Chris invests $2,200 into three accounts that pay 6%, 8% and 9% in annual interest. he has three times as much invested at 9% as he does at 6%. if his total interest for the year is $178, how much is invested at each rate?

in account 1 amount is $200 @ 6%

in account 2 amount is 3×200 is $600 @ 9%

and  in account 3 amount is 2200- (4x) = 2200 - 800 = $1400 @8%

Step-by-step explanation:

Given data

amount =$ 2200

three account rate

r1 = 6% = 0.06

r2 = 8% = 0.08

r3 = 9% = 0.09

total interest = $178

to find out

invested at each rate

solution

we know account 1 +account 2  + account 3  = $2200

amount x in account 1 interest of of 6%                   1

amount 3x in account 2 interest of of 9%               2

and amount 2200-(3x+x) in account 3 interest of of 8%        3

by equation 1, 2 and 3 we can say

total interest = sum of 1, 2 and 3 equation

total interest = 0.06 of x + 0.09 of 3x + 0.08 of 2200-(3x+x)

178 =  0.06 × (x) + 0.09 × (3x)  + 0.08 × 2200-(3x+x)

178 = 0.06x + 0.27x + 176 - 0.32x

178- 176 = 0.01x

x = 200

so in account 1 amount is $200 @ 6%

and in account 2 amount is 3×200 = $600 @ 9%

and  in account 3 amount is 2200- (4x) = 2200 - 800 = $1400 @8%

a is 8.b is 10.c is 10.5. d is -.5

step-by-step explanation:

just minus the smallest number from the biggest.

the answer is no solution, glad i could .