# Math Help - Another Word Problem

1. ## Another Word Problem

Hi again once again I need help setting up the equation for this, it is similar in spirit to my last question but still I can't figure out how to set up the equation , please help.

Bob has made $1980 annual interest from his two bank accounts. One of his bank accounts yields 6% interest and the other yields 9% interest , how much money would he have to have in each account? 2. I tried to work this out all I can think of is $0.06x + 0.09y = 1980$ but than that just seems useless, cause there’s nothing you can do any further. Anyone else? 3. Unless there's something subtle that we've missed, this question can't be solved. Unless it means he made$1980 from each of his accounts, in which case the equations are:

0.06x = 1980
0.09y = 1980

A solution to either of these would be a valid solution for the question where $1980 is the total, you'd assume that either x or y would be zero. Or the amount could be anywhere in between. 4. Originally Posted by Matt Westwood Unless there's something subtle that we've missed, this question can't be solved. Unless it means he made$1980 from each of his accounts, in which case the equations are:

0.06x = 1980
0.09y = 1980

A solution to either of these would be a valid solution for the question where $1980 is the total, you'd assume that either x or y would be zero. Or the amount could be anywhere in between. yeah I was thinking the same thing, maybe the op mis read the question 5. Originally Posted by Matt Westwood Unless there's something subtle that we've missed, this question can't be solved. Unless it means he made$1980 from each of his accounts, in which case the equations are:

0.06x = 1980
0.09y = 1980

A solution to either of these would be a valid solution for the question where $1980 is the total, you'd assume that either x or y would be zero. Or the amount could be anywhere in between. Yes I did not include all of the info as I don't have my book with me (at an internet cafe). I'm pretty sure that this was the original question , if Im wrong I apologize in advance for wasting your time and thank u for your help--would this be solvable?? [Paraphrase]Bob invested$25,000 into two bank accounts which gave him a combined annual interest of $1980. One account gave him a 6% annual return and the other gave him a 9%. How much would he have to have put in each of his accounts in order to get this amount of interest? 6. $C_1*1.06 + (25000-C_1)*1.09=1980+25000$ $C_1*1.06 + 25000*1.09-C_1*1.09=26980$ $C_1*0.03=25000*1.09-26980$ $C_1*0.03=270$ $C_1=9000$ $C_2=16000$ 7. Originally Posted by courteous $C_1*1.06 + (25000-C_1)*1.09=1980+25000$ $C_1*1.06 + 25000*1.09-C_1*1.09=26980$ $C_1*0.03=25000*1.09-26980$ $C_1*0.03=270$ $C_1=9000$ $C_2=16000$ Here's another look at the "setup" Let x = amount invested at 6% Let 25000 - x = amount invested at 9% .06x + .09(25000 - x) = 1980 6x + 225000 - 9x = 198000 -3x = -27000 x =$9,000 (amount invested at 6%0

$25,000 -$9,000 = \$16,000 (amount invested at 9%)