
        12 * [(1/4)x] + 12 * [2/3] = 12 * [(5/6)x] - 12 * [1/2]

Step 4: Cancel denominators:
        3x + 8 = 10x - 6

Step 5: Solve the equivalent integer equation:
        8 + 6 = 10x - 3x
        14 = 7x
        x = 2

Notice how multiplying by 12 replaced four messy fractions with clean integers (3x + 8 = 10x - 6), which resolves in two simple steps.

A Critical Caution: Multiply Every Single Term

A frequent mistake when clearing fractions is failing to multiply terms that are already integers. If an equation is (1/3)x + 5 = 7, you must multiply the 5 and the 7 by 3 as well: 3(1/3)x + 3(5) = 3*(7), giving x + 15 = 21. If you skip any term, the equation becomes unbalanced.

Clearing Decimals Using Powers of Ten

Decimals in base-10 represent fractions with denominators of 10, 100, 1000, and so on. To clear decimals from an equation, inspect all terms to determine the maximum number of decimal places present, and multiply the entire equation by the corresponding power of ten.

Decimal Clearing Rules:
- If greatest number of decimal places is 1 (tenths):       Multiply by 10
- If greatest number of decimal places is 2 (hundredths):   Multiply by 100
- If greatest number of decimal places is 3 (thousandths):  Multiply by 1000

Example: Solve 0.04x + 0.3 = 0.54 - 0.08x

Step 1: Maximum number of decimal places is 2 (hundredths in 0.04, 0.54, 0.08).
Step 2: Multiply every term on both sides by 100:
        100*(0.04x) + 100*(0.3) = 100*(0.54) - 100*(0.08x)

Step 3: Shift decimals two places right:
        4x + 30 = 54 - 8x
        (Notice that 100 * 0.3 = 30, NOT 3!)

Step 4: Solve the integer equation:
        4x + 8x = 54 - 30
        12x = 24
        x = 2

Be especially vigilant with terms like 0.3 when multiplying by 100: shifting the decimal point two places to the right requires adding a trailing zero to make 30.

Chapter Exercises

Exercise 1: Solve the fractional equation (2/5)x - 1/3 = (1/2)x + 1/6 by clearing fractions. State the LCD used, show the integer equation produced, and solve for x.

Exercise 2: Solve the decimal equation 0.25*(x + 4) - 0.05x = 1.6 by clearing decimals. Show whether you distribute first or multiply by 100 first, and explain your reasoning in paragraph form.

Exercise 3: A student attempts to clear fractions in the equation (1/2)x + 4 = 3/4 by multiplying by 4. They write 2x + 4 = 3. Identify the student's mistake, explain why it violates algebraic balance, and solve the equation correctly.

Solutions and Explanations

Solution 1: To solve (2/5)x - 1/3 = (1/2)x + 1/6, first identify the denominators: 5, 3, 2, and 6. The least common multiple of these numbers is 30, so the LCD is 30. Multiply every term on both sides by 30: 30 * [(2/5)x] - 30 * [1/3] = 30 * [(1/2)x] + 30 * [1/6]. Simplifying each term: (30 * 2 / 5)x = 12x; 30 * (1/3) = 10; (30 * 1 / 2)x = 15x; and 30 * (1/6) = 5. This produces the equivalent integer equation 12x - 10 = 15x + 5. To solve, subtract 12x from both sides: -10 = 3x + 5. Next, subtract 5 from both sides: -15 = 3x. Finally, divide by 3: x = -15 / 3 = -5.

Solution 2: When solving 0.25(x + 4) - 0.05x = 1.6, the cleanest approach is to distribute first to eliminate parentheses, which gives 0.25x + 1.0 - 0.05x = 1.6. Next, observe that the maximum number of decimal places across all terms is two (in 0.25x and 0.05x). Multiply every term on both sides of the equation by 100: 100(0.25x) + 100(1.0) - 100(0.05x) = 100(1.6). This shifts each decimal point two places to the right, yielding the integer equation 25x + 100 - 5x = 160. Combine the like variable terms on the left: 20x + 100 = 160. Subtract 100 from both sides: 20x = 60. Divide both sides by 20 to get x = 3. Checking: 0.25(3 + 4) - 0.05(3) = 0.25(7) - 0.15 = 1.75 - 0.15 = 1.6, verifying x = 3.

Solution 3: The student made the critical error of multiplying only the terms containing fractions by 4, completely failing to multiply the constant term 4. The Multiplication Property of Equality requires multiplying the entire expression on the left side by 4: 4 * [(1/2)x + 4] = 4 * (1/2)x + 4 * (4) = 2x + 16. By writing 2x + 4, the student left the constant term unscaled, destroying the balance of the equation. The correct integer equation after multiplying every term by 4 is 2x + 16 = 3. Subtracting 16 from both sides yields 2x = -13. Dividing by 2 yields the correct solution x = -13/2, or -6.5.