Multiply the second equation by 2: \(2x + 2y - 2z = 0\) --> \(2x + 2y =2z\).

From III: \(2x + 2y - z = 1\) --> \(2z - z = 1\) --> \(z=1\).

Subtract II from I: \((x + 2y + 3)- (x + y - 1) = 2\) --> \(y=-2\).

Answer: A.

Hope it's clear.
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is there an easier and quicker way to solve this question than my approach? this one took me around 6 minutes to solve, so i guess there must be an easier way.

my approach:

1. from the given system of equations i can make equation II. to be x + y = z

2. now put that in III. :

2x + 2y - (x + y) = 1

= 2x + 2y - x - y = 1

= x + y = 1

that gives us z = 1!

3. put z = 1 in I. and solve:

x + 2y + 3 = 2; calculate minus 3

= x + 2y = -1; calculate minus x

= 2y = -1 - x;

4. now put 2y = -1 - x in III. and solve:

2x + (-1 - x) - 1 = 1

= x - 2 = 1; calculate plus 2

that gives us x = 3

5. now solve for y with x = 3 and z = 1:

x + y = 1

= 3 + y = 1; calculate minus 3

that gives y = -2, and hence answer (A) -2

any quicker way?

my solution of

x+2y+3z=2
x+y-z=0
2x+2y-z=1

1. subtract equation#2 from equation#1 because we do not need x, we look for y

y+4z=2

2. to eliminate x in second pair multiply equation#2 by -2 and add equation#2 to equation#3

z=1

3. substitute z with 1 and get y=-2

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