Question

Reduce equations to a pair of linear equations and find the value of x and y:
$\frac{6}{x}+\frac{1}{y}=31;\frac{2}{x}+\frac{3}{y}=16$

• A. $x=\frac{1}{5},y=1$
• B. $x=\frac{1}{-5},y=-1$
• C. $x=\frac{2}{5},y=1$
• D. $x=\frac{1}{-5},y=1$

Answer 1

The correct option is A $x=\frac{1}{5},y=1$

The given equation can be written as
$6\left(\frac{1}{x}\right)+\left(\frac{1}{y}\right)=31;2\left(\frac{1}{x}\right)+6\left(\frac{1}{y}\right)=16$

Assume $a=\frac{1}{x},b=\frac{1}{y}$

$6a+b=31$ ----- (1)
$2a+6b=16$ ----- (2)
Multiplying the equation (1) by 6 and then subtract the equations,
$36a+6b=186$
$2a+6b=16$
_____________
$34a=170$

$a=\frac{170}{34}=5$
Put the value of a in equation (1)
$6a+b=31$
$6\left(5\right)+6=31$
$30+b=31$
$b=31-30$
$b=1$
Therefore, $\frac{1}{x}=a;\frac{1}{y}=b$
$x=\frac{1}{5};y=1$
The solution is $(\frac{1}{5},1)$