SOLUTION: im not very good at algebra. Joe's dad s 2 times joe's age. 7 years ago, he was 3 times older than joe. find both's ages.

Question 121550: im not very good at algebra. Joe's dad s 2 times joe's age. 7 years ago, he was 3 times older than joe. find both's ages.
Found 3 solutions by awang1996, Fombitz, Earlsdon:
Answer by awang1996(50)   (Show Source): You can put this solution on YOUR website!
Joe's dad's 2 times joe's age. 7 years ago, he was 3 times older than joe. find both's ages.

2j=d
d-7=3(j-7)
2j-7=3j-21
subtract 2j from both sides.
-7=j-21
add 21 to both sides.
j=14
d=28
Answer by Fombitz(32388)   (Show Source): You can put this solution on YOUR website!
Let's use D to stand for Joe's Dad's age.
Let's use J to stnad for Joe's age.
1.
Seven years ago, both Joe and his Dad were 7 years younger (subtract 7).
2.
Substitute 1 into 2 and solve for J.

Distributive property.
Subtract 3J from both sides.

Add 7 to both sides of the equation.

Multiply both sides by (-1)
Now you have Joe's age you can find his Dad's age using either 1 or 2.
1.

Joe is 14 years old, his dad is 28 years old.(D=2J)
Seven years ago, Joe was 7, his Dad 21. (D-7=3(J-7))
Answer by Earlsdon(6294)   (Show Source): You can put this solution on YOUR website!
Let D = Dad's present age and J = Joe's present age.
From the problem description you can write:
D = 2J "Dad is two times Joe's age". Substitute this value of D into the second equation and solve for J.
D-7 = 3(J-7) "Seven years ago (that's -7), Dad was three times Joe's age".
2J-7 = 3(J-7) Simplify.
2J-7 = 3J-21 Subtract 2J from both sides.
-7 = J-21 Add 21 to both sides.
14 = J
Joe's age is 14
D = 2J
D = 2(14)
D = 28
Dad's age is 28
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