Math, asked by Snigdhha, 1 year ago

2y-3<=17y-4<=-4, y belongs to R.
I've got the solution set correct but in answer along with the solution set it says *null* ! . How ?

Answers

Answered by Swarup1998
4
◀HERE'S ⏬ YOUR ANSWER▶

Taking the 1st two terms, we get
2y - 3 \leqslant 17y - 4 \\ or \: \: - 3 + 4 \leqslant 17y - 2y \\ or \: \: 1 \leqslant 15y \\ or \: \: 15y \geqslant 1 \\ or \: \: y \leqslant \frac{1}{15}

Taking the last two terms, we get
17y - 4 \leqslant - 4 \\ or \: \: 17y \leqslant - 4 + 4 \\ or \: \: 17y \leqslant 0 \\ or \: \: y \geqslant 0

So we get a solution set
0 \leqslant y \leqslant \frac{1}{15}

⏏HOPE THIS ⬆ HELPS YOU⏏

Snigdhha: Yes indeed this is the required solution set but along with this it says "Null"... Why??????????
Swarup1998: It has a solution y=0. So, a set of the solution is y={0} or y=phi
Swarup1998: which is a null set
Swarup1998: this is why this says to have a null solution
Snigdhha: Solution set isn't zero ...its
---) 1/15
Swarup1998: Oo... ok
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