Question

Solve: $4x+\frac{6}{y}=15$ and $6x-\frac{8}{y}=14$. Hence find the value of $k$, if $y=kx-2$.

• A. $x=4,y=1$ and $k=\frac{2}{3}$
• B. $x=5,y=3$ and $k=\frac{3}{5}$
• C. $x=3,y=2$ and $k=\frac{4}{3}$
• D. $x=1,y=2$ and $k=\frac{9}{2}$

Answer 1

The correct option is C $x=3,y=2$ and $k=\frac{4}{3}$

The equation $4x+\frac{6}{y}=15$ can be solved as:

$4x+\frac{6}{y}=15\Rightarrow \frac{4xy+6}{y}=15\Rightarrow 4xy+6=15y.........\left(1\right)$

The equation $6x-\frac{8}{y}=14$ can be solved as:

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

Multiply the equation 1 by $3$ and equation 2 by $2$. Then we get the equations:

$12xy+18=45y.........\left(3\right)$

$12xy-16=28y.........\left(4\right)$

Subtract equation 4 from equation 3 to eliminate $xy$, because the coefficients of $xy$ are same. So, we get

$(12xy-12xy)+(18+16)=45y-28y$

i.e. $17y=34$

i.e. $y=2$

Substituting this value of $y$ in the equation 1, we get

$8x+6=30\Rightarrow 8x=24\Rightarrow x=3$

Therefore, the solution of the equations is $x=3,y=2$.

Now Substitute the values of $x$ and $y$ in the equation $y=kx-2$ as follows:

$2=3k-2$

$3k=4$

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

Hence, $k=\frac{4}{3}$.