Q. 1 The pair of equations 𝑥 + 2𝑦 + 5 = 0 and − 3𝑥 − 6𝑦 + 1 = 0 have:

1. A unique solution 
2. Exactly two solutions
3. Infinitely many solutions 
4. No solution

(OR)

Q.2 If $\frac{1}{2}$ is a root of the equation ,  $x^2 +\hspace{0.17em}kx - \frac{5}{4} = 0$, then the value of  k is

1. 2
2. -2
3. $\frac{1}{2}$
4. $\frac{-1}{2}$

Answer : 4

The given system of equations is 𝑥 + 2𝑦 + 5 = 0 and − 3𝑥 − 6𝑦 + 1 = 0.

These equations are of the form 𝑎₁𝑥 + 𝑏₁𝑦 + 𝑐₁ = 0  and 𝑎₂𝑥 + 𝑏₂𝑦 + 𝑐₂ = 0 

By comparison, we can conclude, 𝑎 ₁ = 1, 𝑎₂ =− 3 𝑏₁ = 2, 𝑏₂ =− 6, 𝑐₁ = 5 𝑎𝑛𝑑 𝑐₂ = 1,

$\text{ Here, }\frac{{\overline{a}}_1}{{\overline{a}}_2}=\frac{1}{-3}$

$\frac{b_1}{b_2}=\frac{2}{-6}$

$\text{ Also, }\frac{c_1}{c_2}=\frac{5}{1}$

From the above results, we can infer that, $\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$

Therefore, the given system of equations has no solution.

(OR)

Answer : 1

Solution: The given equation is $x^2+kx-\frac{5}{4}=0$ and we need to find the value of k.
Since $\frac{1}{2}$ is a root, therefore $x =\frac{1}{2}$ it will satisfy the given equation.

$\Rightarrow {\left(\frac{1}{2}\right)}^2+k\left(\frac{1}{2}\right)-\frac{5}{4}=0$

$\Rightarrow \frac{1}{4}+\frac{k}{2}-\frac{5}{4}=0$

$\Rightarrow k=2$

Hence value of k = 2