Given : - 

Two equation 

x + 2y = 5 and 3x + 6y = 15 

Tip : - We know that 

For a system of 2 simultaneous linear equation with 2 unknowns 

(i) If D ≠ 0, then the given system of equations is consistent and has a unique solution given by

  x = \(\frac{D_1}{D}\), y = \(\frac{D_2}{D}\) 

(ii) If D = 0 and D₁ = D₂ = 0, then the system is consistent and has infinitely many solution. 

(iii) If D = 0 and one of D₁ and D₂ is non – zero, then the system is inconsistent. 

Now, 

We have, 

x + 2y = 5 

3x + 6y = 15 

Lets find D

⇒ D = \(\begin{vmatrix} 1 & 2 \\[0.3em] 3 & 6 \\[0.3em] \end{vmatrix}\)

⇒ D = – 6 – 6 

⇒ D = 0

Again, 

D₁ by replacing 1^st column by B 

Here,

B = \(\begin{vmatrix} 5 \\[0.3em] 15\\[0.3em] \end{vmatrix}\)

⇒ D₁ = \(\begin{vmatrix} 5 & 2 \\[0.3em] 15 & 6 \\[0.3em] \end{vmatrix}\)

⇒ D₁ = 30 – 30 

⇒ D₁ = 0

And, 

D₂ by replacing 2^nd column by B 

Here,

B = \(\begin{vmatrix} 5 \\[0.3em] 15\\[0.3em] \end{vmatrix}\)

⇒ D₂ = \(\begin{vmatrix} 1& 5 \\[0.3em] 3 & 15 \\[0.3em] \end{vmatrix}\)

⇒ D₂ = 15 – 15 

⇒ D₂ = 0 

So, here we can see that

D = D₁ = D₂ = 0 

Thus, 

The system is consistent with infinitely many solutions. 

Let 

y = k 

then, 

⇒ x + 2y = 5 

⇒ x = 5 – 2k 

By changing value of k you may get infinite solutions.