math

posted by .

Money is invested at two rates of interest. One rate is 8% and the other is 2%. If there is $1000 more invested at 8% than at 2%. Find the amount invested at each rate if the annual interest from both investments is $310. Let x amount invested at 8% and y = amount invested at 2%. Then the system that models the problem is [x=y+1000
0.08x+0.02y=310 Solve the system using the method of addition?=.

ok this is how my book says solve it
[2] x - y=100 -----> 2x-2y=2000
[-25] 0.08x+0.02y=310 -->-2x-0.5y=-7750

where and how did they get [-25]?

  • math -

    They were trying to eliminate the decimals, but they did it in a very clumsy way
    I would have multiplied the second by 100
    8x + 2y = 31000
    then the first by 2
    2x - 2y = 2000

    trivial from here on ...

    leaving a decimal after you multiplied by something sort of defeats the purpose, why didn't they just multiply the first by 50 and leave the first alone.
    That would have done the trick also.

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. algebra

    James invested some money at 12% interest. James also invested $188 more than 4 times that amount at 13%. How much is invested at each rate if Sang receives $2249.57 in interest after one year?
  2. algebra

    Money is invested at two rates of interest. One rate is 8% and the other is 2%. If there is $1000 more invested at 8% than at 2%. Find the amount invested at each rate if the annual interest from both investments is $310. Let x amount …
  3. Algebra

    I invested $42,000 in three funds paying 5%, 7%, and 9% simple interest. The total annual interest from these investments was $2,600. The amount of money invested at 5% was $200 less than the amount invested at 7% and 9% combined. …
  4. math

    A student invests two sums of money at 3% and 4% interest, receiving a total of $110 in interest after 1 year. Twice as much money is invested at 4% than at 3%. Find the amount invested at each interest rate. Where do I even begin …
  5. Math

    Larry has an annual return of $213.00 from $3000.00 invested at simple interest. One at 5% and the other at 8%. How much is invested at each rate. (Hint, Interest earned = amount invested x rate of interest.)
  6. math

    you invested $4000 in two accounts paying 2% and 9% annual interest, respectively if the total interest earned for the year was $150, how much was invested at each rate. $_was invested at 2% and _$ was invested at 9%
  7. algebra

    A total of $12,000 is invested in two funds paying 9% and 11% simple interest. If the yearly interest is $1,180, how much of the $12,000 is invested at each rate?
  8. algebra

    Larry Mitchel invested part of his $32,000 advance at 7% annual simple interest and the rest at 6% annual simple interest. If this total yearly interest from both accounts was $2,050, find the amount invested at each. The amount invested …
  9. math

    An investment of $3,000 is made at an annual simple interest of 5%. How much additional money must be invested at an annual simple interest rate of 8%, so that the total annual interest earned is 7.5 of the original amount that you …
  10. Math

    A person invested $7900 for one year, part at 5%, part at 11%, and the remainder at 15%. The total annual income from these investments was $965. The amount of money invested at 15% was $500 more than the amounts invested at 5% and …

More Similar Questions