You don't like fractions? - How about fraction busting

Which of the following equations looks easier to you?

\(\frac{3}{4}x + \frac{1}{2} = -\frac{2}{3}x + \frac{5}{6}\)

or

\(9x + 6 = -8x + 10\)

If you, like me, think that the second equation looks easier, you can simply use the fraction busting method to convert the first into the second equation: Multiply all terms by the Least Common Denominator (LCD) of all the fractions, and the fractions miraculously disappear. In this example the LCD is 12.

\((12)(\frac{3}{4})x + (12)(\frac{1}{2}) = (12)(-\frac{2}{3})x + (12)(\frac{5}{6})\)

\((\frac{\enclose{horizontalstrike}{12}}{1})(\frac{3}{\enclose{horizontalstrike}{4}})x + (\frac{\enclose{horizontalstrike}{12}}{1})(\frac{1}{\enclose{horizontalstrike}{2}}) = (\frac{\enclose{horizontalstrike}{12}}{1})(-\frac{2}{\enclose{horizontalstrike}{3}})x + (\frac{\enclose{horizontalstrike}{12}}{1})(\frac{5}{\enclose{horizontalstrike}{6}})\)

\((\frac{3}{1})(\frac{3}{1})x + (\frac{6}{1})(\frac{1}{1}) = (\frac{4}{1})(-\frac{2}{1})x + (\frac{2}{1})(\frac{5}{1})\)

\(9x + 6 = -8x + 10\)

Why does this work? The LCD is a multiple of all the denominators. Therefore, all products of the LCD and the fractions can be cross-simplified, and all denominators can be reduced to 1.

You don’t have fractions, but decimals? No problem, the method works here, too (decimal busting):

\(1.45x - 3 = 0.9x + 2.92\)

You multiply all terms by the same power of 10 so that all decimals disappear.

\((100)(1.45)x - (100)(3) = (100)(0.9)x + (100)(2.92)\)

\(145x - 300 = 90x + 292\)

I know that my students love this method, but let me know how it worked for you □