Question

if x2+xy+x=12 and y2+xy+y=18 find x+y

Answer 1

Hi ,

x² + xy + x = 12 ----( 1 )

y² + xy + y = 18 ----( 2 )

add equations ( 1 ) and ( 2 ) , we get

x² + 2xy + y² + x + y = 30

( x + y )² + ( x + y ) - 30 = 0

( x + y )² + 6( x + y ) - 5( x + y ) - 30 = 0

( x + y ) [ x + y + 6 ] - 5 ( x + y + 6 ) = 0

( x + y + 6 ) ( x + y - 5 ) = 0

Therefore ,

x + y + 6 = 0 or x + y - 5 = 0

x + y = -6 or x + y = 5

I hope this helps you.

: )

Answer 2

$x^2 + xy + x = 12$ $----------------1equation$


$y^2 + xy + y = 18$ $------------2equation$

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Adding both the equations,

$x^2 + xy + x + y^2 + xy + y = 12 + 18$

$x^2 + y^2 + xy + xy + x + y = 30$

$(x + y)^2 + 1(x + y) = 30$


Assume (x + y) as a

$a^2 + a = 30$

$a^2 + a - 30 = 0$

$a^2 + (6 - 5)a -30 = 0$

$a^2 + 6a -5a - 30 = 0$

$a(a + 6) - 5(a + 6) = 0$

$(a + 6) (a -5) = 0$

a = - 6   Or   a = 5 
x + y = -6   Or   x + y = 5 



i hope this will help you


-by ABHAY