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A total of $9000 is invested: part at 6% and the remainder at 10%. How much is invested at each rate if the annual interest is $790?

User Jmichalicek
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2 Answers

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Answer: So we can say: x + y = 9000. We also know that 3% of one of the investments plus 6% of the other is equal to $396. So we can say .03x + .06y = 396.

User Niczak
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We should set up a system of linear equations to solve this. We will use variables, x and y, for the amount invested. We know the total amount invested is $9000. So we can say:

x + y = 9000

We also know that 3% of one of the investments plus 6% of the other is equal to $396. So we can say:

.03x + .06y = 396

This gives us a system of 2 linear equations with 2 variables. The hard part is now done.

Now we need to isolate one of variables in one of the equations and then do back substitution.

Lets isolate x in the first equation. This will give us:

x = 9000–y

We can now substitute 9000-y in the second equation where x is. This gives us:

.03(9000-y) + .06y = 396

Simplifying this equation we get:

270–.03y +.06y = 396

Now to get y by itself lets first subtract 270 from both sides to get:

—.03y + .06y = 126

Now we can add —.03y and .06y to get:

.03y = 126

Now divide both sides by .03 and we will have our y value.

y = 4200

Now that we have y lets go back to the first original equation and plug 4200 in for y in order to get our x value.

x + 4200 = 9000

Subtract both sides by 4200 to get:

x = 4800

So now we have it.

The amout invested that had a return of 3% is $4800 and the amount invested that had a return of 6% is $4200.

User Lupita
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