1. ## System of Equations

I am familair with these type of problems but I am having a hard time setting it up. I would normally use mathematics but I cannot identify the variables. (ie x+y+z+?)

A student has money in three accounts that pay 5%, 7%, and 8%, in annual interest. She has three times as much invested at 8% as she does at 5%. If the total amount she has invested is $1600 and her interest for the year comes to$115, how much money does she have in each account?

2. Originally Posted by slatram
I am familair with these type of problems but I am having a hard time setting it up. I would normally use mathematics but I cannot identify the variables. (ie x+y+z+?)

A student has money in three accounts that pay 5%, 7%, and 8%, in annual interest. She has three times as much invested at 8% as she does at 5%. If the total amount she has invested is $1600 and her interest for the year comes to$115, how much money does she have in each account?

Your description of your trouble leads me to believe that you do not have a consistent and useful methodology to guide you. I'll suggest such a methodology.

Rule #1 Name Stuff.

What shall we name?

"how much money does she have in each account?"

F = Amount at 5%
S = Amount at 7%
E = Amount at 8%

This first step is VERY important. Do not get sloppy or careless. The more formal and deliberate you are right here, the easier the rest will go.

Now translate:

"She has three times as much invested at 8% as she does at 5%."

3F = E <== You must see the relationship between the words and this equation.

"If the total amount she has invested is $1600" F + S + E =$1600 <== Do you see this one? Right out of the words.

"her interest for the year comes to $115," F(0.05) + S(0.07) + E(0.08) =$115 <== Again, right out of the words.

Reminder: You CANNOT get equations out of words without CLEAR definitions. That first step was just so important. I cannot sufficiently stress it.

Now what?

3. ## Equations

Ok. I now have the three equations:

F+S+E=$1600 F(0.05)+S(0.07)+E(0.08)=$115
3F=E
now as a place holder for all variables for the last equation I do this:

3F-S+E(0)=0

I am going to use mathematica...

A={{1,1,1,1600},{0.05,0.07,0.08,115},{3,-1,0,0}}
MatrixForm[RowReduce[A]]

(1 0. 0. 216.667)
(0 1 0. 650.)
(0 0 1 733.33)

It suppose to be one big parentheses but I do not have the function.

Is this correct?

4. The formula: I=prt, p is the principal(money you invest), r-rate(8%,7% and 5%), t- time( in our case it's 1 year)
I1=p1 x 0.05 x 1; I2= p2 x 0.07 x 1; I3= 3p1 x 0.08 x 1.
According to the problem:
p1+p2+3p1=1600.
I1+I2+I3=115
Lets simplify it, 4p1+p2=1600.
0.05p1+0.07p2+0.24p1=115.
Now, solve for p2=1600-4p1 and plug into last equation, solve for p1

5. How did that zero (0) get attached to the "E"?

I do not see anything that suggests to me that you have any idea where those equations come from. Lot's of folks can solve the problem if they have the equations. Don't make me regret giving them to you!

Given

3*F = E

F + S + E = 1600

F*(0.05) + S*(0.07) + E*(0.08) = 115

Find(F,S,E) ==>

And out pops 300, 400, 900

Check it out.

Like setting up the equations, it appears you also need some more consistent and deliberate methods for solving such things. Work WAY more problems. You'll get it.