word problem

[Guest]

Five thousand dollars less was invested in an account earning 9% simple interest than in an account earning 11% simple interest. If the total interest at the end of 1 year is $1950, how much was invested in each account? What is the total amount invested?

Am I starting this right?

Let x-5000.00 represent the 9% account
Let y represent the 11% account.

x-5000+y=1950.00
Thanks.

 

[krisolaw]

Good try, but try setting it up like this with one variable:

.11x +.09(x-5000)=1950

From here, multiply by 100 to get rid of the decimals unless you are comfortable using them. It should look something like this:

11x + 9(x-5000)=195000
11x+9x-45000=195000
20x=240000
x =12000

Since you know that x = 12000 at 11% then 5000 less is 7000 at 9%.
Hope that I explained it right!

 

[tkhunny]

Five thousand dollars less was invested in an account earning 9% simple interest than in an account earning 11% simple interest. If the total interest at the end of 1 year is $1950, how much was invested in each account?

You can do it with Two Variables

x = 9% Account
y = 11% Account

y - 5000 = x
x*(0.09) + y*(0.11) = 1950

You can set it up with One Variable

x = 9% Account
x+5000 = 11% Account

x*(0.09) + (x+5000)*(0.11) = 1950

If you are consistent, it matters not. That's the beauty of a unique answer. It cares not how you find it.

 

[Guest]

Thank you

I appreciate it ! I tried using just the one variable and I got the answer. Thanks to you all.