Question 1194751
<pre>
It may be that your teacher wanted you to do it without using
ANY trial and error.  If so,

{{{drawing(400,200,-1,7,-1,3,
line(0,0,5,0),line(5,0,5,2),line(0,2,5,2),line(0,0,0,2),
locate(1.4,0,matrix(1,3,length,""="",2w+1)),
locate(5.1,1.03,matrix(1,3,width,""="",w))

 )}}}

{{{matrix(9,3,
Area,""="",length*width,

10,""="",(2w+1)w,

10,""="",2w^2+w,

2w^2+w,""="",10,

2w^2-w-10,""="",0,

(2w+5)(w-2),""="",0,

2w+5=0,"",w-2=0,

2w=-5,"",w=2,
w=-5/2,"","")
}}}

We know the width can't be negative, so we ignore
the negative answer.

That's the way to get the width as 2 without
using any trial and error.

Now substitute w=2 in length = 2w+1 to find the length.

Edwin</pre>