Question

The width of a rectangle is 3 less than 2 times the length. If the perimeter is 51 what is the actual length and width of the rectangle

Answer 1

Let $l=length$ be length and $w=width$

$w=2l-3$

To find the perimeter add all the sides of the rectangle

$Perimeter=w+w+l+l$

Replacing $w=2l-3$

$(2l-3)+(2l-3)+l+l=51$

$6l-6=51$

add to both sides 6

$6l=51+6$

$6l=57$

Divide both sides with 6

$l=57/6 \\ l=9.5$

$w=2l-3 \\ w=2(9.5)-3\\ w=19-3\\ w=16$