How do you solve the following system?: $x + y = 4, 2 x - 5 y = 7$ $x+y=4, 2x -5y=7$

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Answer 1 · 2016-02-05T23:53:40

$y = \frac{1}{7}$ $y=1/7$
$x = \frac{27}{7}$ $x=27/7$

a) $x + y = 4$ $x+y=4$

b) $2 x - 5 y = 7$ $2x-5y=7$

First multiply a) by -2:

a) $- 2 x - 2 y = - 8$ $-2x-2y=-8$

b) $2 x - 5 y = 7$ $2x-5y=7$

Then add a) and b)

$- 7 y = - 1 \Rightarrow y = \frac{1}{7}$ $-7y=-1 => y=1/7$

Replace y in a)

$x + \frac{1}{7} = 4$ $x+1/7=4$
$x = 4 - \frac{1}{7} = \frac{27}{7}$ $x=4-1/7=27/7$

How do you solve the following system?: x+y=4, 2x -5y=7 x+y=4,2x−5y=7 x+y=4, 2x -5y=7