.08x+.09y = 39,000

X+Y = 1,000,000

X=?

Y=?

.08x+.09y = 39,000

X+Y = 1,000,000

X=?

Y=?

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Hello, Derek ---

Your second equation is correct, but the first one doesn't match the numbers in the title to your question. The numbers you have -- .08x and .09y -- cannot be correct because either perentage by itself would yield more than 39,000 (the returns would be 80,000 and 90,000).

But if we use ,035 and .045, we can get an answer:

.035x + .045y = 39000

x + y = 1000000

Now multiply the second one all the way through by -.035 so that we can add the two equations and elminate the x variable temporarily:

.035x + .045y = 39000

-.035x - .035y = -35000

___________________________

0 + .01y = 4000

Now multiply through by 100:

y = 400,000

Now substitute back into the first equation to find x:

x + 400,000 = 1,000,000

x = 600,000

**600,000 was invested at 3.5%, returning 21,000. The rest, 400,000 was invested at 4.5%, yielding 18,000. **

You made 2 investments, x and y, totaling $1,000,000. You invested part of it, let's say x, at 3.5% and part of it, let's say y, at 4.5%, from which your interest return was $39,000.

Note that we have to convert the percentages, 3.5% and 4.5%, into decimals to get rid of the percent sign.

3.5% of 100% is: 3.5% / 100% = 0.035

4.5% of 100% is: 4.5% / 100% = 0.045

The problem yields 2 equations:

(1.) x + y = 1,000,000

(2.) 0.035x + 0.045y = 39,000

This is a system of linear equations, which we can solve for using a couple of different methods. Let's use the method of elimination. When using this method, our goal is to eliminate one of the unknown variables, x or y, and solve for the remaining one. Then we use the variable we solved for to find the other unknown variable.

For simplicity, let's eliminate x and solve for y first. To do so, multiply equation (1.) by -0.035 and keep equation (2.) as is, then add them to one another:

-0.035 * ( x + y = 1,000,000 ) ==> -0.035(x) - 0.035(y) = -35,000

-0.035x - 0.035 y = -35,000

+ 0.035x + 0.045y = 39,000

______________________________

-0.035x + 0.035x - 0.035y + 0.045y = -35,000 + 39,000

0x + 0.010y = 4,000

0.010y = 4,000

(0.010y) / 0.010 = (4,000) / 0.010

y = 400,000

Given y, use the original equation (1.) to solve for x:

x + y = 1,000,000 ==> x + 400,000 = 1,000,000

- 400,000 -400,000

__________________________

x = 600,000

You can check your answers for x and y by plugging them back into the original equation (2.):

0.035x + 0.045y = 0.035(600,000) + 0.045(400,000) = 21,000 + 18,000 = 39,000

Thus, since x was invested at 3.5% and y was invested at 4.5%, then we can say that $600,000 was invested at 3.5% and $400,000 was invested at 4.5%.

While the ways my colleagues have explained the question are the correct way to answer this question algebraically, it is often difficult to remember all the steps. I prefer to teach the guess and check as a back up method as it is much more relevant to real life situations.

Example: Using your question we know that $1 million dollars is invested at 3.5% and 4.5% and that you should return (make) $39,000 in interest. Since there are two options for investment divide $1 million by 2 which gives you $500,000 and pretend you invested that amount in each fund.

$500,000 at 3.5%= $17,500

$500,000 at 4.5%= $22,500

Add your answers: 17,500+22,500=40,000

Obviously this answer is just a bit too large so move some money out of the 4.5% into the 3.5% return. Since this is such a large amount of money to work with, I would start with moving $50,000 which would take $50,000 from the 4.5% fund.

So now we have:

550,000 at 3.5%=19,250

450,000 at 4.5%=20,250

Add the answers together and you get 39,500. If you notice, moving 50,000 drops your return $500. At $39,500 you are $500 over the $39,000 so move $50,000 more into the 3.5% fund and your answer should be perfect.

Lets try it: 600,000 at 3.5%= 19,000 400,000 at 4.5%= 20,000 Add your answers and you have $39,000.

While this doesn't teach you how to find the answer using fancy equations and the elimination method, substitution method, or any of the other methods that are typically taught in Algebra 1 or 2, it is a practical application of math to a word problem. Word problems are hard enough without confusing you with difficult (often unnecessary) math skills.

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