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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.
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