Question

answer the question correctly
otherwise reported
please​

Answer 1

Answer:

X = - 6

then P = 7/6

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Answer 2

Given :-

$\sf \dfrac{x+3}{3}-\dfrac{x-2}{2}=1$

To Find :-

Value of p

Solution :-

$\sf \dfrac{x+3}{3}-\dfrac{x-2}{2}=1$

LCM of 2 and 3 is 6. Multiplying both sides by 6

$\sf 6\times\dfrac{x+3}{3}-6\times\dfrac{x-2}{2}=6(1)$

$\sf \cancel{6}\times\dfrac{x+3}{\cancel{3}}-\cancel{6}\times\dfrac{x-2}{\cancel{2}}=6(1)$

$\sf 2(x+3)-3(x-2)=6$

$\sf 2x+6-3x+6=6$

$\sf -x+12=6$

$\sf -x=6-12$

$\sf -x=-6$

$\sf\cancel{-}x=\cancel{-}6$

$\sf x=6$

Since, x = 6

$\sf\dfrac{1}{x}+p=1$

$\sf\dfrac{1}{6}+p=1$

$\sf p=1-\dfrac{1}{6}$

$\sf p=\dfrac{6-1}6$

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