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?
+129852 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
Cross-multiply 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
+37146 Yo, Taco....starts out as 'Friday' then morphs to 'Firdaus' .....
+37146 (a-9) / f = 2/7 f = 7/2 (a-9)
a / (f-9) = 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 ( a-9) = 42 dollars
+237 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
|
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