# Thread: Math - Algebra Word Problem

1. ## Math - Algebra Word Problem

Solved

2. Originally Posted by DivideBy0
I assume you want the sons to have equal amounts of money, then the solution is easier than you might think.

Let x = the value of the father's will, then

The first son receives $\displaystyle 1+\frac{1}{7}(x-1)=\frac{x+6}{7}$ dollars.

The second son receives $\displaystyle 2+\frac{1}{7}\left(x-\left(\frac{x+6}{7}\right)-2\right)=\frac{6x+78}{49}$ dollars.

As these are equal, equate and solve for x:

$\displaystyle \frac{x+6}{7}=\frac{6x+78}{49}$

$\displaystyle x=36$

Now it's easy to find the number of sons. Since now we know the first son receives 6 dollars, and all the sons receive the same amount, the man has $\displaystyle \frac{36}{6}=6$ sons.

I can't believe how long I spend looking at functions of functions and trying to turn everything into massive overcomplicated expressions, all the while overlooking this easy solution.
I'm trying to do this problem and for the second son, do you distribute the negative to $\displaystyle \frac {(x + 6)}{7}$? Because when I do this, I don't get any "x" values, like 6x

3. Originally Posted by Macleef
I'm trying to do this problem and for the second son, do you distribute the negative to $\displaystyle \frac {(x + 6)}{7}$? Because when I do this, I don't get any "x" values, like 6x
Yes, $\displaystyle -\frac{x+6}{7} = \frac{-x-6}{7}$

4. I seen this problem a while back.

Let A=amount each child gets

Let T=total amount of money given among the kids.

The first one gets $\displaystyle A=1000+\frac{1}{10}(T-1000)=\frac{1}{10}T+900$

The second one gets:

$\displaystyle A=2000+\frac{1}{10}(T-2000-\underbrace{(1000+\frac{1}{10}(T-1000))}_{\text{amount given to first kid}})=\frac{9}{100}T+1710$

Now, we have two equations to solve:

$\displaystyle A=\frac{1}{10}T+900$

$\displaystyle A=\frac{9}{100}T+1710$

Using the substitution method or whatever method you prefer.

We get T=81,000 and A=9000

There are 9 kids getting $9,000 each fom a total of$81,000

5. nvrm