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First plug in option a, x+y = 22 it gives IxI negative which is not possible
same with option b, it give IxI is negative
then plug in option c, you get IxI = 6 then consider x= 6, you get y = -4 and put these values in 2nd equation LHS matches with RHS

So that is right answer
Option C
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If |x| + x + y = 8
and x + |y| - y = 14
Then what is the value of (x+y)

A. 22
B. 12
C. 2
D. 6
E. 15

When x>0 and y<0, we have |x|=x and |y|=-y.

(1) |x| + x + y = 8 or 2x+y=8
(2) x + |y| - y = 14 or x-2y=14

Solving eq(1) and (2), we have x=6 and y=-4.

Therefore, x+y=6+(-4)=2

Ans. (C)







If I put in your signs, why is it not

(1) |x| + x + (- y) = 8 or 2x-y=8
(2) x + |y| - (-y) = 14 or x+2y=14 ???

Because |y| will be positive after all.
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nick1816
If |x| + x + y = 8
and x + |y| - y = 14
Then what is the value of (x+y)

A. 22
B. 12
C. 2
D. 6
E. 15

We can PLUG IN THE ANSWERS, which represent the value of x+y.

|x| + x + y = 8
|x| = 8 - (x+y)
Since an absolute value cannot be negative, the greatest possible value for x+y is 8.
Eliminate A, B and E.

C: x+y = 2
Substituting x+y = 2 into |x| = 8 - (x+y), we get:
|x| = 8 - 2
|x| = 6
x = ±6
If x=6, then x+y=2 implies that y=-4.
Plugging x=6 and y=-4 into x + |y| - y = 14, we get:
6 + |-4| - (-4) = 14
14 = 14
Success!

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ronaldoSuiiii


If I put in your signs, why is it not

(1) |x| + x + (- y) = 8 or 2x-y=8
(2) x + |y| - (-y) = 14 or x+2y=14 ???

Because |y| will be positive after all.

By putting the negative sign in front of Y you are changing the operation to a subtraction of Y in the first case and an addition in the second.

Posted from my mobile device
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