Question 949678
PROBLEM 1:
W=width; L=length=W+40cm; P=perimeter=2(L+W)=400cm
400cm=2(L+W) Divide each side by 2.
200cm=L+W Substitute for L
200cm=(W+40cm)+W Subtract 40cm from each side.
160cm=2W Divide each side by 2.
80cm=W  ANSWER 1: The width is 80 cm.
L=w+40cm=80cm+40cm=120cm ANSWER 2: The length is 120 cm. 
CHECK:
P=2(L+W)
400cm=2(120cm + 80cm)
400cm=2(200 cm)
400cm=400cm

PROBLEM 2:
L=Lacy's age; J=Jack's age=2L
If in 3 years the sum of their age will be 63, currently the sum is 57 (63-3years aging for each) 
J+L=57 years Substitute for J
2L+L=57 years
3L=57 years Divide each side by 3.
L=19 years ANSWER 1: Lacy is 19 years old.
J=2L=2(19 yrs)=38yrs ANSWER 2: Jack is 38 years old.
CHECK:
(J+3 years)+(L+3 years)=63 years
38 yrs+3 yrs+19 yrs+3 yrs=63 yrs
63 yrs=63 yrs

PROBLEM 3:
The pool is 85% of original price (100%-15%) so
$1000/0.85=$1176.47 ANSWER: The original price was $1176.47.

PROBLEM 4:
S=smaller number; S+2=larger number
{{{(S)(S+2)=S^2+14}}}
{{{S^2+2S=S^2+14}}} Subtract S^2 from each side.
2S=14 Divide each side by 2.
S=7 ANSWER 1: The smaller number is 7.
S+2=9 ANSWER 2: The larger number is 9.
CHECK
Product of the two is 14 more than square of the smaller
(7)(9)=(7)(7)+14
63=49+14
63=63