Question

you invest $6000 in two accounts paying 7% and 8% annual interest, respectively. If the total interest earned for the year was $440, how much was invested at each rate?

Answer 1

please help!

Answer 2

Let x = the amount invested at 7%
Then 6000-x was invested at 8%
The interest on the x dollars is .07x
the interest on the 6000-x dollars is .08(6000-x)
The total interest is 440 so we have:
\[.07x+.08(6000-x)=440\]

Answer 3

We should probably multiply by 100 to get rid of all the decimals.

Answer 4

\[7x+8(6000-x)=44000\]

Answer 5

Solve that equation.

Answer 6

7x+48000-8x=44000

Answer 7

?

Answer 8

Combine the like terms.

Answer 9

-1x+48000=44000

Answer 10

subtract 48000 from both sides.

Answer 11

i got 4000 as my answer but dont knw what that is for

Answer 12

What did we say x stands for in the beginning when we defined the variables/

Answer 13

ohhh duh

Answer 14

thank you:D

Answer 15

i have more problems can u help?

Answer 16

yw

Answer 17

Posssibly if they are not too hard.

Answer 18

all math is hard for me! haha

Answer 19

the sum of two numbers is 13. If one number is subtracted from the other the result is 3 find the nmbers.

Answer 20

x+y=13
x-y=3

Answer 21

Can you solve that system?

Answer 22

they cancel dont they?

Answer 23

Yes. If you add the 2 equations, the y's cancel and you have 2x=16

Answer 24

8