Question 119916
You can start with the formula for the perimeter of a rectangle:
{{{P = 2(L+W)}}} where L = length and W = width.
The problem states that the width, W, is 5 inches shorter than the length, L.
You write this as:
{{{W = L-5}}} but since we are looking for the width, W, we'll rewrite this as:
{{{L = W+5}}}
The perimeter, P, is given as 50, so {{{P = 50}}}
Now we'll substitute these values into the formula above and solve for W.
{{{P = 2(L+W)}}} Substitute {{{L = W+5}}}
{{{P = 2((W+5)+W)}}} Combine like-terms.
{{{P = 2(2W+5)}}} Apply the distributive property.
{{{P = 4W+10}}} Substitute P = 50.
{{{50 = 4W+10}}} Subtract 10 from both sides.
{{{40 = 4W}}} Finally, divide both sides by 4.
{{{10= W}}}
The width is 10 inches.
{{{L = W+5}}}
{{{L = 10+5}}}
{{{L = 15}}}
The length is 15 inches.

Check:
{{{P = 2(L+W)}}} Substitute L= 15 and W = 10
{{{P = 2(15+10)}}}
{{{P = 2(25)}}}
{{{P = 50}}} The solution is good!