Solve (x+2) 2 = (x+y) 2 =x 2 +2xy+ y 2
and
(x−3) 3 = (x−y) 3 =x 3 −3 x 2 y+3x y 2 − y 3

X² + 2xy − y² + x = 6 

When you implicitly differentiate, you differentiate the entire expression. Differentiate x terms normally, but use the chain rule when you have a term with y in it. Remember to use the chain and product rule on 2xy (think of 2xy as 2x*y). 

2x + 2y + 2x*dy/dx - 2y*dy/dx + 1 = 0 

Now separate dy/dx from everything else: 
dy/dx*(2x - 2y) = -(2x + 2y + 1) 
dy/dx = -(2x + 2y + 1)/(2x - 2y) 

The slope is dy/dx, so this means at the point (2, 4), the slope is: 
dy/dx = -(2x + 2y + 1)/(2x - 2y) 
dy/dx = -(2(2) + 2(4) + 1)/(2(2) - 2(4)) 
= -(4 + 8 + 1)/(4 - 8) 
= 13/4 

Using point slope form, the tangent line is: 
y - y1 = m(x - x1) 
y - 4 = 13/4*(x - 2)

1.17+3.99a. you combine the terms with a to create 3.99 and the 1.17 stays because it has no like terms or a variable