How do you multiply #(x + -1y)(x + y)#?

#color(blue)("Using the shortcut with a bit of explanation")#

It is the case that #1y# is written as just #y#

We also have #+ - #. When two signs are next to each other and they are different the result is -. So #+ - 1y -> -y# giving

#(x+ - 1y)(x+y)" "->" "(x-y)(x+y)#

The shortcut is to know that if you have the form #a^2-b^2# then this works out to be the same as #(a-b)(a+b)#

That is the condition in this question so #(x-y)(x+y)=x^2-y^2#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Demonstration that this is true")#

Consider:#" "color(blue)((x-y))color(brown)((x+y))#

Multiply everything inside the right hand side bracket by everything inside the left hand side bracket.

#color(brown)(color(blue)(x)(x+y) color(blue)(-y)(x+y)#

Notice that the minus follow the #color(blue)(y)# in #color(blue)(-y)#

So we have:
#color(brown)(color(blue)(x)(x+y)" " color(blue)(-y)(x+y)#
#x^2+xy" "-xy -y^2#

#x^2+0-y^2#

#x^2-y^2#

Each term in the 2nd bracket must be multiplied by each term in the 1st bracket.

That is #(color(red)(x-y))(x+y)=color(red)(x)(x+y)color(red)(-y)(x+y)#

now distribute the brackets

#=x^2+cancel(xy)-cancel(xy)-y^2=x^2-y^2#