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a+b=12+2c
and
3*(a+b)+c=22
so
c=-2

A is correct
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Bunuel
If a + b − 2c = 12, and 3a + 3b + c = 22, what is the value of c ?

A. −2
B. 0
C. 10
D. 17
E. 34
\(a + b − 2c = 12\)

Or, \(3a + 3b − 6c = 36\) & \(3a + 3b + c = 22\)

Now, \(( 3a + 3b − 6c ) - ( 3a + 3b + c ) = 36 - 22\)

Or, \(-7c = 14\)

or, \(c = -2\), Answer must be (A)
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Bunuel
If a + b − 2c = 12, and 3a + 3b + c = 22, what is the value of c ?

A. −2
B. 0
C. 10
D. 17
E. 34

a + b − 2c = 12
3a + 3b + c = 22

3 variables 2 equations obviously mean that we will have to modify equations and cancel the undesired variables.

3(a + b − 2c) = 3(12) becomes 3a+3b-6c=36

So now we have
3a+3b-6c=36
3a + 3b + c = 22

Subtract 2 from 1. We get -7c=14. c=-2.

Option A.
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Bunuel
If a + b − 2c = 12, and 3a + 3b + c = 22, what is the value of c ?

A. −2
B. 0
C. 10
D. 17
E. 34

Multiplying the first equation by -3, we have:

-3a - 3b + 6c = -36

Adding the two equations together, we have:

7c = -14

c = -2

Answer: A
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If a + b − 2c = 12, and 3a + 3b + c = 22, what is the value of c ?

A. −2 --> correct: a + b − 2c = 12 => a+b = 12+2c; and 3a + 3b + c = 22 => 3(a+b)+c =22, replacing a+b =12+2c, 3*(12+2c) +c =22 => 7c = -14 => c =-2
B. 0
C. 10
D. 17
E. 34
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