( 3x - 3y )² + ( 3x + 2y )²
Answers
Answer:
Let me show you the algebra here real quick and then I’ll explain in more detail.
(2x+3y)^2=(3x+2y)^2
(2x+3y) = (3x+2y); (2x+3y) = -1(3x + 2y) (Take the square root of both sides)
(2x+3y) = (3x+2y); (2x+3y) = (-3x-2y) (Distribute out the -1)
(2x+3y-2x) = (3x+2y-2x); (2x+3y-2x) = (-3x-2y-2x) (Subtract 2x from everything)
3y = x+2y; 3y = -5x-2y (Simplify)
3y-y = x+2y-y; 3y = -5x-2y (Subtract y from both sides on the left equation)
2y = x+y; 3y = -5x-2y (Simplify the left equation)
2y = x+y; 3y+2y = -5x-2y+2y (Add 2y to both sides of the right equation)
2y = x+y; 3y+2y+5x = -5x-2y+2y+5x (Add 5x to both sides of the right equation)
2y = x + y; 5y+5x = 0 (Simplify the right equation)
2y = x+y; x + y = 0 (Divide both sides of the right equation by 5)
We can see that x+y in this case can actually be one of two possible values: 2y and 0. We have two equations and each one has its own answer. The reason we have two equations is because we took the square root of both sides at the beginning. You see, -2 and 2 are not the same value, but when squared, they both yield 4. As such, the square root of x^2 is not just x but ±x, and we have to take this into account.
Let me conclude by explaining why the possible values of x+y are those particular expressions. If x and y are the same, the equation is essentially (5x)^2 = (5x)^2 which is clearly true. For x + y, the sum x+y is the same thing as just saying 2y (or 2x, for that matter). The other possibility, is that x is -1 * y. In that case, the equation simplifies to (-x)^2 = x^2 which is true because the squaring operation makes the -1 in -x irrelevant. If x = -y, then x + y is x + -x = 0.
Hope this helps!
The problem is that the correct answer is:
Any value if y=x , 0 if y≠x
The idea behind is shown below, I will just briefly touch it:
(2x+3y−3x−2y)(2x+3y+3x+2y)=0⇔(y−x)⋅5⋅(x+y)=0
So either x=y and x+y can have any arbitrary value or x≠y and then indeed x+y=0
Step-by-step explanation:
(3x-3y)2+(3x+2y)2
=9x2-9y2+9x2+4y2
=-5y2
hope it help you
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