Question

X-6 = (Y-6)^2 ..........(1).
3(x+3)=Y ..........(2)
then find the value of x and y

Answer 1

( i ). x - 6 = ( y - 6 ) ^2

( ii ). 3x + 9 = y

Putting value of y from ( ii ),

x - 6 = [ 3 x + 9 - 6 ] ^2

x - 6 = ( 3x +3 ) ^2

x - 6 = 9 x^2 + 9 + 18 x

0 = 9 x^2 + 9 + 18 x - x + 6

0 = 9 x^2 +15 + 17x

0 = 9 x^2 + 17 x + 15

0 = x ( 9x + 17 ) + 15

0 = 9x + 17 + 15

0 = 9x + 32

-32 / 9 = x

X = - 32 / 9

Putting value of x in y,

y = 3x + 9

Y = 3 ( - 32 /9 ) + 9

y = ( - 32 / 3) + 9

y = ( - 32 + 27 ) / 3

y = - 5 / 3.