D _ N _ _ L
_ S
7
Y _ _ R S
_ L D
_ N D
J _ S S _ C _
_ S
4
Y _ _ R S
_ L D .
L _ T
X
B _
T H _
F _ T H _ R ' S
_ G _ .
S _
D _ N _ _ L ' S
_ G _
_ S :
X / 5
I N
2 1
Y _ _ R S :
2 1
+
X
=
2 ( X / 5
+
2 1 )
S _ L V _ N G
F _ R
X
W _ L L
G _ V _
3 5 .
S _ N C _
D _ N _ _ L ' S
_ G _
_ S
X / 5 :
3 5 / 5
=
7
L _ T
Y
B _
J _ S S _ C _ ' S
_ G _ .
S _
M _ T H _ R ' S
_ G _
_ S :
7 Y
I N
8
Y _ _ R S :
7 Y
+
8
=
3 ( Y
+
8 )
S _ L V _ N G
F _ R
Y
W _ L L
G _ V _
4 Clue
18 YEARS OLD. LET MY AGE BE X YEARS NOW. MY AGE AFTER 3 YEARS IS (X+3). MY AGE 3 YEARS AGO IS (X-3). SO, NOW WE CAN COME UP WITH AN EQUATION FOR X: X=3(X+3)-3(X-3) SOLVING FOR X WILL GIVE 18 DANIEL IS 7 YEARS OLD AND JESSICA IS 4 YEARS OLD. LET X BE THE FATHER'S AGE. SO DANIEL'S AGE IS: X/5 IN 21 YEARS: 21 + X = 2(X/5 + 21) SOLVING FOR X WILL GIVE 35. SINCE DANIEL'S AGE IS X/5: 35/5 = 7 LET Y BE JESSICA'S AGE. SO MOTHER'S AGE IS: 7Y IN 8 YEARS: 7Y + 8 = 3(Y + 8) SOLVING FOR Y WILL GIVE 4 JENNIE IS 12 YEARS OLD. LET X BE JENNIE'S AGE. WE CAN COME UP WITH THE EQUATION: X + 2 = 2(X - 5) SOLVINGO FOR X GIVES 12. IN 2 YEARS TIME, JENNIE WILL BE 14 YEARS OLD. FIVE YEARS AGO, JENNIE WAS 7 YEARS OLD - HALF OF 14 I AM 19 YEARS OLD AND MY SISTER IS 9. LET X = MY SISTER'S AGE AND Y = MY AGE. Y = X + 10 Y + 1 = 2(X + 1) Y = 2X + 1 SUBSTITUTING THIS RESULT IN OUR FIRST EQUATION, WE HAVE: 2X + 1 = X + 10 X = 9 SO Y = 19 WHEN MY SISTER WAS 5, I WAS 3 TIMES OLDER THAN SHE WAS