Let
x------> the length of the rectangle
y------> the width of the rectangle
A-----> area of the rectangle
we know that
The area of rectangle is equal to
[tex] A=x*y\\A=40\ units^{2} [/tex]
so
[tex] 40=x*y [/tex] --------> equation [tex] 1 [/tex]
[tex] x=6+y [/tex] --------> equation [tex] 2 [/tex]
Substitute equation [tex] 2 [/tex] in equation [tex] 1 [/tex]
[tex] 40=[6+y]*y\\ 40=6y+y^{2} \\y^{2} +6y-40=0 [/tex]
using a graph tool-----> to resolve the second order equation
see the attached figure
the solution is
[tex] y=4 [/tex]
Find the value of x
[tex] x=6+y\\ \\ x=6+4=10\ units [/tex]
therefore
the answer is
The length of the rectangle is equal to [tex] 10\ units [/tex]
The width of the rectangle is equal to [tex] 4 \ units [/tex]
