b. (x, y, z) = (-8, 10, -6)
Step-by-step explanation:
The easiest way to do this one is to try the answers to see which works.
2(-8) +4(10) +3(-6) = -16 +40 -18 = 6
5(-8) +8(10) +6(-6) = -40 +80 -36 = 4
4(-8) +5(10) +2(-6) = -32 +50 -12 = 6
The answers of choice B work in the given equations.
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In case you don't have answers to select from, you generally solve this sort of problem using elimination. You can also use Cramer's rule, a graphing calculator, an on-line equation solving tool, or any of a variety of other methods.
Here, we can find the variable x by subtracting twice the first equation from the second:
(5x +8y +6z) -2(2x +4y +3z) = (4) -2(6)
x = -8
This is sufficient to identify the correct answer choice.
We can substitute this into the last two equations to get ...
-40 +8y +6z = 4 . . . . 8y +6z = 44
-32 +5y +2z = 6 . . . . 5y +2z = 38
Subtracting the first of these from 3 times the second gives ...
3(5y +2z) -(8y +6z) = 3(38) -(44)
7y = 70 . . . . . . . simplify
y = 10 . . . . . . . . divide by 7
Substituting this into the second of the above equations, we have ...
5(10) +2z = 38
25 +z = 19 . . . . . . divide by 2
z = -6 . . . . . . . . . . subtract 25
The choice of the combinations to use to eliminate variables can be ad hoc (as here), or it can be made according to some rules (as in Gaussian elimination).
My personal choice for solving systems like this is to use the matrix functions of a graphing calculator.
