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Question

If the system of equations x+y=1,(c+2)x+(c+4)y=6,(c+2)2x+(c+4)2y=36
are consistent, then the value of c can be

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 D 4

For system of equations to be consistent:
Δ=0
Δ=111(c+2)(c+4)6(c+2)2(c+4)236=0

By C1C1C2 and C2C2+C3,

001(2)(c2)64(c+3)(c2)(c+10)36=0

Taking (2) common from C1,(c2) common from C2 and (1) common from C3, we get
(2)(c2)(1)0011162(c+3)(c+10)36=0
c=2,4


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