Adam and Friday's each had some money. If Adam spent $9, the ratio of the
remaining amount of money Adam had to the amount of money that Firdaus
had was 2:7. If Firdaus spent $9, the ratio of the amount of money Adam had
to the remaining amount of money that Firdaus had would become 7:11. How
much money did Adam and Firdaus have altogether at first?
Call the amount that Adam had originally = A
Call the amount that Friday had originaly = F
We have these two equations
(A  9) / F = 2/7
A / ( F  9) = 7/11
Crossmultiply in each case
7 (A  9) = 2F
11A = 7 (F  9) simplify both
7A  63 = 2F (1)
11A = 7F  63 rearrange both as
7A  2F = 63
11A  7F =  63 multiply the first equation by 7 and the second by 2
49A + 14F = 441
22A  14F = 126 add these
27A = 567
A = 567 / 27 = 21 ( Adam originally had $21 )
And using (1) to find F
7(21)  63 = 2F
147  63 = 2F
84 = 2F
84/2 = F = 42 ( Friday originally had $42 )
They originally had $63 dollars combined
(a9) / f = 2/7 f = 7/2 (a9)
a / (f9) = 7/11 f = 11/7 a + 9 equate the two 'f's
7/2 ( a  9 ) = 11/7 a + 9
1 13/14 a = 40.5
a = 21 then you can easily solve for f = 7/2 ( a9) = 42 dollars
Adam = $x money
Firdaus = $y
(x  9)/y = 2/7 (1)
2y = 7(x  9) = (1)
2y = 7x  63 (1)
x/(y  9) = 7/11
11x = 7(y  9)
11x = 7y  63
2 (11x  7y = 63)
7 (2y  7x = 63)
14y  49x = 63 * 7
22x  14y = 63 * 2
                  
 27x = 567
 x = 21
 y = 42
 x + y = $63


