math problem help

Question

Solve the following system by using any method:

 

x+2y-3z=-5

2x+5y+z=12

x-4y-3z=-11

Answer 1

x+2y-3z=-5
2x+5y+z=12
x-4y-3z=-11

 

Subtract 3rd equation from the first equation, getting 6y = 6  so y = 1.

 

Now substitute for y in those equtions, getting x+2 -3z = -5 

(new 1st equation) and new 2nd equation       2x +5 +z = 12

 

Moving the constants following the xvalue over to opposite sides of the equations gives us

x - 3z = -7

2x + z = 7

 

In other words, from the top of the two equations just above, x = 3z - 7; substituting this into the last of the two equations just above gives 2(3z - 7) + z = 7

or 7z = 21 so z = 3.

 

Finally, we then have x = 3z - 7 = 3(3) - 7 = 2.

 

The solution seems to be (2, 1, 3).

 

CHECK THE ANSWERS.  If I've made a mistake, it will be revealed. My method is OK, in any case, and that's what they're teaching in schools.

Answer 2

x+2y-3z=-5
2x+5y+z=12
x-4y-3z=-11

subtract twice the first equation from the second

subtract the first equation from the third

 

x+2y-3z=-5

    y +7z=22

   -6y     =-6

y=1 from third equation

use in second equation as

1+7z=22 to get z=3

use in first equation as

x+2-9=-5

x=2