Check carefully

Rearranging the terms, we get

$x^2 - y^2 - 4x + 4y =0$

$(x-y)(x+y) - 4(x-y)=0$

$(x-y)(x+y-4)=0$

So, it may be either

$x = y$

Or

$x + y = 4$

Y = 4 and X = 0, or vice versa.

Transform coordinates to $x' = x -2$ , $y' = y-2$ . Then we get $\left[(y')^2 + 4y' + 4\right] - 4\left[y' + 2\right] = \left[(x')^2 + 4x' + 4\right] - 4\left[x' + 2\right] \\ (y')^2 + 4y' + 4 - 4y' - 8 = (x')^2 + 4x' + 4 - 4x' - 8 \\ (y')^2 - 4 = (x')^2 - 4 \\ (y')^2 = (x')^2 \\ |y'| = |x'|.$ It follows that either $y = x$ or $y-2 = -(x-2)$ , i.e. $y = 4-x$ . When plotted, the solution consists of two crossing lines, at $45^\circ$ to the coordinate axes, intersecting in the point (2,2).

We can prove the given statement $(x=y)$ false by giving any example.

Example 1:- $x=1, y=3$

Example 2:- $x=0,y=4$ .

$\\$

Proof for this:- We can prove it simply by factorising the equation which will provide us with $x=y$ or $x+y=4$ .

Quadratic formula to the rescue!

The quadratic formula states that if $ax^{2} + bx + c = 0$ , then $x = \frac{-b ± \sqrt{b^{2} - 4ac}}{2a}$ .

Since $y^{2} - 4x = x^{2} - 4x$ , there is a number $k$ such that $y^{2} -4y + k = 0$ and $x^{2} -4x + k = 0$

Thus the quadratic formula can be used to solve for $x$ or $y$ in terms of $k$ .

Assigning values of 1, -4, and $k$ to $a$ , $b$ , and $c$ in the quadratic formula, we find that x or y can have either of two values,

$\frac {4 + \sqrt{16 -4k}}{2}$ or $\frac {4 - \sqrt{16 -4k}}{2}$

thus there are two possible values for y and the same two possible values for x.

Add 4 to both sides to get $(y - 2)^{2} = (x - 2)^{2}$ . This is satisfied by any real numbers $x,y$ that are equidistant from 2. $LaTeX$

Another way to look at this equation is to complete the square on both sides, which is possible in this case because we add the same value to both sides of the equation, keeping the equation equal:

 y^2 - 4y + 4 = x^2 - 4x + 4

(y-2)^2 = (x-2)^2

For the next step, we can temporarily imagine the equation: a² = b² and remember that in this case, a and b can be either positive or negative.

|y - 2| = |x - 2|

Once we realize this, it is obvious that the answer to this question is No , even before we identify what the values must be:

We already know that this squared equation is true when (y-2) = (x-2) so let's look at the other equation:

y - 2 = -(x - 2)

y - 2 = 2 - x

y = 4 - x

This allows us to find solution pairs (0, 4) and (2, 2)

BUT if we ignore this step, and instead look at pairs of values where |y-2| = |x-2|, we find an infinite number of solutions:

|y-2| = z       |x-2| = z

2 - y = z        x - 2 = z

y = 2 - z       x = 2 + z

z    x    y    (x² - 4x)    (y² - 4x)
0    2    2        -8              -8
1    3    1        -3              -3
2    4    0         0               0
3    5    -1        ?               ?
4    6    -2        ?               ?
etc.

After all of this work, I realized that there was a very simple method to find these pairs of numbers:

1. Graph *f(x) = x² - 4x *

2. Draw any horizontal line that passes through both legs of the parabola

3. Define the coordinates for each point

4. Your points are (x, ?) and (y, ?)

In this parabola, I have marked two pairs of points that illustrate that there are many answers. x=-1, y=5; and x=0, y=4

We start by giving a COUNTER EXAMPLE. one possible counter example is when y=x+1.

We need to prove that the counter example (x+1,y) will be true to the equation y^2 - 4y = x^2 - 4x.

Let's substitute x+1 to y:

(x+1)^2 - 4(x+1) = x^2 - 4x. Simplify.

x^2 + 2x + 1 - 4x - 4 = x^2 - 4x. Simplify furthermore.

2x - 3 = 0, then we have,

x = 3/2. so,

if x = 3/2, then y=5/2.

Substitute the solution (3/2, 5/2).

(3/2)^2 - 4(3/2) = (5/2)^2 - 4(5/2).

Then simplify one side at a time.

(5/2)^2 - 4(5/2) = -15/4,

(3/2)^2 - 4(3/2) = -15/4

And so, -15/4 = -15/4

We have prove a COUNTER EXAMPLE. Therefore, it doesn't necessarily follow that y=x if y^2 - 4y = x^2 - 4x. I'M DONE. THANKS FOR THIS PROBLEM.

From the quadratic formula we know that $\Delta =b^2-4ac$ and by using this formula we will find out that $\Delta>0$ . Which means that there will be two roots. And therefore, it does not necessarily follow.

$y^2 - 4y = x^2 - 4x \implies$ $(y - 2)^2 = (x - 2)^2 \implies$ $y = \pm(x - 2) + 2 \implies$ $y = x, y = 4 - x$

Checking $y = 4 - x$ we obtain:

$(4 - x)^2 - (4 - x) = 16 - 8x + x^2 - 16 + 4x = x^2 - 4x.$

$\therefore y = 4 - x$ is the other solution.

If we can get an​ example then we have done it. so the example is x=0 and y=4.

y^2 -4y = x^2 -4x

y^2 -x^2 = 4y -4x

(y+x) (y-x) = 4(y-x)

Asuming that y is not equal to x, we can divide each part of the ecuation by (y-x), so that we get the following: y+x = 4

Conclusion: with y=3 and x=1 or y=4 and x=0, the first ecuation is correct so it is not necessary that x is equal to y.

This equation is quadratic i.e. it is degree 2. Hence it must have two answers for x and two answers for y

(x,y)=(0,4) or (x,y)=(4,0) can be satisfied the equation.

This problem just needs to know if a variable squared minus 4 times that variable has more than one solution. So using the quadratic equation determinant formula, 2 real solutions as (b^2-4ac)= 16-4 is positive.

Or, as there are no constants, zero is a solution and four is a solution to make the two terms equal and cancel out each other.

If you factor out a variable from each side you get y(y-4)= x(x-4). If y is 0, then that side is 0, but if x we're 4 then that side is also 0 or vice Versa.

Try the same question with this one x^y = x^y

Completing the square on both sides: $(y-2)^2-4=(x-2)^2-4$

Both sides of the equation are quadratic functions with a vertex at $(2,-4)$

Therefore equations are equal so long as: $2-y = x-2$ or $y = x$

People are getting really bogged down in "algebraic computational nonsense." Factor both sides to get y(y-4) = x(x-4)

There, now you have two products of two numbers set equal to each other. We can clearly see that both sides can be made positive and have the Y be positive and large enough so we have a product of two positives is positive, then make the X negative and get a product of two negatives is also positive.

There. You don't need to use so much number-crunching power to derive closed-form solutions. Just demonstrate a simple property holds that invalidates the necessity argument.