Question

Reduce equations to a pair of linear equations and find the value of x and y:
$\frac{1}{x}+\frac{6}{y}=10;\frac{2}{x}+\frac{5}{y}=13$

• A. $x=\frac{1}{4},y=1$
• B. $x=\frac{1}{-4},y=-1$
• C. $x=\frac{2}{7},y=1$
• D. $x=\frac{1}{-4},y=1$

Answer 1

The correct option is A $x=\frac{1}{4},y=1$

The given equation can be written as
$1\left(\frac{1}{x}\right)+6\left(\frac{1}{y}\right)=10;2\left(\frac{1}{x}\right)+5\left(\frac{1}{y}\right)=13$

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

$a+6b=10$ ----- (1)
$2a+5b=13$ ----- (2)
Multiplying the equation (1) by 2 and then subtract the equations,
$2a+12b=20$
$2a+5b=13$
_____________
$7b=7$

$b=\frac{7}{7}=1$
Put the value of b in equation (1)
$a+6b=10$
$a+6\left(1\right)=10$
$a=10-6$
$a=4$
Therefore, $\frac{1}{x}=a;\frac{1}{y}=b$
$x=\frac{1}{4};y=1$
The solution is $(\frac{1}{4},1)$