Question

please guys slove this fast ​

Answer 1

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,

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

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

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

$\longrightarrow \sf 6p =\: 5$

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

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

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

Answer 2

Answer:

see the image step by step hope that will be helpful