\(\frac{-4}{3}\)
\(\frac{8}{3}\)
Let's solve for x and y using the given equations:
Given:
1)\(3x + 2|y| + y = 7\)
2) \(x + |x| + 3y = 1\)
From the second equation:
\(x + |x| = 1 - 3y\)
Case 1: x ≥ 0 In this case,\(|x| = x\), so: \(x + x = 1 - 3y\) \(2x = 1 - 3y\) \(x = 0.5 - 1.5y\)... (i)
Case 2: x < 0 In this case, \(|x| = -x\), so: \(x - x = 1 - 3y\)
This gives us 0 = 1 - 3y, which is not possible.
Hence, the first case is our valid scenario.
Substitute the value of x from equation (i) into the first equation:
\(3(0.5 - 1.5y) + 2|y| + y = 7\)
Expanding:
\(1.5 - 4.5y + 2|y| + y = 7\)
\(1.5 - 3.5y + 2|y| = 7\)
\(-3.5y + 2|y| = 5.5\)
Now, for y:
Case 1: y ≥ 0
In this case, \(|y| = y\):
\(-3.5y + 2y = 5.5\)
\(-1.5y = 5.5\)
This gives a negative value for y, which is not possible in this case.
Case 2: y < 0 In this case,
\(|y| = -y\): \(-3.5y - 2y = 5.5\)
\(-5.5y = 5.5\)
\(y = -1\)
Substituting this value of y in equation (i):
\(x = 0.5 - 1.5(-1)\)
\(x = 0.5 + 1.5 = 2\)
So,\(x = 2\) and \(y = -1\).
Finally, \(x + 2y = 2 + 2(-1) = 2 - 2 = 0\)
We must check for all regions:
\(x≥0,y≥0\)
\(x≥0,y<0\)
However, once we find the answer for any one of these regions, we don't need to calculate for the others because the options suggest that there will be a single answer.
For \(x≥0,y≥0:\)
\(3x+3y=7\)
\(2x+3y=1\)
On solving,
\(x=6 \) and \(y =\)\(\frac {-11}{3}\).
Since \(y≥0\), this solution does not meet the given criteria.
For \(x≥0,y<0:\)
\( 3x−y=7\)
\(2x+3y=1\)
On solving,
\(x=2\) and \(y=-1\).
This solution satisfies both conditions, so it is the correct point.
Now,
\(x+2y\)
\(= 2+2(−1)\)
\(=2−2\)
\(=0\)
So, the correct option is (A): \(0\)