Given below is a non-linear equation.

$\frac{3\left(x+y\right)+7\left(x-y\right)}{x^2-y^2}=12$

Garima has reduced it to a linear equation in two variables. Shown below are her steps.

Step $1:$ $\frac{3\left(x+y\right)+7\left(x-y\right)}{x^2-y^2}=12$

Step $2: \frac{3}{\left(x+y\right)}+\frac{7}{\left(x-y\right)}=12$

Step $3:$Takes $\frac{1}{\left(x+y\right)}=a$ and $\frac{1}{\left(x-y\right)}=b$

Step $4: 3a+7b=12$

However, she has made a mistake in one step. Identify the erroneous step and complete the solution.

Convert the following equations into simultaneous equations and solve:

$\sqrt{\frac{x}{y}}=4, \frac{1}{x}+\frac{1}{y}=\frac{1}{xy}$

On comparing the ratios $\frac{a_1}{a_2}, \frac{b_1}{b_2}$ and $\frac{c_1}{c_2}$, find out whether the lines representing the following pair of linear equations intersect at a point, are parallel or coincident.

$x+2y-4=0$

$2x+4y-12=0$

The solution set of the given pair of linear equation $5x-5y=-5$ and $\frac{3x}{2}-\frac{3y}{2}+\frac{3}{2}=0$ is _____.

Solve the following pair of equation by reducing them to a pair of linear equations:

$\frac{1}{3x+y}+\frac{1}{3x-y}=\frac{3}{4}$

$\frac{1}{2\left(3x+y\right)}-\frac{1}{2\left(3x-y\right)}=\frac{-1}{8}$

Solve the following pair of equations  by reducing them to a pair of linear equation:

$\frac{10}{x+y}+\frac{2}{x-y}=4 ; \frac{15}{x+y}-\frac{5}{x-y}=-2$

Consider the following pairs of linear equation:

$$\begin{gathered} \left(i\right) 3x+2y=5 ; 2x+3y=5 \\ \left(ii\right) 2x-3y=7; 2x-3y=8 \end{gathered}$$

Choose the correct alternative.

Find $c$ if the system of equations $cx+3y+\left(3-c\right)=0$; $12x+cy-c=0$ has infinitely many solutions.

In Fig. 1, $ABCD$ is a rectangle. Find the values of $x$ and $y$.