Question 1198509
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Let's solve the 2nd equation for x.


-x+4y = 10
-x = 10-4y
x = -10+4y
x = 4y-10


Then plug it into the 1st equation
x-2y+z = -3
4y-10-2y+z = -3
2y+z = -3+10
2y+z = 7


And do the same for the 3rd equation
2x-y+6z = 7
2(4y-10)-y+6z = 7
8y-20-y+6z = 7
7y+6z = 7+20
7y+6z = 27


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We now have this system of equations
2y+z = 7
7y+6z = 27


I'll now isolate z in the 1st equation of this new system.
I'm picking on z because the coefficient here is 1.


2y+z = 7
z = 7-2y


Then substitute it into the other equation and solve for y.
7y+6z = 27
7y+6(7-2y) = 27
7y+42-12y = 27
-5y+42 = 27
-5y = 27-42
-5y = -15
y = -15/(-5)
y = 3
This narrows our choices between choice (B) or choice (E).


Use that value of y to find z
z = 7-2y
z = 7-2*3
z = 7-6
z = 1
We can immediately conclude the answer is <font color=red>choice (B)</font>


If you wanted, let's keep going to find x.
x = 4y-10
x = 4*3-10
x = 12-10
x = 2
This offers further confirmation that (B) is the answer.


If you are in a time crunch, then you can use the multiple choice answers to your advantage. 
Simply plug each numeric value in for the corresponding variable. Then simplify each side.


Let's say we wanted to check if (E) was the answer.
We'd plug in x = -2, y = 3, z = -4 into the 1st equation to get...
x-2y+z = -3
-2-2(3)+(-4) = -3
-2-6-4 = -3
-12 = -3
We get a false statement at the end. We need to get the same thing on both sides for it to be a true equation.
Since the last equation is false, it causes the first equation to be false for those x,y,z values.
Therefore, we can rule out choice (E) fairly quickly.
Choices (C) and (D) are eliminated for similar reasoning.


To confirm choice (B) is the solution, plug (x,y,z) = (2,3,1) into each of the original three equations. 
You should get true results after simplification. 
I'll let you do these confirmations.


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Answer: <font color=red>Choice (B)</font>  x=2, y=3, z=1
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