Question

A rectangle has an area of 40 square units. The length is 6 units greater than the width.

Answer 1

possible dimensions:
40 x 1
20 x 2
4 x 10
5 x 8

since 10 is 6 more than 4, the length is 10 and the width is 4

Answer 2

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

`A=x*y\\A=40\ units^(2)`

so

`40=x*y` --------> equation
`1`

`x=6+y` --------> equation
`2`

Substitute equation
`2` in equation
`1`

`40=[6+y]*y\\ 40=6y+y^(2) \\y^(2) +6y-40=0`

using a graph tool-----> to resolve the second order equation

see the attached figure

the solution is

`y=4`

Find the value of x

`x=6+y\\ \\ x=6+4=10\ units`

therefore

the answer is

The length of the rectangle is equal to
`10\ units`

The width of the rectangle is equal to
`4 \ units`