SOLUTION: This is from my 8th grade honors algebra... Tori invested $24,000 , part at 8% and the rest at 7.2%. How much did she invest at each rate if her income from the 8% investment

Question 21050: This is from my 8th grade honors algebra...
Tori invested $24,000 , part at 8% and the rest at 7.2%. How much did she invest at each rate if her income from the 8% investment is two-thirds of that of the 7.2% investment?
I've made a chart...
Amount invested x rate = interest income

X x 0.08 = 0.08X = y
($24,000-X) x 0.072 = 0.072($24,000-X) = 2/3y
Then I'm totally stuck, we haven't solved with 2 unknowns yet!
Answer by Earlsdon(6294) (Show Source): You can put this solution on YOUR website!
Try this:
Let x = the amount invested at 8% (0.08).
The interest on this amount is written as: 0.08x
Let (24000 - x) = the amount invested at 7.2% (0.072).
The interest on this amount is written as: 0.072(24000 - x)
The interest on the 8% part is 2/3 the interest on the 7.2% part.
So, you can say:
Now you can write the equation to solve this problem:
Simplify.

Add 0.12x to both sides.
Subtract 1728 from both sides.
Finally, divide both sides by 0.128

So, $9,000.00 was invested at 8% and ($24,000.00 - $9,000.00 = $15,000.00) was invested at 7.2%
Check:
0.08($9,000.00) = $720.00 The interest on the amount invested at 8%
0.072($15,000.00) = $1,080.00 The interest on the amount invested at 7.2%.
Checks ok.

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