Ratio Algebra Word Problem

[Deleted member 4993]

We have now proven the two strategies produce the same answer of \(\displaystyle \dfrac{20}{3}\)

Looking at the 1st one.

\(\displaystyle \dfrac{x}{12} + \dfrac{x}{15} = 1\)

\(\displaystyle (60x)\dfrac{x}{12} + (60x)\dfrac{x}{15} = (60x)1\) Why are you multiplying by 60 x? The LCM of (12, 15, 1) is 60 - not 60 x.

\(\displaystyle \dfrac{60x^{2}}{12} + \dfrac{60x^{2}}{15} = 60x\)

\(\displaystyle 5x^{2} + 4x^{2} = 60x\)

\(\displaystyle 9x^{2} = 60x\)

\(\displaystyle \dfrac{9x^{2}}{x}= \dfrac{60x}{x}\)

\(\displaystyle 9x = 60\)

\(\displaystyle x = \dfrac{60}{9}\)

\(\displaystyle x = \dfrac{20}{3}\)

Looking at the 2nd one.

\(\displaystyle \dfrac{1}{12} + \dfrac{1}{15} = \dfrac{1}{x}\)

\(\displaystyle (60x)\dfrac{1}{12} + (60x)\dfrac{1}{15} = (60x)\dfrac{1}{x}\)

\(\displaystyle \dfrac{60x}{12} + \dfrac{60x}{15} = \dfrac{60x}{x}\)

\(\displaystyle 5x + 4x = 60\)

\(\displaystyle 9x = 60\)

\(\displaystyle x = \dfrac{60}{9}\)

\(\displaystyle x = \dfrac{20}{3}\)

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[Jason76]

Now Corrected

We have now proven the two strategies produce the same answer of \(\displaystyle \dfrac{20}{3}\)

Looking at the 1st one.

\(\displaystyle \dfrac{x}{12} + \dfrac{x}{15} = 1\)

\(\displaystyle (60)\dfrac{x}{12} + (60)\dfrac{x}{15} = (60)1\)

\(\displaystyle \dfrac{60x}{12} + \dfrac{60}{15} = 60\)

\(\displaystyle 5x + 4x = 60\)

\(\displaystyle 9x = 60\)

\(\displaystyle x = \dfrac{60}{9}\)

\(\displaystyle x = \dfrac{20}{3}\)

Looking at the 2nd one.

\(\displaystyle \dfrac{1}{12} + \dfrac{1}{15} = \dfrac{1}{x}\)

\(\displaystyle (60x)\dfrac{1}{12} + (60x)\dfrac{1}{15} = (60x)\dfrac{1}{x}\)

\(\displaystyle \dfrac{60x}{12} + \dfrac{60x}{15} = \dfrac{60x}{x}\)

\(\displaystyle 5x + 4x = 60\)

\(\displaystyle 9x = 60\)

\(\displaystyle x = \dfrac{60}{9}\)

\(\displaystyle x = \dfrac{20}{3}\)