SOLUTION: I just need help setting up the equations and I can do the rest. Word problems have never been easy for me! Substitution method or addition method from system of equations Ge

Question 113234This question is from textbook
: I just need help setting up the equations and I can do the rest. Word problems have never been easy for me!
Substitution method or addition method from system of equations
George has two investments that yield a total of $185.60 in annual interest. The amount invested at 8% is $320 less than twice the amount invested at 6%. How much is invested at each rate?
Amount invested at 8%?
Amount invested at 6%? This question is from textbook

Found 2 solutions by bucky, SHUgrad05:
Answer by bucky(2189) (Show Source): You can put this solution on YOUR website!
Let's pick the problem apart as a practice for setting up the equations.
.
"two investments" indicates two unknowns.
.
"8% and 6%" differentiate between the two investments. Just to avoid confusion, let's call
the unknown amount of money invested at 8% "E" [standing for eight] and the unknown amount
of money invested at 6% "S" [standing for six]. So we have as our two unknowns, E and S.
.
If you invest money at a certain annual rate of interest, how do you determine the amount of
money that you earn annually from that interest? Pretty straightforward. You just multiply
the amount of money invested by percent (in decimal form) of interest. As an easy example,
if you invest $100 at 5% the amount of interest you are due at the end of a year is
$100 times 0.05 which is $5.00. Somebody uses your $100 for a year, and at the end of
that year your account is now $105.00.
.
Since you have E dollars invested at 8%, you should expect to be paid 8% of E in interest at
the end of the year or 0.08*E. Similarly, you have S dollars invested at 6% and for that
investment you should get 0.06*S at the end of a year. Your total interest at the end
of the year will be the sum of these two amounts ... 0.08*E plus 0.06*S. Notice that the
problem tells you that this total is to equal $185.60. Set the sum of the amounts equal
to 185.60 and you have your first equation:
.
0.08*E + 0.06*S = 185.60
.
The problem gives you additional information that has not been used yet. It says that the
amount invested at 8% [the amount we called E] is $320 less than twice the amount invested
at 6% which we called S. So we can shorten this statement to E is $320 less than twice S.
Twice S is 2*S so we can now say E is $320 less than 2*S.
.
This last statement says that E is smaller than 2*S. In fact, if you have 2*S you have to
take away $320 to have the result equal E. In equation form you would write this as:
.
2*S - 320 = E
.
This is the second of the two equations you need. In summary, the two equations you need to
solve simultaneously are:
.
0.08*E + 0.06*S = 185.60 and
2*S - 320 = E
.
It might be a little easier to rearrange the bottom equation so that it is in the same
form as the top equation, but if you are going to use substitution as the method of solution,
the bottom equation is already solved for one of the variables in terms of the other.
.
You did not ask for help with how to reach a solution, but here are the answers that you
can use as a check to make sure you work the problem correctly:
.
E = $1600
S = $960
.
Hope this helps you to understand the problem a little better ...
.
Answer by SHUgrad05(58) (Show Source): You can put this solution on YOUR website!
To set up this word problem,it's good to start writing out what we know:
We know George has two investments (let's call them x & y), one at 6% and the other at 8%.
The two investments yield a total of $185.60 in annual interest.
So, 0.06x+0.08y=185.60
Next:
We know the amount invested at 8% is $320 less than twice the amount invested at 6% which means: y=2x-320.
So that leaves us with: 0.06x+0.08y=185.60 and y=2x-320
Use the substitution method to continue solving. I hope this helps you out!

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