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

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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) About Me  (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 x%2By=3000


"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 0.10x%2B0.08y=256


Multiply both sides by 100 to get 10x%2B8y=25600



So we have the system of equations:

system%28x%2By=3000%2C10x%2B8y=25600%29


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

x%2By=3000 Start with the first equation


y=3000-x Subtract x from both sides


y=-x%2B3000 Rearrange the equation



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

Since y=-x%2B3000, we can now replace each y in the second equation with -x%2B3000 to solve for x



10x%2B8highlight%28%28-x%2B3000%29%29=25600 Plug in y=-x%2B3000 into the second equation. In other words, replace each y with -x%2B3000. Notice we've eliminated the y variables. So we now have a simple equation with one unknown.



10x%2B%288%29%28-1%29x%2B%288%29%283000%29=25600 Distribute 8 to -x%2B3000


10x-8x%2B24000=25600 Multiply


2x%2B24000=25600 Combine like terms on the left side


2x=25600-24000Subtract 24000 from both sides


2x=1600 Combine like terms on the right side


x=%281600%29%2F%282%29 Divide both sides by 2 to isolate x



x=800 Divide





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


So the first part of our answer is: x=800









Since we know that x=800 we can plug it into the equation y=-x%2B3000 (remember we previously solved for y in the first equation).



y=-x%2B3000 Start with the equation where y was previously isolated.


y=-%28800%29%2B3000 Plug in x=800


y=-800%2B3000 Multiply


y=2200 Combine like terms



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


So the second part of our answer is: y=2200









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

So our answers are:

x=800 and y=2200


So this means that Pat invested $800 at 10% and $2,200 at 8%

Answer by Mathtut(3670) About Me  (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
:
$highlight%28y=2200%29amount invested at 8%
:
$highlight%28x=3000-2200=800%29amount invested at 10%