In this method, unlike in the other posts of this same name, we will multiply each term by the Least Common Multiple (LCM) of the denominators. It would give the same answer as if we had solved it by one or more of different optional ways.
\(\dfrac{5}{2x} + \dfrac{1}{7x} = 3\)How can you solve this problem by using the LCM?
\(\dfrac{5}{2x} + \dfrac{1}{7x} = 3\)
\(\dfrac{5}{2x} (14x) + \dfrac{1}{7x} (14x) = 3 (14x)\)Find the LCM of 2x and 7x:
" | " sign is not mathematical in this case. It's only to improve readability.
2 * 1 = 2 | 2 * 2 = 4 | 2 * 3 = 6 | 2 * 4 = 8 | 2 * 5 = 10 | 2 * 6 = 12 | 2 * 7 = 14
7 * 1 = 7 | 7 * 2 = 14
LCM = 14x
\(35 + 2 = 42x\)
\(37 = 42x\)
\(\dfrac{37}{42} = \dfrac{42x}{42}\)
\(\dfrac{37}{42} = x\)
\(\dfrac{5}{2x} + \dfrac{1}{7x} = 3\)
\(\dfrac{5}{2x} + \dfrac{1}{7x} = 3\)
\(\dfrac{5x^{-1}}{2} + \dfrac{1x^{-1}}{7} = 3\)
\(\dfrac{5x^{-1}}{2} (x) + \dfrac{1x^{-1}} {7} (x) = 3(x)\)
\(\dfrac{5}{2} + \dfrac{1}{7} = 3x\)
\(\dfrac{37}{14} = 3x\)
\(\dfrac{37}{14}(\dfrac{1}{3}) = 3x(\dfrac{1}{3})\)
\(\dfrac{37}{42} = x\)
\(\dfrac{5}{2x} + \dfrac{1}{7x} = 3\)
\(\dfrac{5}{2x} + \dfrac{1}{7x} = 3\)
\(\dfrac{5x^{-1}}{2} + \dfrac{1x^{-1}}{7} = 3\)
\(\dfrac{37x^{-1}}{14} = 3\)
\(\dfrac{37x^{-1}}{14} (\dfrac{14}{37}) = 3(\dfrac{14}{37})\)
\(x^{-1} = \dfrac{42}{37}\)
\((x^{-1})^{-1} = (\dfrac{42}{37})^{-1}\)
\(x = \dfrac{37}{42}\)
