Answer:
b. x=1, y=2, z=3
Explanation:
The system of equations ...
- 3x +2y +z = 10
- 9x -6y +z = 0
- x -y -3z = -10
has solution (x, y, z) = (1, 2, 3) . . . . matches choice B.
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While it is convenient to solve this using a graphing calculator or web site, one can easily solve the system by hand.
Subtract the second equation from 3 times the first:
3(3x +2y +z) -(9x -6y +z) = 3(10) -(0)
12y + 2z = 30 . . . . simplify
Dividing this result by 2 gives ...
6y +z = 15 . . . . . . [eq4]
Subtract 3 times the third equation from the first:
(3x +2y +z) -3(x -y -3z) = (10) -3(-10)
5y +10z = 40 . . . . simplify
y + 2z = 8 . . . . . . . divide by 5 . . . . . [eq5]
The two equations [eq4] and [eq5] can be solved any of the ways you usually solve two equations in two variables. Here, we'll use the first equation to write an expression for z that we can substitute into the second equation.
z = 15 -6y . . . . . subtract 6y from [eq4]
y + 2(15 -6y) = 8 . . . . . substitute for z in [eq5]
-11y +30 = 8 . . . . . simplify
-11y = -22 . . . . . . . subtract 30
y = 2 . . . . . . . . . . . divide by the coefficient of y
z = 15 -6(2) = 3 . . . . substitute for y in our equation for z
Substituting these values for y and z into the third original equation gives ...
x - 2 -3(3) = -10
x -11 = -10 . . . . . . . . simplify
x = 1 . . . . . . . . . . . . add 11
The solution to the above system of equations is (x, y, z) = (1, 2, 3).
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Comment on the problem statement
Math is generally unforgiving of imprecision. The given system of equations has no variable "z", and some other typos are apparently involved. That is why we rewrote the system to the equations shown above.
It is very easy to mistake z for 2, or g for 9, or o for 0, or 1 for 7. There are other confusions that are possible, as well. Letters I (eye) and l (ell) are easily confused, and may be confused with 1 (one) as well. Sometimes y and 4, or 4 and 9, can also be written so as to be difficult to tell apart. Great care must be taken when handwriting these symbols.