Solve graphically the pair of linear equations : 

3𝑥 − 4𝑦 + 3 = 0 and 3𝑥 + 4𝑦 − 21 = 0.

Find the coordinates of the vertices of the triangular region formed by these lines and the x-axis. 

Also, calculate the area of this triangle.

                                                (OR)

$\text{ Given, Solve for }x:{\left(\frac{2x}{x-5}\right)}^2+5\left(\frac{2x}{x-5}\right)-24=0,x+5\text{. }$

We need to graphically solve the given pair of linear equations and find the coordinates and area of the triangle region. 

In order to draw the graphs, we need to find out two points with their 𝑥 and 𝑦 coordinates that satisfy the equations respectively. We will do this by putting random value of 𝑥 and finding corresponding value of 𝑦. 

For 3𝑥 − 4𝑦 + 3 = 0 or 3𝑥 − 4𝑦 =− 3, it can be written as

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

For 3𝑥 + 4𝑦 − 21 = 0 or 3𝑥 + 4𝑦 = 21, it can be written as

$y=\frac{21-3x}{4}$

Now, putting these coordinates on a graph and joining and extending them will give us the equation. 

On plotting the point, we get the graph as

Since the curves intersect at the point (3, 3), the solution of the equations

3𝑥 − 4𝑦 + 3 = 0 and 3𝑥 + 4𝑦 − 21 = 0 𝑖𝑠 (3, 3). 

Hence, 𝑥 = 3 𝑎𝑛𝑑 𝑦 = 3. 

From the graph, the vertices of the triangular region are (3, 3), (− 1, 0) and (7, 0). 

The area of the triangular region is

$A=\frac{1}{2}\times \text{ base }\times \text{ height }$

On substituting the values, we get the area as

$A=\frac{1}{2}\times \text{8}\times \text{ 3}$

⇒ 𝐴 = 12 𝑠𝑞. 𝑢𝑛𝑖𝑡𝑠

                                  (OR)

$\text{ We need to find the value of }x\text{ for }{\left(\frac{2x}{x-5}\right)}^2+5\left(\frac{2x}{x-5}\right)-24=0\text{ and }x\ne 5\text{. }$$Let \left(\frac{2x}{x-5}\right)=y$

On replacing, we get 

$y^2+5y-24=0$

on solving for y, we get

$\begin{array}{l}\Rightarrow y^2+5y-24=0\\ \Rightarrow y^2+8y-3y-24=0\\ \Rightarrow (y+8)-3(y+8)=0\\ \Rightarrow (y+8)(y-3)-0\\ \Rightarrow y=3\text{ or }-8\end{array}$

On substituting 𝑦 = 3 in (𝑖), we get

$\Rightarrow \left(\frac{2x}{x-5}\right)=3$

⇒ 2𝑥 = 3(𝑥 − 5) 

⇒ 2𝑥 = 3𝑥 − 15 

⇒ 𝑥 = 15

Similarly, on substituting 𝑦 =− 8 in (𝑖), we get

$\Rightarrow \left(\frac{2x}{x-5}\right)=-8$

⇒ 2𝑥 =− 8(𝑥 − 5) 

⇒ 2𝑥 =− 8𝑥 + 40 

⇒ 10𝑥 = 40 

⇒ 𝑥 = 4 

Hence, 𝑥 = 4 𝑜𝑟 15.