How do you solve the following system?: $\frac{1}{4} x - \frac{3}{7} y = 1, \frac{3}{2} x - 8 y = 2$ $1/4x-3/7y=1 , 3/2x - 8y = 2$

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How do you solve the following system?: $\frac{1}{4} x - \frac{3}{7} y = 1, \frac{3}{2} x - 8 y = 2$ $1/4x-3/7y=1 , 3/2x - 8y = 2$

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Answer 1 · 2016-05-14T12:07:03

$x = \frac{24}{19}$ $x = 24/19$ and $y = \frac{14}{19}$ $y = 14/19$

Explanation:

Given the system:

$\frac{1}{4} x - \frac{3}{7} y = 1$ $1/4x - 3/7y = 1$

$\frac{3}{2} x - 8 y = 2$ $3/2x - 8y = 2$

Multiply the first equation by $4$ $4$ to get:

$x - \frac{12}{7} y = 4$ $x - 12/7y = 4$

Add $\frac{12}{7} y$ $12/7y$ to both sides to get:

$x = \frac{12}{7} y + 4$ $x = 12/7y+4$

Substitute this expression for $x$ $x$ in the second equation to get:

$2 = \frac{3}{2} (\frac{12}{7} y + 4) - 8 y$ $2 = 3/2(12/7y+4)-8y$

$= \frac{18}{7} y + 6 - 8 y$ $=18/7y+6-8y$

$= \frac{18 - 56}{7} y + 6$ $=(18-56)/7y+6$

$= - \frac{38}{7} y + 6$ $=-38/7y+6$

Add $\frac{38}{7} y - 2$ $38/7y-2$ to both ends to get:

$\frac{38}{7} y = 4$ $38/7y = 4$

Multiply both sides by $\frac{7}{38}$ $7/38$ to get:

$y = \frac{14}{19}$ $y = 14/19$

Then we find:

$x = \frac{12}{7} y + 4 = \frac{12}{7} (\frac{14}{19}) + 4 = \frac{24}{19}$ $x = 12/7y+4 = 12/7(14/19)+4 = 24/19$

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How do you solve the following system?: $\frac{1}{4} x - \frac{3}{7} y = 1, \frac{3}{2} x - 8 y = 2$ $1/4x-3/7y=1 , 3/2x - 8y = 2$

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How do you solve the following system?: $\frac{1}{4} x - \frac{3}{7} y = 1, \frac{3}{2} x - 8 y = 2$ $1/4x-3/7y=1 , 3/2x - 8y = 2$