Question

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

Answer 1

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

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

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

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

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

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

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

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

Answer 2

Answer:

Question:-

$\frac{y - (y - 1)}{7} = \frac{1 - (y - 2)}{5}$

Solution:-

First cross multiply numbers.

$5(y - (y - 1) \: ) \: = 7(1 - (y - 2) \: )$

$5y - (5y - 5) = 7 - (7y - 14)$

$5y - 5y + 5 = 7 - 7y + 14$

$7y = 7 + 14 - 5$

$7y = 21 - 5$

$7y \: = 16$

$y = \frac{16}{7}$

Step-by-step explanation:

Hope this helps you.

# By Sparkly Princess