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If \(|x+1| + (y+2)^2 = 0\) and \(ax - 3ay = 1\), then the value of \(a\) is

Question and options

If ∣x+1∣+(y+2)2=0|x+1| + (y+2)^2 = 0 and ax−3ay=1ax - 3ay = 1, then the value of aa is
  1. 15\frac{1}{5}
  2. 12\frac{1}{2}
  3. 17\frac{1}{7}
  4. 2

Answer and explanation

Correct answer: A - \(\frac{1}{5}\)
Sum of non-negative terms is 0, so x=-1, y=-2. Then a(-1+6)=1, a=1/5.

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