Interest problem ?????????????? 10pts within the hour???/

Walt mad an extra $9000 last year from a part-time job. He invested part of the money 8% and the rest at 10%. He made a total of $780 In interest. How much was invested at 10%.

If somebody could please explain how to do this that would be great. I will give 10 pts to the best answer when I can award them.

Thanks in advance.

Let the amount invested at 10% be x. Then the amount invested at 8% is (9000 - x).

.08(9000 - x) + .10x = 780

Distribute the .08:

720 - .08x + .10x = 780

Subtract 720 from each side:

.10x - .08x = 780 - 720

Simplify both sides:

.02x = 60

Divide both sides by .02:

x = 60/.02 = 3000

Walt invested $3,000 at 10%.

let he be investing x amount in 8% and y amount in 10%.
Obviously,
x+y=9000
y=9000-x

from the second data(8% of x=0.08x),
(0. 08x)+(0.1y)=780
On sub y=9000-x,
(0. 08x)+(0.1(9000-x))=780
(0. 08x)+(900-0.1x)=780
900-0.02x=780
120=0.02x
x=120/0.02

x=6000
and,
y=9000-x
y=3000
The answer is,
He invested 6000 in 8% interest and 3000 in 10% interest.

Hope this was helpful..

a - part of money invested with 8%
b part of money invested with 10%

a+b=$9000
ax8/100+bx10/100=780

a+b=9000
0.08a+0.1b=780

a+b=9000
8a+10b=78000

a=9000-b
8(9000-b)+10b=78000

a=9000-b
72000-8b+10b=78000

a=9000-b
2b=6000 => b=$3000
a=$6000

proof

$6000x8/100+$3000x10/100=480+300=780

$3000 @ 10%=$300 $6000@8%=$480 $300 +480=$780