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by sl750 » Mon Dec 05, 2011 10:45 pm
x + 2y + 3z = 2 --- i

x + y - z = 0 ----ii

2x + 2y - z = 1 ----iii


Multiply equation ii by 2

2x+2y=2z. Substitute value of 2x+2y in equation iii
2z-z=1 or z=1

x+2y=-1 --- iv
x+y =1 ---v

Solving , we get, y=-2
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by Anurag@Gurome » Mon Dec 05, 2011 10:48 pm
karthikpandian19 wrote:What is the value of y in the system below?


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

(A) -2 (B) -1(C) 0(D) 1(E) 2
Multiplying 2nd equation by 2, 2x + 2y - 2z = 0
Subtracting 2x + 2y - 2z = 0 and 2x + 2y - z = 1, we get,
-z = -1 or z = 1
So, x + 2y = -1 (1st equation) and x + y = 1 (2nd equation)
Subtracting the above equations, we get y = -2

The correct answer is A.
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by CollegeKart » Tue Dec 06, 2011 7:59 pm
x + 2y + 3z = 2 - (1)

x + y - z = 0 - (2)

2x + 2y - z = 1 - (3)


eq 2 gives x +y=z - (4)

put value of eq4 in eq 3

2z - z =1 => z = 1

putting value of z in eq 1 we get x + 2y= -1 - (5)

putting value of z in eq 4 we get x + y= 1 - (6)

eq(5) - eq (6)
y = -2
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by GMATGuruNY » Tue Dec 06, 2011 9:44 pm
karthikpandian19 wrote:What is the value of y in the system below?


x + 2y + 3z = 2

x + y - z = 0

2x + 2y - z = 1


(A) -2 (B) -1(C) 0(D) 1(E) 2
Subtracting the second equation from the third, we get:
(2x + 2y - z) - (x + y - z) = 1-0.
x+y = 1.

Since the second equation implies that x+y = z, z=1.

The first equation can be rewritten as y + x + y + 3z = 2.
Substituting x+y=1 and z=1 into this equation, we get:
y + 1 + 3(1) = 2.
y = -2.

The correct answer is A.
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