Question 134042
These are all really the same sort of problems, so I'll do one of them.


Width of a rectangle: W
Three times the width: 3W
Length of a rectangle: L
Twice the length: 2L
Three inches more than twice the length: 2L + 3
So:  3W = 2L + 3


The perimeter of a rectangle:  P = 2L + 2W, so the perimeter plus 12 would be 2L + 2W + 12, and that is equal to four times the length, or 4L, so:


4L = 2L + 2W + 12


Take the first equation and solve for W:  {{{W=(2L+3)/3}}}


Substitute this expression for W into the second equation:
{{{4L = 2L + 2((2L+3)/3) + 12}}}


Multiply both sides by 3:
{{{12L=6L+2(2L+3)+36}}}


Distribute:
{{{12L=6L+4L+6+36}}}


Collect terms:
{{{12L-6L-4L=42}}}
{{{2L=42}}}


Divide:
{{{L=21}}}


Since {{{W=(2L+3)/3}}}


{{{W=(2(21)+3)/3=45/3=15}}}


For the other two problems, define your variables, set up the given relationships, solve one of them for one variable in terms of the other, substitute, and finally solve for one of the variables.


Hint for the second problem: Let x be the smaller integer, and then the larger integer is {{{5x/2}}}.


Hint for the third problem:  If x and y are the tens and ones digits of a number, the number is 10x + y and the sum of the digits is x + y.