
Let us take ,
- First integer be x .
- Second integer be y .
Also let us consider x > y .
<u>Acco</u><u>rding</u><u> to</u><u> </u><u>first</u><u> </u><u>Condition</u><u> </u><u>:</u><u>-</u>
⇒ x - y = 1.
⇒ x = y + 1. .............(i)
<u>Acco</u><u>rding</u><u> to</u><u> </u><u>seco</u><u>nd</u><u> </u><u>Condition</u><u> </u><u>:</u><u>-</u>
⇒ x × y = 30 .
⇒ ( y + 1 )y = 30 . [ From (i) ]
⇒ y² + y = 30.
⇒ y² + y - 30 = 0 .
⇒ y² + 6y - 5y -30 = 0.
⇒ y ( y + 6 ) -5 ( y + 6 ) = 0 .
⇒ ( y + 6 ) ( y - 5 ) = 0 .

Here y can sustain both values 5 and minus 6 as y is an integer.
So , x = -6 , 5 .
Hence possible answers are ,



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