Question

a rectangle has a perimeter of 76 cm. Its length is 5 more that twice its width. The system of equation below can be use to fine the length,l, and the width,w, of the rectangle 2l+2w=76,l=2w+5

Answer 1

`l=27cm` and
`w=11cm`

Explanation:

the perimeter of a rectangle is given by the following equation:

`p=2(l+w)=2l+2w`

if a rectangle has a perimeter of 76cm and its length is 5 plus the double of its width we can use a system of equations to find the length
`l`, and the width
`w`.

For the perimeter of the rectangle

`2l+2w=76` where
`l` is its length, and
`w` is its width, 76 is the perimeter of the rectangle in cm.

We know that the lengh of the rectangle is 5 plus the double of its width, writing the equation:

`l=5+2w`

Substituting the equation above in
`2l+2w=76`

`2(5+2w)+2w=76\\10+4w+2w=76\\6w=76-10\\w=66/6\\w=11`

We got that the rectangle width is w=11cm

Substituting w in the equation
`2l+2w=76`

`2l+2(11)=76\\2l+22=76\\l=(76-22)/2=54/2\\l=27`

We got that the rectangle length is l=27cm

Verifying the values ​​obtained

`2l+2w=76` with l=27 and w=11

`2(27)+2(11)=76`

`54+22=76`

Satisfying the equation