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

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Answer 1 · 2016-02-07T17:23:10

$x = - \frac{51}{26}$ $x=-51/26$ and $y = \frac{1}{13}$ $y=1/13$

(a) $\frac{x}{2} - \frac{y}{4} = - 1$ $x/2-y/4=-1$

(b) $\frac{2}{3} x + 4 y = - 1$ $2/3x+4y=-1$

First eliminate the fractions, multiplying (a) by 4 and (b) by 3:

(a) $2 x - y = - 4$ $2x-y=-4$

(b) $2 x + 12 y = - 3$ $2x+12y=-3$

Do (b)-(a):

$13 y = 1 \Rightarrow y = \frac{1}{13}$ $13y=1 =>y=1/13$

For x, replace y in (a)

$2 x - \frac{1}{13} = - 4$ $2x-1/13=-4$
$2 x = - 4 + \frac{1}{13}$ $2x=-4+1/13$
$2 x = - \frac{51}{13}$ $2x=-51/13$
$x = - \frac{51}{26}$ $x=-51/26$

How do solve the following linear system?: x/2 - y/4 = -1 , 2/3x + 4y = -1 x2−y4=−1,23x+4y=−1 x/2 - y/4 = -1 , 2/3x + 4y = -1 ?