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Question

# Find the value of k for which the given system of equations has infinite number of solutions.5x+2y=2k and 2(k+1)x+ky=(3k+4)

A
4
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B
7
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C
3
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D
6
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Solution

## The correct option is A 4Comparing 5x+2y=2k with a1x+b1y+c1=0, we get,a1=5,b1=2,c1=−2kComparing 2(k+1)x+ky=3k+4 with a2x+b2y+c2=0, we get, a2=2(k+1),b2=k,c2=−(3k+4)Now, Given: the equations has infinite no of solutions.∴a1a2=b1b2=c1c252(k+1)=2k=−2k−(3k+4)When,52(k+1)=2k⟹5k=4(k+1)⟹5k−4k=4⟹k=4and 2k=−2k−(3k+4)⟹2(3k+4)=2k2⟹3k+4=k2⟹k2−3k−4=0⟹k2−4k+k−4=0⟹k(k−4)+(k−4)=0⟹(k−4)(k+1)=0Either, k=4 or k=−1k=4 is the value that satisfies both the conditions.So, Option A is correct

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