1.

Find the value of k for which the following pair of linear equations have infinitely many solutions:2x + 3y = 7,(k –1)x + (k + 2)y = 3k.

Answer»

Given system of equation is

2x + 3y = 7 

And (k – 1)x + (k + 2)y = 3k. 

By comparing given system of equation with standard form of system of equations, 

we get a1 = 2, b1 = 3, c1 = 7 

And a2 = k – 1, b2 = k + 2 and c2 = 3k. 

Given that system of equations have infinitely many solutions.

∴ \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}.\)

Now, \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\)

⇒ \(\frac{2}{k-1}=\frac{3}{k+2}\)

⇒ 2k + 4 = 3k – 3 

⇒ 3k – 2k = 4 + 3 

⇒ k = 7.

And \(\frac{b_1}{b_2}=\frac{c_1}{c_2}\)

⇒ \(\frac{3}{k+2}=\frac{7}{3k}\)

⇒ 9k = 7k + 14 

⇒ 2k = 14 

⇒ k = 7.

Hence for k = 7, given system of equations have infinitely many solutions.


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