How do you solve the system $x-y+z=3$ , $2x+y+z=8$ , and $3x+y-z=1$ ?

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Answer 1 · 2017-01-27T03:01:25

Solution is $x=1$ , $y=2$ and $z=4$

We are given three equations in three variables, which are

$x-y+z=3$ ..........................................(1)

$2x+y+z=8$ ..........................................(2)

and $3x+y-z=1$ ..........................................(3)

To solve them, what we need is to eliminate some vatiables to get the value of others.

It is apparent from (1) nd (3) that as signs of $y$ and $z$ are opposite, they will cancel out on adding. Hence adding them, we get

$x-y+z+3x+y-z=3+1$ or $4x=4$ i.e. $x=1$

Now subtracting (2) from (1), we get

$x-y+z-2x-y-z=3-8$ or $-x-2y=-5$ .

But as $x=1$ , we have

$-1-2y=-5$ i.e. $-2y=-4$ or $y=(-4)/-2=2$

Putting values of $x$ and $y$ in (1), we get

$1-2+z=3$ or $z=3-1+2=4$

Hence, solution is $x=1$ , $y=2$ and $z=4$

How do you solve the system x-y+z=3x-y+z=3, 2x+y+z=82x+y+z=8, and 3x+y-z=13x+y-z=1?