Question 483452
A rectangular table can be be set against one wall of a room, as shown, so that it is 5 feet from the two side walls and 8 feet from the fourth wall.
If the perimeter of the room is 58 feet
:
let x = length of the table against the wall
let y = the width of the table
the room perimeter equation
2(x+10) + 2(y+8) = 58
2x + 20 + 2y + 16 = 58
2x + 2y + 36 = 58
2x + 2y = 58 - 36
2x + 2y = 22
simplify, divide by 2
x + y = 11
:
"and the combined length of the three exposed sides of the table is 17.5 feet,"
x + 2y = 17.5
:
Use elimination with these two equations
x + 2y = 17.5
x + 2 = 11
---------------subtraction eliminates x, find y
y = 6.5 ft is the width of the table
:
Find x: 
x + y = 11
x + 6.5 = 11
x = 11 - 6.5
x = 4.5 ft is the length of the table against the wall
:
then what are the dimensions of the table?  6.5' by 4.5'
;
:
Check solution in the room perimeter equation
2(4.5+10) + 2(6.5+8) =