Question

Solve : $4x+\frac{6}{y}=15$ and $3x+\frac{4}{y}=7$ Hence , find ${}^{\prime }{a}^{\prime }$ if $y=ax-2$.

Answer 1

The given expressions can be written as

$4xy+6=15y$ ......(1)

$3xy+4=7y$ ......(2)

Multiplying equation (1) by $3$ and equation (2) by $4$, we get

$12xy+18=45y$ ......(1)

$12xy+16=28y$ ......(2)

On subtracting equation (2) from equation (1)

$17y=2$

$y=\frac{2}{17}$and $\frac{1}{y}=\frac{17}{2}$

Substituting this value in equation (1), we get

$4x+6\times \frac{17}{2}=15$

$4x+51=15$

$4x=-36$

$x=-9$

Now,

$y=ax-2$

Substituting the derived values of $x$ and $y$,

$\frac{2}{17}=-9a-2$

$2=-9\times 17\times a-34$

$9\times 17\times a=-36$

$a=-\frac{4}{17}.$

Hence, the value of $a$ is $-\frac{4}{17}.$