1Cademy - Solving $$rac{4q+3}{2} + 6 = rac{3q+5}{4}$$ by Clearing Fractions

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Strategy for Solving Equations with Fraction or Decimal Coefficients

Example

Solving $rac{4q+3}{2} + 6 = rac{3q+5}{4}$ by Clearing Fractions

To solve the linear equation $rac{4q+3}{2} + 6 = rac{3q+5}{4}$ by clearing fractions, first determine the least common denominator (LCD) of all terms. The denominators are $2$ and $4$ , so the LCD is $4$ . Multiply both sides of the equation by $4$ to clear the fractions. It is important to remember that every term must be multiplied by the LCD, including constant terms that do not contain a fraction.

ight) = 4\left( rac{3q+5}{4} ight)$$ Distribute the $$4$$ to both terms on the left: $$4\left( rac{4q+3}{2} ight) + 4(6) = 4\left( rac{3q+5}{4} ight)$$ Simplify by canceling common factors: $$2(4q+3) + 24 = 3q+5$$ Distribute to remove the parentheses: $$8q+6+24=3q+5$$ Combine like terms: $$8q+30=3q+5$$ Collect the variable terms to the left by subtracting $$3q$$ from both sides: $$5q+30=5$$ Collect the constant terms to the right by subtracting $$30$$ from both sides: $$5q=-25$$ Divide both sides by $$5$$: $$q=-5$$ Finally, check the solution by substituting $$-5$$ for $$q$$ in the original equation.

Updated 2026-04-22

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OpenStax Intermediate Algebra

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