Q.1 Solve the pair of linear equations graphically 𝑥 + 3𝑦 = 6 and 2𝑥 − 3𝑦 = 12, Also shade the region bounded by the line 2𝑥 − 3𝑦 = 12 and both the coordinates. 

 

(OR)

 

Q.2 The denominator of the fraction is one more than twice its numerator. If the sum of the fraction and its reciprocal is $2\frac{16}{21}$. find the fraction.

Given equations are

 𝑥 + 3𝑦 = 6 …(𝑖) 

2𝑥 − 3𝑦 = 12 …(𝑖𝑖)

 From (𝑖), 𝑥 = 6 − 3𝑦

From (𝑖𝑖), 𝑥 = $From (𝑖𝑖), 𝑥 = \frac{12+3y}{2}$

By plotting the points in the graph, we get

Both lines intersect at (6, 0)

(OR)

Let us assume that, 

Numerator of the fraction be 𝑥 

The denominator of the fraction is one more than twice its numerator 

⇒Denominator = 2𝑥 + 1 

Then the fraction be, …$\frac{ 𝑥}{ 2𝑥+1}$ ........(𝑖) 

And its reciprocal is $\frac{2𝑥+1}{x}$

But the sum of the fraction and its reciprocal is $2\frac{16}{21}$

$\begin{array}{r}\Rightarrow \frac{x}{2x+1}+\frac{2x+1}{x}=2\frac{16}{21} \\ \Rightarrow \frac{x\left(x\right)+2x+1(2x+1)}{(2x+1)\left(x\right)}=\frac{58 }{21} \\ \Rightarrow 21\left(x^2+4x^2+2x+2x+1\right)=58\left(2x^2+x\right) \\ \Rightarrow 21\left(5x^2+4x+1\right)=116x^2+58x \\ \Rightarrow 105x^2+84x+21=116x^2+58x \\ \Rightarrow 11x^2-26x-21=0 \\ \Rightarrow 11x^2-33x+7x-21=0 \\ \Rightarrow 11x(x-3)+7(x-3)=0 \\ \Rightarrow (x-3)(11x+7)=0 \\ \Rightarrow x=3\text{ and }x=\frac{-7}{11} \end{array}$

The denominator of the fraction should not be negative So, we take 𝑥 = 3

Substitute 𝑥 = 3 in $\frac{x}{2x+1}$ 

⇒ $\frac{3}{2\left(3\right)+1}$ = $\frac{3}{7}$ 

Hence, the fraction is $\frac{3}{7}$