The annual interest on an $18,000 investment exceeds the interest earned on a $4000 investment by $860. The $18,000 is invested at a 0.5% higher rate of interest than the $4000.

Question

The annual interest on an $18,000 investment exceeds the interest earned on a $4000 investment by $860.  The $18,000 is invested at a 0.5% higher rate of interest than the $4000.  What is the interest rate of each investment?

I put the final 'What is the interest rate of each investment?' here in the description as it wouldn't fit in the title, but I would really appreciate any help.

Answer 1

x= y + .86

x/18 = y/4 + .005

x= interest on $18000, y = interest on $4000

converted to 1/1000 value for simplifying the calculations

x=18, y=4

multiply 2nd equation by (18)(4) = 72

divide by 2

multiply 1st by 2

subtract

4x = 18y +.005(72)= 18y + .36

2x = 9y +.18

2x = 2y + 1.72

7y = 1.54

y = .22= $220 interest on $4,000 investment

x = 220 +860 = $1080 interest on $18000 investment

1080/18000 = .54/9= = 6% interest rate on $18000

220/4000= 22/400 = 5.5% interest rate on $4000