How do you solve the following system of equations?: $x + y = - 2, x - y = 10$ $x+y= -2 , x - y= 10$ ?

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Answer 1 · 2016-01-28T23:36:32

$x = 4$ $x=4$
$y = - 6$ $y=-6$

$x = - 2 - y$ $x= -2-y$

subsitute $x = - 2 - y$ $x=-2-y$ into $x - y = 10$ $x-y=10$

$\to - 2 - y - y = 10$ $rarr-2-y-y=10$

$\to - 2 - 2 y = 10$ $rarr-2-2y=10$

$\to - 2 y = 12$ $rarr-2y=12$

$y = - 6$ $y=-6$

substitute $y = - 6$ $y=-6$ into $x - y = 10$ $x-y=10$

$\to x - 6 - (- 6) = 10$ $rarrx-6-(-6)=10$

$\to x + 6 = 10$ $rarrx+6=10$

$x = 4$ $x=4$

This is just substitution, you can also use elimination

$\to x + y = - 2$ $rarrx+y=-2$

$\to x - y = 10$ $rarrx-y=10$

add the two equations
(the y's cancel each other out)

$\to 2 x = 8$ $rarr2x= 8$

$x = 4$ $x=4$

substitute $x = 4$ $x=4$ into $x - y = 10$ $x-y=10$

$\to 4 - y = 10$ $rarr4-y=10$

$\to - y = 6$ $rarr-y=6$

$y = - 6$ $y=-6$

How do you solve the following system of equations?: x+y=−2,x−y=10x+y= -2 , x - y= 10?