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anonymous
 5 years ago
need help with algebra word problems
anonymous
 5 years ago
need help with algebra word problems

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok do you have an example?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0These are generally specific to the word problem, and probably best learned by example

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0A $6000 investment was $12 less than a $10000 investment at 1% less interest. what are the two rates

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hmmm, is that the exact problem?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and is it after one year?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that is how much less the interest earned is on $6000

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so the amount of interest earned in a year is .% x the investment

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so for the first part of the word problem we know this: 1. 6000 * x = 10000 * y  120

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that is, 6000 * the rate of interest for the first amount/rate is 120 less than the 10000 amount times its rate of interest

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now, they also give you the second interest rate  for the 10000 value  is 1%

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no $10000 is one percent less than the rate for $6000

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok 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?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0partially.. the $10000 investment is at a rate that is 1% less than the rate of the $6000 investment which yeilded $120 less.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok, 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so now you need to substitute one equation into the other

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how about we take the value of y from the first equation, and substitute it in place of y in the second equation

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so 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%

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now given that x = .5%, and y = x  .01, we can solve for y

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I messed up in step 4

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0120 = 4000x  100 means 220 = 4000x

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I got confused at step 2 lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so x = 220/4000 = .055 = 5.5%

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.01% should equal the $120 dollars is what I was thinking

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0y = x  .01 means y = .055  .01 = .045 = 4.5%

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well, 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I guess I have more of a problem setting up the equation. Once I seen it I could solve it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well, practice is key with these. and honestly  that was a tricky one

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0took me a long time to understand what they wanted

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0key 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ok thanks a lot! I haven't had math in over 15 years.
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