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For solution to have infinite solutions, ratio of x coefficients=ratio of y coefficients=ratio of constants
=>(k+1)/k=8/(k+3)=4k/(3k-1)
solve (k+1)/k=8/(k+3)
=>k^2+4k+3=8k
=>k^2–4k+3=0
=>k=1,3
Now,8/(k+3)=4k/(3k-1)
=>6k-2=k^2+3k
=>k^2–3k+2=0
=>k=1,2
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