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Question 920759: Please help me solve this equation: Sue invested $10,000 in three parts. With one part she bought mutual funds that offered a return of 4% per year. The second part which amounted to twice the first was used to buy government bonds paying 4.5% per year. She put the rest into an account paying 2.5% annual interest. During the first year the total interest was $415. How much did she invest at each rate?
I set up a table, but am not sure how to put the numbers together into an equation that I can work through. I have no idea how to set this up.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! sue invested a total of $10,000
let x + y + z equal the amount of money invested in the three parts.
x is for the first part
y is for the second part
z is for the third part
with the first part she bought mutual funds that returned 4% per year.
that would be x and the return would be equal to .04 * x.
with the second part she bought governmen
t bonds paying 4.5% per year.
that would be y and the return would be equal to .045 * y.
you are also given that the second part was twice the size of the first part.
that would make y = 2x.
hold on to that thought because you will use it later to reduce the number of variables involved.
with the third part she placed it into an account paying 2.5% annual interest.
that would be z and the return would be equal to .025 * z.
during the first year she made $415 in interest.
you have 2 equations to work with and those equations need to be solved simultaneously.
the first equation is:
x + y + z = $10,000
this equation gives you the total amount invested by each part.
the second equation is:
.04 * x + .045 * y + .025 * z = $415.
this equation gives you the total interest earned by each part.
you have 2 equations in 3 variables which will not allow you to solve for a specific solution.
in order to solve for a specific solution, the number of equations needs to be less than or equal to the number of variables.
the preferred situation is that the number of equations is equal to the number of variables.
the equations that you want to solve simultaneously are:
x + y + z = $10,000
.04 * x + .045 * y + .025 * z = $415
fortunately, you can eliminate one of the variables, because you are told that the second part is twice the size of the first part which leads to y = 2x.
this means you can replace y with 2x in x + y + z = $10,000 to get x + 2x + z = $10,000 which becomes 3x + z = $10,000.
this also means you can replace y with 2x in .04x + .045y + .025z = $415 to get .04x + .045*2x + .025*z = $415k which becomes .04x + .09x + .025z = $415 which becomes .13x + .025z = $415.
your 2 equations now only have 2 variables that need to be solved and you can solve for a specific solution.
those equations are:
3x + z = $10,000
.13x + .025z = 415
in the first equation, solve for z to get:
z = $10,000 - 3x
in the second equation, replace z with $10,000 - 3x to get:
.13x + .025 * ($10,000 - 3x) = $415.
solve for x as follows:
simplify to get:
.13x + .025*10,000 - .025*3x = 415
simplify further to get:
.13x + 250 - .075x = 415
simplify to get:
.055x + 250 = 415
subtract 250 from both sides of this equation to get:
.055x = 165
divide both sides of this equation by .055 to get:
x = 3000.
since y = 2x, then y = 6000.
since the total is 10,000, then z = 1000.
you have:
x = 3,000
y = 6,000
z = 1,000
x + y + z = $10,000 so that part's good.
.04x + .045y + .025z = 120 + 270 + 25 = $415 so that part's good.
solution looks good.
solution is:
she invested $3,000 at 4% and $6,000 at 4.5% and $1,000 at 2.5%.
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