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Answered by
4
[
x + y = lal
ax - y = 1
Equations solved simultaneously
x = lal + 1 ÷ 1 + a, y = a lal -1 ÷ 1 + a
Both equations fall in the first quadrant so both are greater than zero.
1 + a > 0 and a lal - 1 >0
After solving the answer is a > 1
So all possible values of a are in the interval
a ∈ (1, ∞)
Answered by
1
Answer:
Step-by-step explanation:
Given x+y=∣a∣ and ax−y=1
solving these equations simultaneously,
we et,
As both are in the 1st
Quadrant both should be greater than 0,
$$ 1+a> 0$$ and $$ a\left | a \right |-1> 0$$
$$a>-1 $$ and $$a|a|>1$$
After solving this, we get
$$ a> 1,$$ i.e aϵ(1,∞)
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