Question

(x-y)^2 + (x+3)(x+3)=0
What is value of x and y?

Answer 1

Answer: Question error. Two variables should have two equations to solve.

• (x^2-2xy+y^2)+(x^2+6x+9)=0
• By expanding (x-y)^2 as x^2-2xy+9.
• Then taking (x+3)(x+3) as (x*x+3*x+3*x+3*3)
• It now becomes:
• x^2-2xy+y^2+x^2+6x+9=0
• By removing brackets. You have to multiply with the symbol appearing before bracket to get the signs right.
• 2x^2+6x-2xy=-y^2-9

Step-by-step explanation: