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antiseptic1488 [7]
3 years ago
6

Solve for y. y/m + y/n = p

Mathematics
1 answer:
RUDIKE [14]3 years ago
7 0
<h3>Given :</h3>

\boxed{ \rm  \frac{y}{m} +  \frac{y}{n}  = p}

\\  \\

<h3>To find:</h3>

\\

  • Value of y

\\  \\

<h3>Step by step expansion:</h3>

\dashrightarrow \sf\dfrac{y}{m} +  \dfrac{y}{n}  = p

write the equation

\\  \\

\dashrightarrow \sf \dfrac{\frac{y \times mn}{m} +  \frac{y \times mn}{n}}{mn}  = p

Add y/n with y/n and we will get Lcm as mn

\\  \\

\dashrightarrow \sf \dfrac{\frac{y \times  \cancel mn}{\cancel m} +  \frac{y \times m\cancel n}{\cancel n}}{mn}  = p

Cancel the similar terms

\\  \\

\dashrightarrow \sf \dfrac{\frac{y \times  n}{1} +  \frac{y \times m}{1}}{mn}  = p

\\  \\

\dashrightarrow \sf \dfrac{yn + ym}{mn}  = p

remove 1 as denominator

\\  \\

\dashrightarrow \sf \dfrac{y(n + m)}{mn}  = p

take y as common

\\  \\

\dashrightarrow \sf \dfrac{(n + m)}{mn}  =  \dfrac{p}{y}

transfer y to other side as denominator of p

\\  \\

\dashrightarrow \sf \dfrac{mn}{(n + m)}  =  \dfrac{y}{p}

reciprocal the whole equation

\\  \\

\dashrightarrow \sf \dfrac{mnp}{(n + m)}  =  \dfrac{y}{1}

transfer p other side as numerator

\\  \\

\dashrightarrow \boxed{ \bf y = \dfrac{mnp}{(n + m)} }

\\  \\

\therefore \red {\underline {\textbf{\textsf{Option 2 i.e \{y= mnp/(m+n) \} is correct }}}}

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Jim was thinking of a number.

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