Question 205643
First we'll start with what formula to use
{{{A = LW}}}
1. Since the length of a rectangle is 5 inches more than twice the width we can write the following equation
{{{L=2W+5}}}
2. Now we have two equations in a system of equations
{{{LW=102}}} and 
{{{L=2W+5}}}
3. First substitute the 2nd equation into the 1st
{{{(2W+5)W=102}}}
- then simplify
{{{2W^2+5W=102}}}
{{{2W^2+5w-102 = 0}}}
- next factor the equation
{{{(2W+17)(W-6) = 0}}}
- then set each factor equal to 0 and solve for W
{{{2W+17 = 0}}} and {{{W-6 =0}}}
{{{2W=-17}}} and {{{w=6}}}
{{{W = -17/2}}} amd {{{w=6}}}
- Because we can't have a negative side length the only answer is 6
4. So W = 6 and now plug W into the 2nd equation
{{{L = 2W+5}}}
{{{L = 2(6) + 5}}}
{{{L = 12+5}}}
{{{L=17}}}
5. So L = 17
The final answer is W = 6, L = 17