SOLUTION: Solve the following word problem. Money is invested at two rates of interest. One rate is 5% and the other is 4%. If there is $800 more invested at 5% than at 4%, find the amoun

Question 912112: Solve the following word problem.
Money is invested at two rates of interest. One rate is 5% and the other is 4%. If there is $800 more invested at 5% than at 4%, find the amount invested at each rate if the total annual interest received is $670. Let x = amount invested at 5% and y = amount invested at 4%. Then the system that models the problem is {x = y + 800 and 0.05x + 0.04y = 670}. Solve the system by using the method of addition.
Answer by richwmiller(17219) (Show Source): You can put this solution on YOUR website!
We don't know the total invested.
We know the difference in the two accounts.
We know that the interest for the two accounts is $670
0.05*x+0.04*y=670
We know that the account at 5% has $800 more.
x=800+y
We substitute for x
0.05*(800+y)+0.04*y=670
We multiply out
40+0.05y+0.04*y=670
We combine like terms.
0.09*y=630
Isolate y
y=$7000 at 4%
x=800+y
Calculate x
x=$7800 at 5%
Now,we know the total invested is: 14800
Total invested $7800+$7000=$14800
We check
0.05*7800+0.04*7000=670
390+280=670
670=670
Since this statement is TRUE and neither x nor y is negative all is well.
codeintmt

But you want it solved using elimination or addition
0.05*x+0.04*y=670
x=800+y
x-y=800
0.05*x+0.04*y=670
multiply by 100
5*x+4*y=67000
x-y=800
multiply by 6
4x-4y=3200
5*x+4*y=67000
add
9x=70200
x=7800 at 5%
y=7000 at 4%
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