How do you solve the following system?: $x + y = 1, 2 x - y = - 2$ $x + y =1 ,2x-y=-2$

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How do you solve the following system?: $x + y = 1, 2 x - y = - 2$ $x + y =1 ,2x-y=-2$

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Answer 1 · 2015-12-07T12:41:57

$x + y = 1$ $x+y=1$ (Consider this equation as equation 1)
$2 x - y = - 2$ $2x−y=−2$ (Consider this equation as equation 2)

Adding equation 1 & 2 we get
$3 x = - 1$ $3x=-1$
This implies $x = - \frac{1}{3}$ $x=-1/3$
Substitute this in equation 1
$- \frac{1}{3} + y = 1$ $-1/3+y=1$
$y = 1 + \frac{1}{3}$ $y=1+1/3$
$y = 3 + \frac{1}{3}$ $y=3+1/3$
$y = \frac{4}{3}$ $y=4/3$

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How do you solve the following system?: $x + y = 1, 2 x - y = - 2$ $x + y =1 ,2x-y=-2$

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How do you solve the following system?: $x + y = 1, 2 x - y = - 2$ $x + y =1 ,2x-y=-2$