Math, asked by Anonymous, 11 months ago

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Answered by ItsShree
4

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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 Anonymous
1

Answer:

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Step-by-step explanation:

Given x+y=∣a∣ and ax−y=1

solving these equations simultaneously,

we et,

x = ( \frac{ |a| + 1}{1 + a} ) \:  \:  \\  y = ( \frac{a |a| - 1}{1 + a} )

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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