Question

Find the value of k for which the system of linear equations has an infinite number of solutions.
2x + 3y=9,
6x + (k – 2)y =(3k – 2

Answer 1

The given pair of linear equations are
2x + 3y – 9 = 0                            ……(i)
6x + (k – 2)y – (3k – 2) = 0           ……(ii)
Which is of the form ${\displaystyle a_{1}x+b_{1}y+c_1=0\text{and}{a}_{2}x+b_{2}y+c_2=0}$, where
${\displaystyle a_1=2,b_1=3,c_1=-9,a_2=6,b_2=k–2\text{and}{c}_2=-(3k–2)}$
For the given pair of linear equations to have infinitely many solutions, we must have
${\displaystyle \frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}}$
${\displaystyle \Rightarrow \frac{2}{6}=\frac{3}{k-2}\ne \frac{-9}{-(3k-2)}}$
${\displaystyle \Rightarrow \frac{2}{6}=\frac{3}{k-2},\frac{3}{k-2}\ne \frac{-9}{-(3k-2)}}$
${\displaystyle \Rightarrow k=11,\frac{3}{k-2}\ne \frac{9}{(3k-2)}}$
⇒ k = 11, 3(3k – 2) ≠ 9(k – 2)
⇒ k = 11, 1 ≠ 3 (true)
Hence, k = 11.