Find the width of the rectangle if the length of the rectangle is 5 more than it’s width, and the perimeter of the rectangle is 20 units

where w is width and l is length. This can then be solved as a system of equatios. I'll solve by substitution, and as the [tex]l[/tex] is already solved for at the top:

[tex]2w+2l=20\\2w+2(w+5)=20\\2w+2w+10=20\\4w+10=20\\4w=10\\w=2.5[/tex]

If you just need the width, then you're already done there. I'll find the other variable too, and then check it to be sure it's correct.

[tex]l=w+5\\l=(2.5)+5\\l=7.5[/tex]

Now, confirm that with the second equation:

[tex]2w+2l=20\\2(2.5)+2(7.5)=20\\5+15=20\\20=20[/tex]

MrRoyalid MrRoyalid · Answer 2

The perimeter of a rectangle is the sum of its side lengths.

The width of the rectangle is 2.5 units

Represent the length with l and the width with w.

So, we have:

[tex]P =2 \times (l + w)[/tex] --- the formula of perimeter

The perimeter is 20 units.

So, we have:

[tex]20 =2 \times (l + w)[/tex]

Divide both sides of the equation by 2

[tex]10 =l + w[/tex]

The length is said to be 5 more than its width.

So, we have:

[tex]l =5 + w[/tex]

Substitute 5 + w for l in [tex]10 =l + w[/tex]

[tex]10 = 5 + w + w[/tex]

[tex]10 = 5 + 2w[/tex]

Subtract 5 from both sides

[tex]5 = 2w[/tex]

Divide both sides by 2

[tex]2.5 = w[/tex]

Rewrite as:

[tex]w =2.5[/tex]

Hence, the width of the rectangle is 2.5 units