need help with algebra word problems

- anonymous

need help with algebra word problems

- Stacey Warren - Expert brainly.com

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- chestercat

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- chris

ok do you have an example?

- chris

These are generally specific to the word problem, and probably best learned by example

- anonymous

A $6000 investment was $12- less than a $10000 investment at 1% less interest. what are the two rates

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## More answers

- anonymous

that should be $120

- chris

hmmm, is that the exact problem?

- anonymous

yes

- chris

what should be $120?

- chris

and is it after one year?

- anonymous

that is how much less the interest earned is on $6000

- anonymous

yes annual interest

- chris

ok got it

- chris

so the amount of interest earned in a year is .% x the investment

- chris

so for the first part of the word problem we know this:
1. 6000 * x = 10000 * y - 120

- chris

that is, 6000 * the rate of interest for the first amount/rate is 120 less than the 10000 amount times its rate of interest

- chris

now, they also give you the second interest rate - for the 10000 value - is 1%

- chris

which is .01

- anonymous

no $10000 is one percent less than the rate for $6000

- chris

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?

- anonymous

partially.. the $10000 investment is at a rate that is 1% less than the rate of the $6000 investment which yeilded $120 less.

- chris

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

- chris

so now you need to substitute one equation into the other

- chris

how about we take the value of y from the first equation, and substitute it in place of y in the second equation

- chris

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%

- chris

now given that x = .5%, and y = x - .01, we can solve for y

- chris

ack sorry

- chris

I messed up in step 4

- chris

120 = 4000x - 100 means 220 = 4000x

- anonymous

I got confused at step 2 lol

- chris

so x = 220/4000 = .055 = 5.5%

- anonymous

1% should equal the $120 dollars is what I was thinking

- chris

y = x - .01 means y = .055 - .01 = .045 = 4.5%

- chris

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

- anonymous

I guess I have more of a problem setting up the equation. Once I seen it I could solve it

- chris

well, practice is key with these. and honestly - that was a tricky one

- chris

took me a long time to understand what they wanted

- chris

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

- anonymous

Ok thanks a lot! I haven't had math in over 15 years.

- chris

np =)

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