Question

Reduce equations to a pair of linear equations and find the value of p and q: $3p+\frac{7}{q}=14;2p+\frac{7}{q}=21$

• A. $p=\frac{2}{-7},q=\frac{1}{5}$
• B. $p=\frac{1}{-7},q=\frac{1}{5}$
• C. $p=\frac{2}{-7},q=\frac{-1}{5}$
• D. $p=\frac{1}{-7},q=\frac{-1}{5}$

Answer 1

The correct option is B $p=\frac{1}{-7},q=\frac{1}{5}$

The given equation can be written as
$3\left(p\right)+7\left(\frac{1}{q}\right)=14;2\left(p\right)+7\left(\frac{1}{q}\right)=21$

Assume $p=x,\frac{1}{q}=y$
$3x+7y=14$ ----- (1)
$2x+7y=21$ ----- (2)
Multiplying the equation (1) by 2 and equation (2) by 3 and then subtract the equations,
$6x+14y=28$
$6x+21y=63$
______________
$-7y=-35$

$y=\frac{-35}{-7}=5$
Put the value of y in equation (1)
$3x+7\left(5\right)=14$
$3x=1435$
$3x=-21$
$x=-7$
Therefore, $\frac{1}{p}=x;\frac{1}{q}=y$

$p=\frac{1}{-7};q=\frac{1}{5}$
The solution is $(\frac{1}{-7},\frac{1}{5})$