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. |
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