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

Answer:

$l=27cm$ and $w=11cm$

Step-by-step 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