# Thread: Word Problem for Algebra.

1. ## Word Problem for Algebra.

Shawn Hetrick has an annual interest income of $3290 from two investments. He has$7000 more invested at 8% than he has invested at 6%. find the amount invested at each rate.
Can anyone solve this for me? I'm a bit confused.. Well not really, more like I'm stuck. The problem I have is that this question doesn't give to total he invested on the 8% and 6%. I just can't think of an equation for this.

2. one equation could be
x=amount invested in 8%
y=amount invested in 6%
$0.08x+0.06y=3290$
find the other equation then use elimination method to eliminate x and find y then use one of the equations to plug in y and find x
other equation is something like
$x+7000=y$
which equals to:
$x-y=7000$
then just use simple elimination and u shall find the answer

3. $.08(x+7000) + .0 6(x) = 3290$

4. Originally Posted by 11rdc11
$.8(x+7000) + .6(x) = 3290$
0.8 is 80% while 0.08 is 8%
so you were wrong the the right equation is still
$

0.08x+0.06y=3290
$

and
$

x-y=7000
$

now just use those 2 to solve it with elimination

5. Originally Posted by saar4ever
0.8 is 80% while 0.08 is 8%
so you were wrong the the right equation is still
$

0.08x+0.06y=3290
$

and
$

x-y=7000
$

now just use those 2 to solve it with elimination
Thanks for spotting that. I used the equation I did incase he or she does not know how to work with 2 variables

6. Sweet, thanks guys! I can see the picture of it I still need more practice.
The thing is that I have is making an equation out of a word problem

A freight train travels 60 miles in the same time that a passenger train travels 100miles. If the passenger train goes 30miles per hour faster than the freight train, find the rate of each.

Honestly, I don't have an idea of what this question is trying to ask.

8. Originally Posted by saar4ever
0.8 is 80% while 0.08 is 8%
so you were wrong the the right equation is still
$

0.08x+0.06y=3290
$

and
$

x-y=7000
$

now just use those 2 to solve it with elimination
Don't solve it for me but can you give me a brief explain of x and y using in an equation? I'm only use to working with x Yet, I'm in a community college learning algebra.
Sorry for the trouble and thanks for helping me out, I'm still a newb.

9. multiply this by 100
$

0.08x+0.06y=3290
$

=
$

8x+6y=329000
$

then divide by 6 and add equation 1 and 2
then u have x all alone and divide it to find what it is
then plug in x into the second equation to find out what y is

10. Originally Posted by BrilliantLegacy
A freight train travels 60 miles in the same time that a passenger train travels 100miles. If the passenger train goes 30miles per hour faster than the freight train, find the rate of each.

Honestly, I don't have an idea of what this question is trying to ask.
60x=100y
x=y+30

then

60x-100y=0
x-y=30

then

-0.6x+y=0
x-y=30

0.4x=30

x=75mph
y=45mph
just double checked pretty sure its right answer
hope it makes sense

11. Originally Posted by saar4ever
60x=100y
x=y+30

then

60x-100y=0
x-y=30

then

-0.6x+y=0
x-y=30

0.4x=30

x=75mph
y=45mph
just double checked pretty sure its right answer
hope it makes sense
Thanks a lot dude. I have a pre-test tomorrow and you have just saved me!

12. your welcome no prob
see ya
gl on the test hope u ace it

13. Originally Posted by saar4ever
your welcome no prob
see ya
gl on the test hope u ace it
Do you mind telling me the the answer for the first one? I want to make sure I have it right.

14. Originally Posted by saar4ever
60x=100y
x=y+30

then

60x-100y=0
x-y=30

then

-0.6x+y=0
x-y=30

0.4x=30

x=75mph
y=45mph
just double checked pretty sure its right answer
hope it makes sense
I got a different answer. Speeds are 75 and 105. The passenger should be faster

x=y-30 or x+30=y

15. Originally Posted by 11rdc11
I got a different answer. Speeds are 75 and 105. The passenger should be faster

x=y-30 or x+30=y
could be need to check it again

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