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

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

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Answer 1 · 2016-01-27T08:33:24

$x = \frac{9}{28}$ $x = 9/28$

$y = \frac{17}{45}$ $y =17/45$

Explanation:

Rearrange one of the equations to get a variable on its own on the left hand side, then substitute it into the other equation and solve.

$5 x + 2 y = 1$ $5x+2y =1$
$\frac{2}{3} x + 4 y = - 1$ $2/3x +4y = -1$
$2 y = 1 - 5 x$ $2y = 1-5x$
$y = \frac{1}{2} - \frac{5}{2} x$ $y = 1/2 - 5/2x$

Substituting into he second equation
$\frac{2}{3} x + 4 (\frac{1}{2} - \frac{5}{2} x) = - 1$ $2/3x +4(1/2-5/2x) =-1$
$\frac{2}{3} x + 2 - 10 x = - 1$ $2/3x +2 -10x = -1$
$2 x - 30 x = - 9$ $2x-30x = -9$
$x = \frac{- 9}{- 28} = \frac{9}{28}$ $x=(-9)/(-28) = 9/28$

Then $y = \frac{1}{2} - \frac{5}{2} \cdot (\frac{9}{28})$ $y=1/2-5/2*(9/28)$
$y = \frac{28 - 45}{56} = \frac{17}{45}$ $y = (28-45)/56=17/45$

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

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Explanation:

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