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 1st 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 2nd 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.