Question

. Write the value of k for which the system of equation 3 X + k Y = 0, 2 x – 5=0 has a unique solution. ​

Answer 1

Answer:

The system of equations is:

3x + Ky = 0

2x - y = 0

For a unique solution,

\begin{gathered}\frac{a_1}{a_2} \neq \frac{b_1}{b_2} \\\end{gathered}a2a1=b2b1

here, a1 = 3, b1 = k, a2 = 2, b2 = -1

\begin{gathered}\frac{a_1}{a_2} \neq \frac{b_1}{b_2} \\ \\ \frac{3}{2} \neq \frac{k}{ - 1} \\ \\ \frac{3}{2} \times ( - 1) \neq k \\ \\ k \neq \frac{ - 3}{2}\end{gathered}a2a1=b2b123=−1k23×(−1)=kk=2−3

Thus k ≠ -3/2. k can be any value except -3/2.

So possible values of k are 2, 3, 5, -100 etc.

You can take anything. Just do not take k as -3/2.

Step-by-step explanation:

Please mark me as brainliest

Answer 2

mark as brainliest plz

i will follow u