SOLUTION: Algebra-linear equation with 3 variables. Hans decided to divide $14000 into 3 investments: a savings account paying 6% annual interest, a bond paying 9%, and a mutual fund paying

Question 752593: Algebra-linear equation with 3 variables.
Hans decided to divide $14000 into 3 investments: a savings account paying 6% annual interest, a bond paying 9%, and a mutual fund paying 13%. His annual interest frm the investments was $1100 and he had twice as much invested in the bond as in the mutual fund. What amount did he invest in each type?
I have 2 figured out: x+y+z=$14,000 and
0.06x + 0.09Y + 0.13z=1100
But beyond this, I am stuck.
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
you need to use the fact that he had twice as much invested in the bond as in the mutual fund to reduce the number of unknowns to 2.
then you will have 2 equations with 2 unknowns which can be solved for a specific value.
let x = amount invested in savings
let y = amount invested in bonds
let z = amount invested in mutual funds

since the amount invested in bonds is twice the amount he invested in mutual funds, you get y = 2z.

as you stated, you have 2 equations.
they are:
x + y + z = 14000
.06x + .09y + .13z = 1100

not you have to replace y with 2z to get:

x + 2z + z = 14000
.06x + .18z + .13z = 1100

combine like terms to get:

x + 3z = 14000
.06x + .31z = 1100

you can solve this system of equation through substitution as follows:

solve for x in the first equation to get x = 14000 - 3z.
substitute for x in the second equation to get:
.06 * (14000 - 3z) + .31z = 1100
simplify to get:
.06 * 14000 - .06 * 3z + .31z = 1100
simplify further to get:
840 + .13z = 1100
subtract 840 from both sides of the equation to get:
.13z = 1100 - 840 = 260
solve for z to get:
z = 260/.13 = 2000

since y = 2z, then:
y = 4000

since x + y + z = 14000, then:
x + 2000 + 4000 = 14000
solve for x to get:
x = 8000

you have:
x = 8000
y = 4000
z = 2000

x + y + z = 8000 + 4000 + 2000 = 14000 (this is good)

.06x + .09y + .13z equals:
.06*8000 + .09*4000 + .13*2000 = 1100 (this is also good).

the solutions are confirmed to be good.

the key was using y = 2z to reduce the number of unknowns to 2.
2 equations in 2 unknowns can be solved for specific values.
2 equations in 3 unknowns cannot.

RELATED QUESTIONS
Anthony invested a certain amount of money into a savings account, and invested 3 times... (answered by rfer)

Please if someone can help me solve this problem! thank you! Anthony invested... (answered by Theo)

1,000 is placed into a savings account paying 4.5% for 3 yrs. the interest earn... (answered by swincher4391)

Katherine Chong invests $10,000 received from her grandmother in 3 ways. With one part,... (answered by lwsshak3)

You can afford monthly deposits of $160 into an account that pays 3.6% compounded... (answered by Theo)

Maria invested a total of $26000 into three accounts: a saving account paying 3\%... (answered by ikleyn,greenestamps,math_tutor2020)

i have 10,000.00 and I decided to invest some money in a savings account that pays 2%... (answered by mananth)

A worker received a $10,000 bonus and decided to split it among three different... (answered by ikleyn)

Could someone please help me soldve this question, I am really trying to assistanace my... (answered by josmiceli)