Math, asked by KingSrikar, 6 months ago

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

Answers

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

Step-by-step explanation:

−(y−1)/7=1−(y−2)/5

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

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

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

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

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

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

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

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