Question

How to I solve x+y+z=0, -x+2y+5y=3, and 3x-y=6. The three equations are in brackets. I need help!!!!

Answer 1

I don't understand exactly what you mean by they are in brackets..are they like this: [x+y+z = 0] or all together in one bracket?

Answer 2

yes. Why?

Answer 3

\[\left[x+y+z=0 \right]\]
-x+2y+5z=3
3x-y=6.

Equations are in brackets. What do I do to solve the problem? I need to solve your z,y,z

Answer 4

I need help!!!

Answer 5

[x+y+z = 0][-x+2y + 5z = 3][3x-y=6]. If you aren't multiplying them times each other, then first you would get rid of the brackets by taking them times 1. Making them x+y+z =0, -x+2y + 5z = 3, 3x-y=6.

Answer 6

And then? What's next?

Answer 7

Hang on.have to see if my idea works first

Answer 8

Ok. I will waiting for your idea. :D

Answer 9

Okay if I did this right, I'm getting Z = 6, Y = -6, and X = 0.

Answer 10

3x - y = 6, subtracted 3x from both sides to make (-y = 6 - 3x)-1 = y = 3x - 6. Filled this equation into the same equation 3x - y = 6. So 3x - 3x - 6 = 6. Combine like terms, 0x - 6 = 6. Add six both sides, X = 12 *Caught my own mistake there lol*

Answer 11

3(12) - y = 6. 36 - y = 6. Subtract 36 from both sides, (-y = -30)-1. Y = 30

Answer 12

Fill X and Y into both sides. 12 + 30 + z = 0, combine like terms --> 42 + z = 0, subtract 42 from both sides, z = -42. Not entirely sure if I did it right but I caught my own mistake so please disregard my first post of answers :)

Answer 13

you multiply the first equation with (-5) and add the obtained result to the second equation.then you get -6x-3y=3.now,multiply the third equation with -3.now,add the obtained result to -6x-3y=3.then you get -15x=-15.therefore x=1.by substituting x=1 in the third equation,you get y=-3.by substituting x=1 & y=-3 in the first equation,you get z=2.