Question

If the equations $x+y=1$,$(c+2)x+(c+4)y=6$, $(c+2{)}^{2}x+(c+4{)}^{2}y=36$
are consistent, then $c=?$

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (B), (D)

4
2

Given, $x+y=1$ ............(1)
$(c+2)x+(c+4)y=6$ ............(2)
$(c+2{)}^{2}x+(c+4{)}^{2}y=36$ ............(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 $y=\frac{4-c}{2}$,

substituting the above value in equation (5) we get,

So, $2c(4-c)+6(4-c)=36-c^2-4-4c$
$8c-2c^2+24-6c=32-c^2-4c$
$c^2+2c-8c+32-24=0$
$c^2-6c+8=0$
$c=\frac{6\pm \sqrt{36-32}}{2}=\frac{6\pm 2}{2}$
$c=4,2$