word problem help, please!! I stink at these.....?

Question

Homer has three investments totaling $110,000. These investments earn interest at 4%,6%,and 8% respectively. Homer's total income from these investments is $7200. The income from the 8% investment exceeds the total income from the other two investments by $800. Find how much Homer has invested at 4%?

I got this wrong on a test earlier, according to the test the answer should've been $20000.

Answer 1

Let Investment at 4% be x, 6% be y, 8% be z.

x+y+z= 110000
interest = [principle*years*rate]/100 ... Let rate be per annum. so no. of years = 1

7200 = 1/100 [ 4x + 6y + 8z]

but, 8z/100 = 4x/100 + 6y/100 +800
8z = 4x + 6y + 80000 ..substitute this in the above eq.

7200 = 1/100 [ 4x + 6y + 4x + 6y + 80000]
720000 - 80000 = 8x + 12y
8x - 12y = 640000
2x + 3y = 160000 .......... equation A

x + y + z = 110000
but we got z = 1/2x + 3/4y + 10000 ...substitute this below,
x + y + [1/2x + 3/4y + 10000] = 110000
6x + 7y = 440000 - 40000
6x + 7y = 400000 ............equation B

now u got 2 equations A and B,
2x + 3y = 160000
6x + 7y = 400000

multiply equation A by 3 and subract the two equations to cancel x. Now u'll get,

2y = 80000
y = 40000
substitute the value of y in eq. A,
2x + 3(40000) = 160000
2x = 160000 - 120000
x = 20000.

Homer invests $ 20000 at 4% .

you are right the answer is $ 20000.

Answer 2

x= amount @4%, y=amount @6%, z = amount @ 8%
x+y+z = 110,000
.04x+.06y+.08z = 7200
.08z=.04x+.06y +800
-.04x -.06y +.08z= 800
.16z =8000
z= $50,000
x+y =110,000-50000 = 60,000
.04x +.06y +.08(50000).=7200
.04x+.06y = 3200
4x+6y = 320000
-4x-4y = -240000
2y= 80000
y = $40,000
x= 110000-(50,000+40,000)= $20,000 invested at 4%

Answer 3

Solve linear equations

x + y + z = 110000

0.04 * x + 0.06 * y + 0.08 * z = 7200

0.08 * z = 0.04 * x + 0.06 * y + 800

and find x.