Proof of $ x^2 + y = y^2 + x$ when $ x+ y =1$ and $x$ is larger than $y$

Question

I know that whatever numbers you choose for x and y and their sum equals to 1 will satisfy the equation $x^2 + y = y^2 + x$

Algebraic proof:

Given: $x + y = 1$

$$LS = x^2+ y = (1-y)^2 + y = 1 - 2y+y^2 + y = y^2 - y + 1$$

$$RS = y^2 + x = y^2 + (1-y) = y^2 - y + 1$$

Therefore,$$ LS = RS $$

How can this be proved geometrically? (Ex. in a diagram of rectangular areas)

I tried to add a square piece with side lengths y with a rectangle with side lengths x and x+y but I can't seem to prove it geometrically.

Can someone help?

Answer 1

Here is a picture. The left shows $y^2+x$, the right $x^2+y$.

Answer 2

Sorry for the description of the lacking shapes...

You have to consider for the LHS a square of side $x$ and a rectangle of sides $y$ and $x+y=1$.

This can be decomposed into the "big" square $x^2$, the "little" square $y^2$ and the remaining rectangle $xy$.

For the RHS, a square of side $y$ and a rectangle of sides $x$ and $x+y=1$.

In turn, this can be decomposed into the "little" square $y^2$, the "big" square $x^2$ and the remaining rectangle $yx$.

Then rotate $yx$.

Answer 3

Because $$0=x^2+y-(y^2+x)=(x-y)(x+y)-(x-y)=(x-y)(x+y-1).$$