SOLUTION: Please help me solve this equation: Pat invested a total of $3,000. Part of the money yields 10 percent interest per year, and the rest

Question 175658: Please help me solve this equation: Pat invested a total of $3,000. Part of the money yields 10 percent interest
per year, and the rest yields 8 percent interest per year. If the total yearly
interest from this investment is $256, how much did Pat invest at 10 percent
and how much at 8 percent?
Found 2 solutions by jim_thompson5910, Mathtut:
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
Let x=amount invested at 10% and y=amount invested at 8%

"Pat invested a total of $3,000" translates to

"Part of the money yields 10 percent interest
per year, and the rest yields 8 percent interest per year. If the total yearly
interest from this investment is $256" translates to

Multiply both sides by 100 to get

So we have the system of equations:

Let's solve by substitution

Now in order to solve this system by using substitution, we need to solve (or isolate) one variable. I'm going to solve for y.

So let's isolate y in the first equation

Start with the first equation

Subtract from both sides

Rearrange the equation

---------------------

Since , we can now replace each in the second equation with to solve for

Plug in into the second equation. In other words, replace each with . Notice we've eliminated the variables. So we now have a simple equation with one unknown.

Distribute to

Multiply

Combine like terms on the left side

Subtract 24000 from both sides

Combine like terms on the right side

Divide both sides by 2 to isolate x

Divide

-----------------First Answer------------------------------

So the first part of our answer is:

Since we know that we can plug it into the equation (remember we previously solved for in the first equation).

Start with the equation where was previously isolated.

Plug in

Multiply

Combine like terms

-----------------Second Answer------------------------------

So the second part of our answer is:

-----------------Summary------------------------------

So our answers are:

and

So this means that Pat invested $800 at 10% and $2,200 at 8%
Answer by Mathtut(3670) (Show Source): You can put this solution on YOUR website!
let x and y be the amounts invested at 10% and 8%
:
x+y=3000........eq 1
.1x+.08y=256....eq 2
:
rewrite eq 1 to x=3000-y and plug that value into eq2
:
.1(3000-y)+.08y=256
:
300-.1y+.08y=256
:
-.02y=-44
:
$amount invested at 8%
:
$amount invested at 10%
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