Question

Solve: $4x+\frac{6}{y}=15$ and $6x-\frac{8}{y}=14$. Hence, find $a$ if $y=ax-2$

• A. $x=2;y=26$ and $a=2\frac{5}{3}$
• B. $x=3;y=2$ and $a=1\frac{1}{3}$
• C. $x=1;y=5$ and $a=2\frac{7}{3}$
• D. $x=5;y=1$ and $a=5\frac{1}{8}$

Answer 1

The correct option is B $x=3;y=2$ and $a=1\frac{1}{3}$

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:

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

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

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

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

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

Now substitute the values of $x$ and $y$ in $y=ax-2$ as shown below:

$2=3a-2$

$3a=2+2$

$3a=4$

$a=\frac{4}{3}=1\frac{1}{3}$

Hence, $a=1\frac{1}{3}$.