Question

need help with algebra word problems

Answer 1

ok do you have an example?

Answer 2

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

Answer 3

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

Answer 4

that should be $120

Answer 5

hmmm, is that the exact problem?

Answer 6

yes

Answer 7

what should be $120?

Answer 8

and is it after one year?

Answer 9

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

Answer 10

yes annual interest

Answer 11

ok got it

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

which is .01

Answer 17

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

Answer 18

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?

Answer 19

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

Answer 20

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

Answer 21

so now you need to substitute one equation into the other

Answer 22

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

Answer 23

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%

Answer 24

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

Answer 25

ack sorry

Answer 26

I messed up in step 4

Answer 27

120 = 4000x - 100 means 220 = 4000x

Answer 28

I got confused at step 2 lol

Answer 29

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

Answer 30

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

Answer 31

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

Answer 32

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

Answer 33

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

Answer 34

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

Answer 35

took me a long time to understand what they wanted

Answer 36

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

Answer 37

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

Answer 38

np =)