Math, asked by KingSrikar, 13 days ago

Solve THIS Question : \rm{y-(y-1)/7=1-(y-2)/5}

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Answered by ExploringMathematics
0

\rm{y-(y-1)/7=1-(y-2)/5}

\longrightarrow\rm{7y/7-(y-1)/7=5/5-(y-2)/5}

\longrightarrow\rm{(7y-(y-1))/7=(5-(y-2))/5}

\longrightarrow\rm{(7y-y+1)/7=(5-y+2)/5}

\longrightarrow\rm{(6y+1)/7=(7-y)/5}\longrightarrow\rm{5(6y+1)=7(7-y)}

\longrightarrow\rm{30y+5=49-7y}\longrightarrow\rm{30y+5-5=49-7y-5}

\longrightarrow\rm{30y=-7y+44}\longrightarrow\rm{30y+7y=-7y+44+7y}

\longrightarrow\rm{37y=44}\longrightarrow\rm{37y/37=44/37}\longrightarrow\rm{y=44/37}

Answered by Anonymous
0

Hope this will help you.

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