Let
Length of rectangle be "x"
Width of rectangle be "y"
Given, length is 3 more than twice width, we can write:
[tex]x=2y+3[/tex]Also, given the perimeter is 102.
Recall, perimeter is sum of all the sides of a rectangle. Thus, we can write:
[tex]\begin{gathered} x+y+x+y=102 \\ 2x+2y=102 \end{gathered}[/tex]Substituting, equation 1 into equation 2, we get:
[tex]\begin{gathered} 2x+2y=102 \\ 2(2y+3)+2y=102 \end{gathered}[/tex]Multiplying out and solving for y:
[tex]\begin{gathered} 2(2y+3)+2y=102 \\ 4y+6+2y=102 \\ 6y=102-6 \\ 6y=96 \\ y=16 \end{gathered}[/tex]Now, x is:
[tex]\begin{gathered} x=2y+3 \\ x=2(16)+3 \\ x=32+3 \\ x=35 \end{gathered}[/tex]Answer:
The length is 35 and width is 16