Math, asked by bantu6431, 5 hours ago

please guys slove this fast ​

Attachments:

Answers

Answered by Anonymous
48

Answer:

\boxed{\bold{\underline{\mathscr{Question\: :-}}}}

\bullet\: \: \tt{\dfrac{x + 3}{3} - \dfrac{x - 2}{2} =\: 1. \: Hence,\: find\: p,\: if\: \dfrac{1}{x} + p =\: 1\: .}\\

\boxed{\bold{\underline{\mathscr{To\: Find\: :-}}}}

  • What is the value of p.

\boxed{\bold{\underline{\mathscr{Solution\: :-}}}}

First, we have to find the value of x :

\bigstar\: \: \:  \sf\bold{\purple{\dfrac{x + 3}{3} - \dfrac{x - 2}{2} =\: 1}}

\implies \bf \dfrac{x + 3}{3} - \dfrac{x - 2}{2} =\: 1

\implies \sf \dfrac{2(x + 3) - 3(x - 2)}{6} =\: 1

By doing cross multiplication we get,

\implies \sf 2(x + 3) - 3(x - 2) =\: 6(1)

\implies \sf 2x + 6 - 3x + 6 =\: 6

\implies \sf 2x - 3x + 6 + 6 =\: 6

\implies \sf - x + 12 =\: 6

\implies \sf - x =\: 6 - 12

\implies \sf {\cancel{- x}} =\: {\cancel{-}} 6

\implies \sf \bold{\purple{x =\: 6}}

Now, we have to find the value of p, if :

\bigstar \: \: \sf\bold{\green{\dfrac{1}{x} + p =\: 1}}

\longrightarrow \bf \dfrac{1}{x} + p =\: 1

By putting the value of x = 6 we get,

\longrightarrow \sf \dfrac{1}{6} + p =\: 1

\longrightarrow \sf \dfrac{1 + 6p}{6} =\: 1

By doing cross multiplication we get,

\longrightarrow \sf 1 + 6p =\: 6(1)

\longrightarrow \sf 1 + 6p =\: 6

\longrightarrow \sf 6p =\: 6 - 1

\longrightarrow \sf 6p =\: 5

\longrightarrow \sf p =\: \dfrac{5}{6}

\longrightarrow \sf\bold{\red{p =\: \dfrac{5}{6}}}

{\small{\bold{\underline{\therefore\: The\: value\: of\: p\: is \: \dfrac{5}{6}\: .}}}}

Answered by diyakajaljha0812
1

Answer:

see the image step by step hope that will be helpful

Attachments:
Similar questions