# Thread: Solving Applied Problems (word problem)

1. ## Solving Applied Problems (word problem)

Hi, If you could show how you solve this, that would be great. Thanks.

Mr. Silvester is 5yr older than his wife. Five years
ago his age was 4/3 her age. What are their ages now?

2. ## Re: Solving Applied Problems (word problem)

Let s be the age of Mr. Silvester and w be the age of his wife

s-5= w
s-5 = (4/3)(w-5)

Hence
w = (4/3)w - 20/3
-(1/3)w = -20/3

w = 20, s=25

3. ## Re: Solving Applied Problems (word problem)

I thought I was over this, but apparently not. I have completed all the rest of the hw without a bother, but when it comes to these word problems, I don't know how to form the problem. Here's another one: Chris is 10yr older than Josh. Next year, Chris will be twice as old as Josh. What are their ages?

So then...
Chris = (J+10) -1
Josh = (C/2) -1

How do I write the equation?

4. ## Re: Solving Applied Problems (word problem)

Hello, Sprinkledozer!

Chris is 10 years older than Josh.
Next year, Chris will be twice as old as Josh.
What are their present ages?

We can solve this with one variable, one equation.

Let $J$ = Josh's age (now).
Then Chris' age is $J + 10$

Next year, they both will be one year older.
Chris will be $(J +10) + 1 \,=\,J+11$
Josh will be: $J + 1$

$\text{At that time: }\:\underbrace{\text{Chris}}_{J+11}\;\underbrace{ \text{will be}}_{=}\;\underbrace{\text{twice as old as Josh}}_{2(J+1)}$

Solve the equation: . $J + 11 \:=\:2(J+1)$
. . . . . and we get: . $J = 9$

Then Chris $= J+10 = 19$