Question

36x-15y+50z=-10
2x+25y=40
54x-5y+30z=-160

Answer 1

are you solving for x, y, and z ?
what method have you learned ?

I would change the second equation to
2x = 40 - 25y

now think of the first equation,
36x-15y+50z=-10 as
18(2x) - 15y + 50z = -10
18(40 - 25y) - 15y + 50z = -10

simplify and do the same thing to the third equation.
Now you have 2 equations in y and z, which can be readily solved.

Let me know what you got.

Answer 2

The method we learned is like multiplying an equation and then add it to another equation to cancel a variable. I did:

36x-15y+50z=-10
-162x+15y-90z=480 (3rd eq. multiplied by 3)
which gave me -126x-40z=470

then I did:

2x+25y=40
270x-25y+150z=-800 (3rd eq. multiplied by 5)
which gave me 272x+150z=-760

This is the part where I'm stuck on because I usually would combine the 2 new equations (after multiplying one of them to cancel out a variable) to cancel out a variable. But in this case I can't find a number that helps?

Answer 3

ok, I would now work on the z's
you have -40z and +150z

the LCM of 40 and 150 is 600, so multiply your -126x-40z=470 by 15 and
272x+150z=-760 by 4

You would then add to eliminate the z's

I did not check your arithmetic, so good luck

Answer 4

Oh okay I forgot to use the LCM. Thanks :]