The number of values of k for which the system of linear equations, (k+2)x+10y=kkx+(k+3)y=k–1
has no soution, is :
A
1
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B
2
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C
3
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D
4
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Solution
The correct option is A1 For no solution, a1a2=b1b2≠c1c2⇒k+2k=10k+3≠kk−1⇒(k+2)(k+3)=10k⇒k2−5k+6=0⇒(k−2)(k−3)=0⇒k=2,3If k=2,then both the lines becomes identical. So, k≠2∴k=3 is the only one value of k.