question_answer

For what value of k, will the following pair of linear equations have infinitely many solutions? \[2x+3y=4\]and \[\left( k+2 \right)x+6y=3k+2\]

A) 1

B) 2

C) 3

D) 4

Correct Answer: B

Solution :

Given, pair of equations is \[2x+3y-4=0\]

and \[\left( k+2 \right)x+6y-\left( 3k+2 \right)=0\]

On comparing the given equations with standard form i.e.

\[{{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0\] and \[{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0\],

we get

\[{{a}_{1}}=2,\,\,{{b}_{1}}=3,\,\,{{c}_{1}}=-4\]

and      \[{{a}_{2}}=k+2,\,{{b}_{2}}=6\],

\[{{c}_{2}}=-\left( 3k+2 \right)\]

For infinitely many solutions,

\[\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{{{b}_{1}}}{{{b}_{2}}}=\frac{{{c}_{1}}}{{{c}_{2}}}\]

(i)

On taking I and II terms, we get

\[\frac{2}{k+2}=\frac{3}{6}\,\,\,\Rightarrow \frac{2}{k+2}=\frac{1}{2}\]

\[\Rightarrow \,\,\,\,k+2=4\,\,\,\,\Rightarrow k=2\]

which also satisfies the last two terms of Eq. (i).

Hence, the required value of k is 2.