Show as an equation and solve!!?

Question

U invest $17,000 for 1yr. part at 10% part at 12% and the rest at 15%. The total annual was $2110. The amount invested at 12% was $1000 less than the amount invested at 10% and 15% combined. Find the amount invested at each rate>

Answer 1

let a represent the amount invested at 10% and b represent the amount invested at 15%

a+b+(a+b-1000)=17000 solve for a
2a+2b-1000=17000
2a+2b=18000
2a=18000-2b
a=9000-b

.10a+.15b+.12(a+b-1000)=2110 expand the bracket
.10a+.15b+.12a+.12b-120=2110 collect like terms
.22a+.27b=2230 substitute value of a
.22(9000-b)+.27b=2230 expand the bracket
1980-.22b + .27b =2230 collect like terms
.05b=250 divide bot sides by .05
b=5000

a=9000-b substitute value of b
a= 9000-5000
a= 4000

Therefore $4000 was invested at 10% $5000 was invested at 15% and $8000 was invested at 12%

Answer 2

solve for x,y,z
@ 12% = x
@ 10% = y
@ 15% = z

A [total investment is 17k]
(x+y+z = 17k)

B [investment at 12% is 1k less than the 10%+15%]
(x-1k = y+z)

A becomes (y = 17k - x - z)
sub in the y in B
gives
(x-1k = 17k - x - z + z)
the z cancel, the 1k adds to the 17k and the -x is moved to the other side of the equals)

2x = 18k
!!x = 9k
interest made due to x = 1.08k

total interest = 2.11k

remaining interest needed = 1.03

if the full remaining 8k was put in the 10% we would make 0.8k but we need 1.03k
we are short by 0.23k

15% offers 5% more interest on the investment than 10% therefore 5% of what will give us the required 0.23k

(z)(5/100) = 0.23k
z = 4.6k

!!z = 4.6k

y = 8k - 4.6 = 3.4k

!!y = 3.4k

time for simple test
x+y+z = 17k?
9k + 4.6k + 3.4k = 17k investment is correct

9k@ 12% = 1.08k
3.4k@ 10% = 0.34k
4.6k@ 15% = 0.69k
1.08k+ 0.34k + 0.69k = 2.11k profit is correct

i now conclude with
x = 9k
y = 3.4k
z = 4.6k
where
12% = x
10% = y
15% = z

Answer 3

x = amount invested at 12%
Then 17,000 - x = invested at 10%and 15% combined.
So x = 17,000 -x - 1000
2x = 16,000 --> x= $8000 = amount invested at 12%
So $9000 was invested at 10% and 15% combined
Let y = amount invested at 10%
Then 9000-y = amount invested at 15%
So .1y +.15(9000-y) +.12(8000) = 2110
.1y + 1350 -.15y + 960 = 2110-.05y = -200
y = $4000 = amount invested at 10%
9000-y = $5000 = amount invested at 15%

Answer 4

part at 10% = x
15% = y
12% = x + y - 1000

x + y + (x + y - 1000) = 17000
2x + 2y = 18000
x + y = 9000

(0.10)x + (0.15)y + (0.12)(x + y - 1000) = 2110
(0.22)x + (0.27)y - 120 = 2110
(0.22)x + (0.27)y = 2230

So, now you have two equations...
x + y = 9000
(0.22)x + (0.27)y = 2230

Solve the first one for one variable...
x = 9000 - y
Put that in the other equation...
(0.22)(9000-y) + (0.27)y = 2230
1980 - (0.22)y + (0.27)y = 2230
(0.05)y = 250
y = 5000

x = 9000 - y = 9000 - 5000 = 4000

12%:
x + y - 1000
= 5000 + 4000 - 1000
= 8000

So:
10% = $5000
12% = $8000
15% = $4000

Answer 5

the amount invested at each rate is $1400

Answer 6

Maybe you should learn something and try to do the question yourself.

Answer 7

im tired of math today!