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in Determinants by (27.4k points)
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Show that each of the following systems of linear equations has infinite number of solutions and solve : 

x + 2y = 5 

3x + 6y = 15

1 Answer

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Best answer

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.

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