Question

Consider set of linear equations :$4x+7y=p$ and $kx+28y=120$, Where $k$ and $p$ are constant number. If these equation are having Infinitely many solution. Then, What will be the value of $k$ and $p$?

• A. $k=16$
• B. $p=30$
• C. $k=30$
• D. $p=16$

Answer 1

Answer: (A), (B)

$p=30$

Given,
$4x+7y=p$ $......\left(1\right)$
$kx+28y=120$ $......\left(2\right)$
These equations are having infinitely many solutions
$\therefore$ slope and y-intercept of both the equations will be same.
Now, converting both equations in slope intercept form we get,
$4x+7y=p$ [From equation (1)]
$\Rightarrow 7y=-4x+p\Rightarrow y=\frac{-4}{7}x+\frac{p}{7}$

$\text{Slope}=\frac{-4}{7}\&\text{y-intercept}=\frac{p}{7}$

Similarly,
$kx+28y=120$ [from equation (2)]
$\Rightarrow 28y=-kx+120\Rightarrow y=\frac{-k}{28}x+\frac{120}{28}$

$\text{Slope}=\frac{-k}{28}\&\text{y-intercept}=\frac{120}{28}=\frac{30}{7}$


Now, according to question
slope of equation (1)= slope of equation (2)
$\Rightarrow \frac{-4}{7}=\frac{-k}{28}$
On cross multiplication and simplification we get,
$\Rightarrow k=16$.
Again,
y-intercept of (1) = y-intercept of (2)
$\Rightarrow \frac{p}{7}=\frac{30}{7}$
$\Rightarrow p=30$
So, the values of $k$ and $p$ are $16$ and $30$ respectively.