Question 824026
The Pythagorean theorem says
{{{one}}}{{{leg^2+the}}}{{{other}}}{{{leg^2=hypotenuse^2}}} .
Applied to this case, we get the equation
{{{x^2+(2x-2)^2=(x+2)^2}}} .
 
Solving:
{{{x^2+(2x-2)^2=(x+2)^2}}}
{{{x^2+4x^2-8x+4=x^2+4x+4}}}
{{{5x^2-8x+4=x^2+4x+4}}}
{{{5x^2-8x+cross(4)=x^2+4x+cross(4)}}}
{{{5x^2-x^2-8x-4x=0}}}
{{{4x^2-12x=0}}}
{{{4x(x-3)=0}}}
{{{4x=0}}}<-->{{{x=0}}} does not make sense.
The only solution that makes sense is
{{{x-3=0}}}<-->{{{x=3}}}
So, if {{{x=3}}}
{{{x+2=3+2}}} --> {{{x+2=5}}} and
{{{2x-2=2*3-2}}} --> {{{2x-2=6-2}}} --> {{{2x-2=4}}}