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How old was each when they got married?

2007-09-14 04:33:34 · 7 answers · asked by nutellafella 1 in Science & Mathematics Mathematics

7 answers

Let m = Mother's age, f = Father's age
m = f-8 =======> mother is 8 years younger
m = (5/7)f ======> mother's age is 5/7 of father's

Substitute the value for m in the second eqn into the first
f-8 = 5f/7

Multiply both sides by 7
7f - 56 = 5f

Add 56 to each side
7f = 5f + 56

Subtract 5f from each side
2f = 56

Divide each side by 2 to find the father's age
f = 28

Put this value back in the first equation to find mom's age
m = f-8 = 20

When they got married, Patricia's mother was 20 and her father was 28.

:)

2007-09-14 04:37:48 · answer #1 · answered by MamaMia © 7 · 0 0

Let's make x Patricia's mother. That means Patricia's father would be x+8. So the equation would be:

x/x+8 = 5/7

If you cross multiply:

7x = 5(x+8), or 7x = 5x + 40

Subtract 5x from both sides and you get:

2x = 40

x = 20 (Patricia's mother's age)

and x+8 = 28 (Patricia's father's age)

To check:

Does 20/28 = 5/7? Divide the numerator and the denominator of 20/28 by 4, and you get 5/7.

2007-09-14 11:40:39 · answer #2 · answered by Anonymous · 0 0

Age of mother = x
Age of father = y

Patricia's mother is eight years younger than her father
x = y - 8

When they were married, she was 5/7 as old as he
x = (5/7)y

(5/7) y = y - 8
8 = y - (5/7)y
8 = (2/7)y
y = 8(7/2)
y = 28

x = (5/7)(28)
x = 20

Age of mother = 20
Age of father = 28

2007-09-14 11:39:22 · answer #3 · answered by coolesteugene 2 · 0 0

Let x and y be the ages of Patricia's parents (mother and father resp).

x = y - 8

x = 5y / 7

7x = 5y

7y - 56 = 5y

2y = 56

y = 28

x = y - 8 = 20

They were 20 and 28 respectively when they got married.

2007-09-14 11:39:58 · answer #4 · answered by Swamy 7 · 0 0

M=F-8
M=(5 divided by 7)multiply F
F-8=(5 divided by 7)multiply F
F-(5 divided by 7)multiply F=8
(2 divided by 7)multiply F=8
F=8 multiply 7 divide by 2
=28
M=F-8
=28-8=20

2007-09-14 11:42:40 · answer #5 · answered by Ricky 2 · 0 0

mom was 20 and dad was 28

mom = x, dad = x + 8
5/7(x+ 8) = x, and solve
after I distributed and got rid of the fraction I got
5x + 40 = 7x, so x = 20

2007-09-14 11:45:44 · answer #6 · answered by sfroggy5 6 · 0 0

x = mother's age
y = father's age

y = x + 8

5/7y = x
y = 7/5x

7/5x = x + 8
2/5x = 8
x = 20
The mother was 20 years old.

y = x +8 = 20 + 8 = 28
The father was 28 years old.

2007-09-14 11:41:17 · answer #7 · answered by Chameleon 2 · 0 0

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