How do you solve the system $x + y - z = 2$ $x+y-z=2$ , $3 x + 5 y - 2 z = - 5$ $3x+5y-2z=-5$ , and $5 x + 4 y - 7 z = - 7$ $5x+4y-7z=-7$ ?

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Answer 1 · 2016-09-05T05:46:49

The soln. : $x = 30, y = - 13, z = 15$ $x=30, y=-13, z=15$ .

Let us solve the system of eqns. using Elimination Method.

From the First eqn., $z = x + y - 2$ $z=x+y-2$

Using this $z$ $z$ in the second and the third eqns., we get,

$3 x + 5 y - 2 (x + y - 2) = - 5, i.e., x + 3 y = - 9$ $3x+5y-2(x+y-2)=-5", i.e., "x+3y=-9$

$5 x + 4 y - 7 (x + y - 2) = - 7, or, - 2 x - 3 y = - 21$ $5x+4y-7(x+y-2)=-7", or, "-2x-3y=-21$

Adding these last eqns., $x = 30$ $x=30$ .

Then from, $x + 3 y = - 9, y = - 13$ $x+3y=-9, y=-13$ .

Finally, $z = x + y - 2 = 15$ $z=x+y-2=15$ .

These roots satisfy the given eqns.

The soln. : $x = 30, y = - 13, z = 15$ $x=30, y=-13, z=15$ .

How do you solve the system x+y−z=2x+y-z=2, 3x+5y−2z=−53x+5y-2z=-5, and 5x+4y−7z=−75x+4y-7z=-7?