Question

Solve for x and y:
$\begin{align}
  & 2x-\dfrac{3}{y}=9 \\
 & 3x+\dfrac{7}{y}=2\text{ }\left( y\ne 0 \right) \\
\end{align}$

Answer 1

Hint: We solve this question by simultaneously solving both the equations. We do this by making the coefficient of x same by multiplying first equation by 3 and second equation by 2 and subtracting the two equations and obtaining an equation only in terms of y. Then we calculate the value of y and use it in any one of the equations above and get the value of x.

Complete step by step solution:
In order to solve this question, we need to make the coefficients of x same in both the equations. This can be done by multiplying the first equation by 3 and the second equation by 2 and subtracting the two equations.
Multiplying the first equation by 3 on both sides,
$\Rightarrow \left( 2x-\dfrac{3}{y} \right)\times 3=9\times 3$
Expanding and multiplying, we get
$\Rightarrow 6x-\dfrac{9}{y}=27\ldots \left( 1 \right)$
Now, we multiply the second equation by 2,
$\Rightarrow \left( 3x+\dfrac{7}{y} \right)\times 2=2\times 2$
Expanding and multiplying,
$\Rightarrow 6x+\dfrac{14}{y}=4\ldots \left( 2 \right)$
Now, we subtract equation 2 from equation 1 as,
$\begin{align}
  & \Rightarrow 6x-\dfrac{9}{y}=27 \\
 & \text{ }-6x-\dfrac{14}{y}=-4 \\
 & \text{ }\overline{\begin{align}
  & \dfrac{-9-14}{y}=27-4\ldots \left( 3 \right) \\
 & \\
\end{align}} \\
\end{align}$
Using equation 3, we calculate the value of y.
$\Rightarrow \dfrac{-9-14}{y}=27-4$
Adding the terms on the left-hand side and subtracting the terms on the right-hand side,
$\Rightarrow \dfrac{-23}{y}=23$
Multiplying both sides by y,
$\Rightarrow -23=23y$
Dividing both sides by 23,
$\Rightarrow -1=y$
We now substitute this value in equation 1 to obtain the value of x.
$\Rightarrow 2x-\dfrac{3}{-1}=9$
Dividing -3 by -1 gives us +3 and the subtracting both sides by 3,
$\Rightarrow 2x+3-3=9-3$
Subtracting on the right-hand side,
$\Rightarrow 2x=6$
Dividing both sides by 2,
$\Rightarrow x=3$
Hence, by solving the two equations simultaneously, we get $x=3,y=-1.$

Note: We need to be careful while dividing with -1 in the third equation. The sign of the numerator term changes by doing so. We can also solve this sum by taking y as the LCM and solving for both the equations.