Example

Solving rac{3r+5}{6} + 1 = rac{4r+3}{3} by Clearing Fractions

To solve the linear equation rac{3r+5}{6} + 1 = rac{4r+3}{3} by clearing fractions, identify the LCD of the denominators 66 and 33, which is 66. Multiply every term on both sides of the equation by 66 to eliminate the fractions:

ight) + 6(1) = 6\left( rac{4r+3}{3} ight)$$ Simplify each term to clear the denominators: $$3r+5+6=2(4r+3)$$ Combine like terms on the left and distribute on the right: $$3r+11=8r+6$$ Collect the variable terms on the right by subtracting $$3r$$ from both sides: $$11=5r+6$$ Collect the constant terms on the left by subtracting $$6$$ from both sides: $$5=5r$$ Divide both sides by $$5$$: $$r=1$$ Checking this solution by substituting $$1$$ back into the original equation verifies that both sides match, so the solution is $$r=1$$.

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Updated 2026-04-22

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