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October 23, 2014

October 23, 2014

Posted by **Kara** on Wednesday, August 29, 2007 at 10:10am.

(x/3+3/4)(3x/4-3/5)

- Algebra -
**Kara**, Wednesday, August 29, 2007 at 11:10amknow I am supposed to use the FOIL method.But is this right? I am really not sure I did this right, something just seems off about it.

So i would Multiply

x/3 * 3x/4=x^2/4

x/3 *-3/5= -x/5

3/4 *3x/4=9x/4

3/4 * -3/5

I know to add both of my bottom numbers have to be the same so the common factors of them would be 20

so I get x^2-4x/20+45x/20-9/20=

x^2+41x/20-9/20

- Algebra -
**DrBob222**, Wednesday, August 29, 2007 at 11:31amknow I am supposed to use the FOIL method.But is this right? I am really not sure I did this right, something just seems off about it.

So i would Multiply

x/3 * 3x/4=x^2/4

x/3 *-3/5= -x/5

3/4 *3x/4=9x/4**wouldn't this be 9x/16?**

3/4 * -3/5

I know to add both of my bottom numbers have to be the same so the common factors of them would be 20

so I get x^2-4x/20+45x/20-9/20=

x^2+41x/20-9/20

- Algebra -
**Kara**, Wednesday, August 29, 2007 at 11:36amOhh, I knew something looked off...so it should be x^2-5x/80-9/20

- Algebra -
- Algebra -
**DrBob222**, Wednesday, August 29, 2007 at 12:24pmWhat happened to the 4 (x^2/4) from the first term? You started with that but I think your dropped it.

For the middle term, it is -x/5 + 9x/16 and that part becomes (-16x + 45x)/80 which is 29x/80.

The last term is 9/20.

Then you can get rid of the 4, 80, and 20 as denominators by multiplying through by 80.

**Check my thinking. Check my arithmetic.**

**Answer this Question**

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