Question

solve (y+7)/7=1+(3y-2)/5​

Answer 1

$\huge\underline\mathrm\purple{Question}$

solve (y+7)/7=1+(3y-2)/5

$\huge\underline\mathrm\purple{Find\:out}$

Value of y

$\huge\underline\mathrm\purple{Solution}$

$\implies\sf \frac{(y+7)}{7}=1+\frac{(3y-2)}{5}$

$\implies\sf \frac{(y+7)}{7}-\frac{(3y-2)}{5}=1$

$\implies\sf \frac{5(y+7)-7(3y-2)}{35}=1$

$\implies\sf \frac{(5y+35)-21y+14}{35}=1$

$\implies\sf \frac{-16y+49}{35}=1$

$\implies\sf -16y+49=35$

$\implies\sf -16y = -49+35$

$\implies\sf y=\cancel\frac{14}{16}=\frac{7}{8}$

$\huge\underline\mathrm\purple{NoTE}$

• Same sign = addition
• opposite sign = subtraction

Answer 2

$\bold{\huge{\fbox{\color{Black}{Question}}}}$

$\frac{y + 7}{7} = 1 + \frac{3y - 2}{5}$

$\bold{\huge{\fbox{\color{Black}{Find}}}}$

Value of y

$\bold{\huge{\fbox{\color{Black}{Solution}}}}$

=> $\frac{y + 7}{7} = 1 + \frac{3y - 2}{5} \\=> \frac{y + 7}{7} - \frac{3y - 2}{5} = 1 \\ => \frac{5(y + 7)}{7 \times 5} - \frac{7(3y - 2)}{5 \times 7} = 1 \\ => \frac{5y + 35 - 21y + 14}{35} = 1 \\ => \frac{5y - 21y + 35 + 14}{35} = 1 \\ => \frac{ - 16y + 49}{35} = 1 \\ => - 16y + 49 = 35 \times 1 \\=> - 16y = 35 - 49 \\=> - 16y = - 14 \\=> y = \frac{ - 14}{ - 16} \\=> y = \frac{7}{8}$

Hence,

=> y = 7/8