If the equations x+y=1,(c+2)x+(c+4)y=6, (c+2)2x+(c+4)2y=36 are consistent, then c=?
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 options are C 4 D 2 Given, x+y=1 ............(1) (c+2)x+(c+4)y=6 ............(2) (c+2)2x+(c+4)2y=36 ............(3) this equation are consistent so, solution exist. from (1) & (2), 2y=6−c−2=4−c ............(4) and from (1) & (3), 4cy+12y=36−(c+2)2 ........(5) By y=4−c2,
substituting the above value in equation (5) we get,
So, 2c(4−c)+6(4−c)=36−c2−4−4c 8c−2c2+24−6c=32−c2−4c c2+2c−8c+32−24=0 c2−6c+8=0 c=6±√36−322=6±22 c=4,2