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# Linear Algebra

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If e=2 1/2 and f=3 1/3, then what is 4e-2f equal

Mar 25, 2015

#1
+2353
+10

Here's how I solved it

$$\begin{array}{lcl} e = 2\frac{1}{2} = \frac{4}{2} + \frac{1}{2} = \frac{5}{2}\\ f = 3\frac{1}{3} = \frac{9}{3} + \frac{1}{3} = \frac{10}{3}\\ 4e-2f = 4*\frac{5}{2} - 2*\frac{10}{3}\\ = \frac{20}{2} - \frac{20}{3}\\ = \frac{20}{2} * 1 - \frac{20}{3} * 1\\ = \frac{20}{2} * \frac{3}{3} - \frac{20}{3} * \frac{2}{2}\\ = \frac{60}{6} - \frac{40}{6}\\ = \frac{60-40}{6}\\ = \frac{20}{6}\\ = \frac{10}{3}\\ = 3 \frac{1}{3} \end{array}$$

Reinout

Mar 25, 2015

#1
+2353
+10
$$\begin{array}{lcl} e = 2\frac{1}{2} = \frac{4}{2} + \frac{1}{2} = \frac{5}{2}\\ f = 3\frac{1}{3} = \frac{9}{3} + \frac{1}{3} = \frac{10}{3}\\ 4e-2f = 4*\frac{5}{2} - 2*\frac{10}{3}\\ = \frac{20}{2} - \frac{20}{3}\\ = \frac{20}{2} * 1 - \frac{20}{3} * 1\\ = \frac{20}{2} * \frac{3}{3} - \frac{20}{3} * \frac{2}{2}\\ = \frac{60}{6} - \frac{40}{6}\\ = \frac{60-40}{6}\\ = \frac{20}{6}\\ = \frac{10}{3}\\ = 3 \frac{1}{3} \end{array}$$