How do you solve $\frac{3}{x} + \frac{2}{y} = - \frac{5}{6}$ $3/x +2/y = - 5/6$ and $\frac{1}{x} - \frac{1}{y} = - \frac{5}{6}$ $1/x - 1/y = -5/6$ using substitution?

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Answer 1 · 2018-01-09T16:17:11

$x = - 2$ $x=-2$ and $y = 3$ $y=3$

As $\frac{1}{x} - \frac{1}{y} = - \frac{5}{6}$ $1/x-1/y=-5/6$ , we have $\frac{1}{x} = \frac{1}{y} - \frac{5}{6}$ $1/x=1/y-5/6$

substituting $\frac{1}{x}$ $1/x$ in first equation we get

$3 (\frac{1}{y} - \frac{5}{6}) + \frac{2}{y} = - \frac{5}{6}$ $3(1/y-5/6)+2/y=-5/6$

or $\frac{3}{y} - \frac{15}{6} + \frac{2}{y} = - \frac{5}{6}$ $3/y-15/6+2/y=-5/6$

or $\frac{5}{y} = \frac{10}{6}$ $5/y=10/6$

or $10 y = 30$ $10y=30$ and $y = 3$ $y=3$

Hence $\frac{1}{x} = \frac{1}{3} - \frac{5}{6} = \frac{2 - 5}{6} = - \frac{3}{6} = - \frac{1}{2}$ $1/x=1/3-5/6=(2-5)/6=-3/6=-1/2$

i.e. $x = - 2$ $x=-2$

How do you solve 3x+2y=−563/x +2/y = - 5/6 and 1x−1y=−561/x - 1/y = -5/6 using substitution?