SOLUTION: an amount of $5000 is put into three investments at rates 6%, 7%, and %8 per annum, respectively. The total annual income is $358. The income from the first two investments is $70

Question 147579: an amount of $5000 is put into three investments at rates 6%, 7%, and %8 per annum, respectively. The total annual income is $358. The income from the first two investments is $70 more than the income from the third investment. Find the amount of each investment.
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
n amount of $5000 is put into three investments at rates 6%, 7%, and %8 per annum, respectively. The total annual income is $358. The income from the first two investments is $70 more than the income from the third investment. Find the amount of each investment
:
Let x = 6% amt (1st investment)
Let y = 7% amt (2nd investment)
Let z = 8% amt (3rd investment)
:
The total invested equation:
x + y + z = 5000
:
The total annual income equation:
.06x + .07y + .08z = 358
:
It says,"The income from the first two investments is $70 more than the income from the third investment." equation for this:
.06x + .07y = .08z + 70
.06x + .07y - .08z = 70
:
Use elimination by subtracting the above equation from the total income equation
.06x + .07y + .08z = 358
.06x + .07y - .08z = 70
------------------------- subtraction eliminates x and y
0x + 0y + .16z = 268
z =
z = $1800 invested at 8%
:
Using the total equation; z = 1800
x + y + 1800 = 5000
x + y = 5000 - 1800
x + y = 3200
y = (3200-x); use for substitution
:
Using the total income equation: z = 1800
.06x + .07y + .08(1800)
.06x + .07y + 144 = 358
.06x + .07y = 358 - 144
.06x + .07y = 214
:
Substitute (3200-x) for y, find x
.06x + .07(3200-x) = 214
.06x + 224 - .07x = 214
.06x -.07x = 214 - 224
-.01x = -10
x =
x = $1000 amt invested at 6%
:
I'll let you find y, the 7% investment

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