If the perimeter of a right triangle is 42 units and the length of the hypotenuse is 20 units, find the length of the other two sides of the triangle.
can you show work

So
a^2+b^2=c^2
c=hypotenuse
a^2+b^2=20^2=400

if the permiter is 42 then
p=hypotonuse+side+side
42=20+side+side
subtract 20 from both sides
22=side+side

a+b=22
subtract b from both sides
a=22-b
subsitute for a in equation below
a+2+b^2=c^2
(22-b)^2+b^2=400
b^2-44x+484+b^2=400
2b^2-44x+484=400
divide both sides by 2
b^2-22x+242=200
subtract 200 from both sides
b^2-22x+42=0
factor
(b+/-number)(b+/-number)
to find the numbers, find which 2 number multiply to 42 and add to get -22
the numbers must be neative since same signs multiplied=positive and negative+negative=negative
(b-number)(b-number)
facrors of 42=2,3,6,7,14,21
2,21
3,14
6,7
it's hard to find the number, but the numbers are -19.8882 and -2.11181
(b-19.8882)(b-2.11181)=0
if xy=0 then assume that x and y=0

b-19.8882=0
add 19.8882 to both sides
b=19.8882

b-2.11181=0
add 2.11181 to both sides
b=2.11181

the side legnths are 19.8882 units and 2.11181 units