Question

Question 9 (iii)
Solve the following pair of equations
$4x+\frac{6}{y}=15,6x-\frac{8}{y}=14$

Answer 1

Given pair of linear equations is
$4x+\frac{6}{y}=15and6x-\frac{8}{y}=14y\ne 0$
Taking $\frac{1}{y}=u$
We get the equations as
$4x+6u=15$......(i)
$6x-8u=14$......(ii)
On multiplying Eq. (i) by 8 and Eq. (ii) by 6 and then adding both of them, we get
$32x+48u=120$
$36x–48u=84$
$\underset{–}{}$
$68x=204$
$\Rightarrow$ $x=3$
Now, put the value of $x$ in Eq. (i), we get
4 $\times$ 3 + 6u = 15
6u = 15 – 12 $\Rightarrow$ 6u = 3
$\Rightarrow u=\frac{1}{2}\Rightarrow \frac{1}{y}=\frac{1}{2}[\because u=\frac{1}{y}]\Rightarrow y=2$
Hence, the required values of $x$ and $y$ are 3 and 2, respectively.