How do you solve the following system?: $\frac{1}{6} x - \frac{3}{5} y = 1, \frac{4}{5} x + \frac{2}{3} y = 2$ $1/6x-3/5y=1 , 4/5x+2/3y = 2$

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How do you solve the following system?: $\frac{1}{6} x - \frac{3}{5} y = 1, \frac{4}{5} x + \frac{2}{3} y = 2$ $1/6x-3/5y=1 , 4/5x+2/3y = 2$

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Answer 1 · 2018-07-10T19:23:39

$x = \frac{60}{19}, y = - \frac{15}{19}$ $x=60/19 , y= -15/19$

Explanation:

Multiply the first equation by 10 $\Rightarrow \frac{10}{6} x - 6 y = 10$ $=> 10/6x-6y=10$

Multiply the second equation by 9 $\Rightarrow \frac{36}{5} x + 6 y = 18$ $=> 36/5x+6y=18$

Add these two equations together to give $\frac{133}{15} x = 28$ $133/15x=28$

Divide by $\frac{133}{15} \Rightarrow x = \frac{60}{19}$ $133/15 => x=60/19$

Substitute this into the first equation

$\Rightarrow \frac{1}{6} \times \frac{60}{19} - \frac{3}{5} y = 1$ $=> 1/6xx60/19-3/5y=1$

$\Rightarrow \frac{10}{19} - \frac{3}{5} y = 1$ $=> 10/19-3/5y=1$

$\Rightarrow - \frac{3}{5} y = \frac{9}{19}$ $=>-3/5y=9/19$

$\Rightarrow y = - \frac{15}{19}$ $=> y= -15/19$

Answer 2 · 2018-07-10T19:24:29

$x = \frac{60}{19}, y = - \frac{15}{19}$ $x=60/19,y=-15/19$

Explanation:

Solving the first equation for $x$ $x$ :

$\frac{1}{6} \cdot x = 1 + \frac{3}{5} \cdot y$ $1/6*x=1+3/5*y$
so
$x = 6 + \frac{18}{5} y$ $x=6+18/5y$
substituting this in the second equation:

$\frac{4}{5} \cdot (6 + \frac{18}{5} y) + \frac{2}{3} y = 2$ $4/5*(6+18/5y)+2/3y=2$

expanding

$\frac{24}{5} + \frac{72}{25} y + \frac{2}{3} y = 2$ $24/5+72/25y+2/3y=2$
adding $- \frac{24}{5}$ $-24/5$

$\frac{72}{25} y + \frac{2}{3} y = \frac{10 - 24}{5}$ $72/25y+2/3y=(10-24)/5$

note that $\frac{72}{25} + \frac{2}{3} = \frac{266}{75}$ $72/25+2/3=266/75$

so we get

$\frac{266}{75} y = - \frac{14}{5}$ $266/75y=-14/5$

multiplying by $\frac{75}{266}$ $75/266$

we get $y = - \frac{15}{19}$ $y=-15/19$

so $x = 6 + \frac{18}{5} \cdot (- \frac{15}{19})$ $x=6+18/5*(-15/19)$

$x = \frac{60}{19}$ $x=60/19$

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How do you solve the following system?: $\frac{1}{6} x - \frac{3}{5} y = 1, \frac{4}{5} x + \frac{2}{3} y = 2$ $1/6x-3/5y=1 , 4/5x+2/3y = 2$

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How do you solve the following system?: $\frac{1}{6} x - \frac{3}{5} y = 1, \frac{4}{5} x + \frac{2}{3} y = 2$ $1/6x-3/5y=1 , 4/5x+2/3y = 2$