Question 386758: Mona's age is 4 times Bailey's age. The sum of their ages is 15. Find the age of each.
Found 2 solutions by stanbon, gwendolyn:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Mona's age is 4 times Bailey's age.
The sum of their ages is 15.
Find the age of each.
---------------------
Equations:
m = 4b
m+b = 15
----
Substitute for "m" and solve for "b":
4b + b = 15
5b = 15
b = 3 (Bailey's age now)
---
Since m = 3b, m = 3*3 = 9 (Mona's age now)
===========================
Cheers,
Stan H.
Answer by gwendolyn(128)  (Show Source):
You can put this solution on YOUR website! First, we'll assign variables to both ages.
let M=Mona's age
let B=Bailey's age
Then, we'll come up with two equations to solve our two variables.
Mona's age is 4 times Bailey's age. This can be shown algebraically as below:
M=4B
We also know that Mona's age plus Bailey's age is 15. This, in an algebraic statement, can be written as below:
M+B=15
We can substitute the value of M in terms of B from the first equation into the second equation:
4B+B=15
5B=15
Then, we can divide both sides of the w equation by 5:
B=3
Therefore, Bailey is 3 years old. We also want to know Mona's age, so we can simply plug in Bailey's age to either equation and find it.
M+B=15
M+3=15
M=12
Therefore, Mona is 12 and Bailey is 3.
|
|
|
