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anonymous

  • 5 years ago

need help with algebra word problems

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  1. Chris
    • 5 years ago
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    ok do you have an example?

  2. Chris
    • 5 years ago
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    These are generally specific to the word problem, and probably best learned by example

  3. anonymous
    • 5 years ago
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    A $6000 investment was $12- less than a $10000 investment at 1% less interest. what are the two rates

  4. anonymous
    • 5 years ago
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    that should be $120

  5. Chris
    • 5 years ago
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    hmmm, is that the exact problem?

  6. anonymous
    • 5 years ago
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    yes

  7. Chris
    • 5 years ago
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    what should be $120?

  8. Chris
    • 5 years ago
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    and is it after one year?

  9. anonymous
    • 5 years ago
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    that is how much less the interest earned is on $6000

  10. anonymous
    • 5 years ago
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    yes annual interest

  11. Chris
    • 5 years ago
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    ok got it

  12. Chris
    • 5 years ago
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    so the amount of interest earned in a year is .% x the investment

  13. Chris
    • 5 years ago
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    so for the first part of the word problem we know this: 1. 6000 * x = 10000 * y - 120

  14. Chris
    • 5 years ago
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    that is, 6000 * the rate of interest for the first amount/rate is 120 less than the 10000 amount times its rate of interest

  15. Chris
    • 5 years ago
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    now, they also give you the second interest rate - for the 10000 value - is 1%

  16. Chris
    • 5 years ago
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    which is .01

  17. anonymous
    • 5 years ago
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    no $10000 is one percent less than the rate for $6000

  18. Chris
    • 5 years ago
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    ok so they're saying after one year, the 6000 investment was 120 less than the 10000 investment, and the 10000 investment was at a 1% interest rate?

  19. anonymous
    • 5 years ago
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    partially.. the $10000 investment is at a rate that is 1% less than the rate of the $6000 investment which yeilded $120 less.

  20. Chris
    • 5 years ago
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    ok, so there are two equations for two variables, x and y - with y being the % interest for the 10k investment, and x being the % interest for the second investment of 6k 1. y = x - .01 - the rate of investment for the 10k is 1 percent less than the 6k 2. 120 = 10000 *y - 6000 * x - the 10k investment at its rate yielded 120 more than the 6k at its rate

  21. Chris
    • 5 years ago
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    so now you need to substitute one equation into the other

  22. Chris
    • 5 years ago
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    how about we take the value of y from the first equation, and substitute it in place of y in the second equation

  23. Chris
    • 5 years ago
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    so substituting y = x - .01 into the second equation we get: 1. 120 = 10000 * (x - .01) - 6000x 2. 120 = 10000x - 100 - 6000x 3. 120 = 4000x - 100 4. 20 = 4000x 5. x = 20/4000 6. x = .005 = .5%

  24. Chris
    • 5 years ago
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    now given that x = .5%, and y = x - .01, we can solve for y

  25. Chris
    • 5 years ago
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    ack sorry

  26. Chris
    • 5 years ago
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    I messed up in step 4

  27. Chris
    • 5 years ago
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    120 = 4000x - 100 means 220 = 4000x

  28. anonymous
    • 5 years ago
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    I got confused at step 2 lol

  29. Chris
    • 5 years ago
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    so x = 220/4000 = .055 = 5.5%

  30. anonymous
    • 5 years ago
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    1% should equal the $120 dollars is what I was thinking

  31. Chris
    • 5 years ago
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    y = x - .01 means y = .055 - .01 = .045 = 4.5%

  32. Chris
    • 5 years ago
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    well, step 2 means that , at their annual investment rates (the amount they earn each year), we know that the 10k investment earns 120 more than the 6k investment. The amount the 10k investment earns after the first year, is 10k * rate of investment (which we're saying is "y") The mount the 6k investment earns after the first year, is 6k * rate of investment , which we're calling "x" so we know that, the amount earned by the 10k investment is 120 more than the 6k investment. so 120 = 10k*x - 6k*y

  33. anonymous
    • 5 years ago
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    I guess I have more of a problem setting up the equation. Once I seen it I could solve it

  34. Chris
    • 5 years ago
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    well, practice is key with these. and honestly - that was a tricky one

  35. Chris
    • 5 years ago
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    took me a long time to understand what they wanted

  36. Chris
    • 5 years ago
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    key is to keep trying to put down equations for the things you think you know - and see if they work out. the worst thing to do is get writers block. it's hard to visualize these problems all at once, just start writing potential equations and then double check that they really do represent the word problem

  37. anonymous
    • 5 years ago
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    Ok thanks a lot! I haven't had math in over 15 years.

  38. Chris
    • 5 years ago
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    np =)

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