How do i solve these equations

4x + y + z = 1
8x – 4y – 7z = 2
5y + 9z = 3
A. (0.5, -3, 2)
B. (1, 2, 3)
C. (-3, 0.5, 2)
D. (0.5, 3, 2)






3x – y + 2z = 6
-x + y = 2
x – 2z = -5
A. (-1, -3, -3)
B. (1, 3, 3)
C. (1, -3, -3)
D. (1, 3, -3)

What is your math level, have you ever worked with Linear Algebra working with Matrices,more specific reduced row echelon form...have you done Calculus yet????? It would be so much simpler if you have done either one of these maths, more preferably the Linear Algebra.

If you haven't, I will assume you haven't, I would simply plug the values in knowing the values for A through D are give in the form of (x,y,z) plug them into your equations and see if they work..
for instance A for the top one is correct
because ;
4(.5)-3+2=1 and it works for the other two equations as well
for the second set of equations I would go with B

Select a pair and solve to get an equation in only two variables. Combine this with the third equation to get the answers for all 3 variables.

4x + y + z = 1
8x – 4y – 7z = 2
5y + 9z = 3

There are many ways to do this. Here's one...
y = 1 - z - 4x. substitute in the other equations
8x - 4 - 4z + 16x - 7z = 2, giving 24x - 11z = 6

5 - 5z - 20x + 9z = 3. giving - 20x - 4z = - 2, reducing to -10x - 2z = -1

Solve these two for x and z and that will enable you to see your answer.
Alternatively substitute the given answers into any of the three equations and see if it fits
subtract this from the previous equation to get 22x = 4 from which x = 2/11

Looks like A to me, but I'm too busy to do it all

First one : A (x=0.5, y=-3 and z=2)
second one: B (x=1, y=3 and z=3)

1) x=61/116 or 0.5258
y=-363/319 or -1.379
z=308/319 or 0.9655
2) x= 1.25
y= -12
z= 8