Question

Given:
$4x+\frac{6}{y}=15$
$6x-\frac{8}{y}=14$.
Find 'p' if, $y=px-2$

• A. $\frac{5}{3}$
• B. $\frac{7}{3}$
• C. $\frac{4}{3}$
• D. Insufficient data
• E. None of the above

Answer 1

The correct option is C $\frac{4}{3}$

We have, $4x+\frac{6}{y}=15\dots \dots \left(1\right)$

$6x-\frac{8}{y}=14\dots \dots \left(2\right)$

Multiplying equation (1) by 3 and equation (2) by 2, we get

$12x+\frac{18}{y}=45\dots \dots \left(3\right)$

$12x-\frac{16}{y}=28\dots \dots \left(4\right)$

Subtracting equation (4) from equation (3), we get

$\frac{34}{y}=17\Rightarrow y=\frac{34}{17}=2$

Putting y = 2 in equation (1), we get

$4x+\frac{6}{2}=15\Rightarrow 4x+3=15\Rightarrow 4x=15-3$

$\Rightarrow 4x=12\Rightarrow x=\frac{12}{14}=3$

Hence, the solution is x = 3, y = 2 Now

$y=px-2\Rightarrow 2\Rightarrow p\left(3\right)-2$

$\Rightarrow 3p=4\Rightarrow p=\frac{4}{3}$