ASK ME MATH QUESTION

Nice trick. Two variables but only one equation. So there's no fixed solution.
But if you specify that x and y are both integers then it works for x=130 and y=11.
Are there other higher solutions with integers?
Yes. It looks like it'll work for all integer values of y that are 11 or greater and they also end in one. Then you can very quickly find x. There's an infinite set of solutions.
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But it also appears that there's an even 'bigger infinity' of solutions for positive real number values.
And it appears there would be a graph that is not symmetrical around the origin of coordinates.

168, i gave 2

162 and 163

Pf

Sry for that msg, 😐 😕 😔

I meant prove x = x+1 is impossible

Assume x DOES equal x + 1.

x = x + 1

Subtract x from both sides.

x - x = x + 1 - x

0 = 1

False

Proof by contradiction

QED

Now can u solve 162 and 163?

What if x is ±0.5 ?
(roots of equations and square roots are often ±)
If you were answering a question in an examination and you put
0.5 = -0.5 + 1 .... then the teacher would probably mark it wrong in this context.
And say something like 'yes ± often occurs in square roots and roots of equations but you have to use one or the other - you can't 'switch within'.

Are they a system, or are they separate problems?

2.

That would mean y=z^4 but y is also the cube root of z.
That would work for y and z both equal to one.
And x= 1 would work for both of those.
And all three of them equal to zero seems to work too.
Doesn't seem to work for minus one.