Question

Given that p=x^2-y^2/x^2+xy

Answer 1

Answer:

p = - $\frac{1}{2}$

Step-by-step explanation:

Given

p = $\frac{x^2-y^2}{x^2+xy}$ ( factorise numerator and denominator )

x² - y² ← is a difference of squares and factors as (x - y)(x + y)

x² + xy ← factor out x from each term

= x(x + y) , thus

p = $\frac{(x-y)(x+y)}{x(x+y)}$ ← cancel (x + y) on numerator/ denominator

   = $\frac{x-y}{x}$ ← substitute x = - 4, y = - 6

   = $\frac{-4-(-6)}{-4}$

   = $\frac{-4+6}{-4}$

    = $\frac{2}{-4}$ = - $\frac{1}{2}$