Question 534076
Call S = sides of square 1 (square S).  We'll make this one the square with the larger perimeter.
Call T = sides of square 2 (square T)

The perimeter of square 1 is S + S + S + S = 4S.  The area of square 1 is S^2

The perimeter of square 2 is T + T + T + T = 4T.  The area of square 2 is T^2

We called S the square with the larger perimeter, and we know the difference is 24, so:

4S = 4T + 24

We know the sum of their areas is 468.  

S^2 + T^2 = 468

Two equations, two unknowns.  So let's solve the first equation for S in terms of T.

4S = 4T + 24  divide both sides by 4 --->  S = 4T/4 + 24 = T+6

Substitute that for S in the area equation, so we can solve it for T.

S^2 + T^2 = 468  (substitute T+6 for S)

(T+6)^2 + T^2 = 468  Use FOIL to square T+6

T^2 + 12T + 36 +T^2 = 468  Combine the T^2 terms, subtract 468 from each side.

2T^2 + 12T - 432 = 0

Factor 2 from each term.

T^2 + 6T - 216 = 0  Factor using reverse FOIL.

(T+18)(T-12)  solve each for T

T+18=0  (subtract 18 from each side)  T=-18  This solution is valid, but not in the real world because the side of a square can't be a negative number.

T-12=0  (add 12 to both sides)  T = 12

Sides of T = 12.  It's perimeter is 4*12 = 48m

Remember earlier we found that S = T+6 = 12 + 6 = 18.  Thus the perimeter of S is 4*18 = 72m

Now check if these are the correct answers.

The perimeter of S should be 24m more than the perimeter of T.

72-48 = 24  CHECK

The combined areas should be 468 m^2.

12^2 + 18^2 = 144+324 = 468  CHECK

HURRAH...we found the answer.