Question

If $\frac{a}{3y}+\frac{3b}{x}=7$ and $a+1=2b+1=x=5,$ find the value of ${}^{\prime }{y}^{\prime }.$

• A. $\frac{10}{77}$
• B. $\frac{22}{69}$
• C. $\frac{20}{87}$
• D. $\frac{14}{93}$

Answer 1

Answer: (C)

$\frac{20}{87}$

Since, $a+1=2b+1=x=5$
$=>a=5-1=4$
$2b=5-1=4=>b=2$
$x=5$
Substituting these values in $\frac{a}{3y}+\frac{3b}{x}=7$, we get $\frac{4}{3y}+\frac{6}{5}=7$
$=>\frac{4}{3y}=7-\frac{6}{5}$
$=>\frac{4}{3y}=\frac{29}{5}$
$=>\frac{3y}{4}=\frac{5}{29}$
$=>y=\frac{20}{87}$