Juan Said:

any tips in solving algebra age problems?

We Answered:

Yes. First, examine the problem and find the exact question being asked.

"What is Ann's age now?"

"What will Achbar's age be three years from now?"

Let that value be x.

Then, express the other parts of the problem as equations using x:

"Fred is three times as old as Ann is now." Fred's age is thus 3x.

"Two years from now, Fred will be twice as old as Ann is then."

How old will Fred be two years from now? 3x + 2
How old will Ann be two years from now? x + 2

"Two years from now, Fred will be twice as old as Ann is then."

3x + 2 = 2(x + 2)

Solve that equation and you have Ann's age now.

Valerie Said:

Age Algebra problems help!!?

We Answered:

Substitutions, make the equations with the data you have.

1/2x = y
1/3(x-12) = y-12

Isolate a variable, then plug in that infomation into the second exquation, in this case, plug 1/2x everywhere you have a y in the second equation.

1/3(x-12) = 1/2x-12

Solve for x

1/3x -4 = 1/2x - 12
1/3x = 1/2x -8
8 = 1/6x
48 = x

Now plug this into your first equation in place of x

1/2(48) = y
24 = y

Father is 48, son is 24

Do this for the second problem as well, here are the equations to work with using your data.

x+y = 50
x+8 = 2(y+8)

Teresa Said:

algebra 1- age problems. How do i do this?

We Answered:

Let

B = Barney's age
F = Fred's age

B = F + 8 (Barney is 8 years older than Fred)

F = B/2 or B = 2F (Fred is half as old as Barney)

Substituting B = 2F in the first equation,

2F = F + 8

F = 8 and solving for B,

B = 8 + 8 = 16

ANSWER:

Barney is 16 years old and Fred is 8 years old.

Jamie Said:

How to solve Algebra age problems?

We Answered:

The man's age during the 6 years was
n, n+1, n+2, n+3, n+4, n+5
and the girls age was
m, m+1, m+2, m+3, m+4, m+5
with each n being a multiple of the corresponding m.

We want to avoid prime numbers:
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101,
And grandpa is probably at least 40, so we trim the list to
41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101
Gaps of at least 5 are
61 to 67, 73 to 79, 83 to 89, 89 to 97.

None of these have more than 5 consecutive composite numbers,
so the girl's initial age is 1, and we need to find consecutive multiples of 2,3, 4, 5, 6

61, 62, 63, 64, 65, 66 works as those are multiples of
1, 2, 3, 4, 5, 6 respectively.

Let A = Al's age now,
let N = Ann's age now
A = 3/4 * N (now)
(A - 6) = (1/2) (N - 6) (six years ago)

Solve by substitution:
3N/4 - 6 = N/2 - 3
1/4 N = 3
N = 12
A = 9
Six years ago they were 6 and 3.

Olga Said:

what site contains many algebra (high school) problems.. age motion, digit, work, etc with solution. thanks?

We Answered:

www.coolmath.com
www.purplemath.com
www.algebra.com

Just to name a few

Google "Algebra" and you should get a ton of results

Jessica Said:

Help with these algebra age problems! PLEASE!?

We Answered:

1) Shawn is 4/5ths as old now as he will be in 7 years.

If x is his age now, in 7 years he will be x + 7

Now he is 4/5 of (x + 7)

x = 4(x + 7)/5
5x = 4x + 28
x = 28

check
28 = 4(28 + 7)/5 = 4(35)/5 = 4(7) = 28


2) Jeff is two years younger than Carrie.

a) j = c - 2

12 years ago, Carrie was twice as old as Jeff.

b) c - 12 = 2(j - 12)

Solve the system


3) Steve is twice as old as Sylvie.

Let v = steve, y = sylvie

a) v = 2y

Sylvie is three years older than Jacob.

b) y = j + 3
j = y - 3

4 years ago, Sylvie was twice Jacob’s age.

c) y - 4 = 2(j - 4)
y - 4 = 2j - 8
y - 4 = 2(y - 3) - 8 {substitute for j from equation b}
y - 4 = 2y - 6 - 8
y - 4 = 2y - 14
y = 2y - 10
-y = -10
y = 10 {sylvie's age}


Note that "Steve is twice as old as Sylvie" is redundant information - you have to watch for information not needed to answer the question and just ignore it