How do you solve the system $x+y=1$ and $2x-y=2$ ?

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Answer 1 · 2016-02-28T05:06:16

$x=1$
$y=0$

$x+y=1$ --------------------- (1)
$2x-y=2$ ------------------(2)

Solving the 1st equation for $y$ , we get

$y=1-x$

Substitute $y=1-x$ in equation (2)

$2x-(1-x)=2$

Simplify and solve for $x$

$2x-1+x=2$
$3x=2+1$
$x=3/3=1$
$x=1$

Substitute the value $x=1$ in equation (1) and solve for $y$

$1+y=1$
$y=1-1$
$y=0$

How do you solve the system x+y=1x+y=1 and 2x-y=22x-y=2?