I need help on Grade 11 math.
- Thread starter needshelp
- Start date
- Joined
- Feb 4, 2004
- Messages
- 16,582
You have posted:
. . . . .2x<sup>2</sup> - 5xy + (2y<sup>2</sup>)/(y<sup>2</sup> - 3xy + 2x<sup>2</sup>
...but I suspect you actually mean:
. . . . .(2x<sup>2</sup> - 5xy + 2y<sup>2</sup>) / (y<sup>2</sup> - 3xy + 2x<sup>2</sup>)
If so, then you need to factor the numerator and denominator, and then cancel the common factor.
Do you know how to factor quadratics?
Thank you.
Eliz.
. . . . .2x<sup>2</sup> - 5xy + (2y<sup>2</sup>)/(y<sup>2</sup> - 3xy + 2x<sup>2</sup>
...but I suspect you actually mean:
. . . . .(2x<sup>2</sup> - 5xy + 2y<sup>2</sup>) / (y<sup>2</sup> - 3xy + 2x<sup>2</sup>)
If so, then you need to factor the numerator and denominator, and then cancel the common factor.
Do you know how to factor quadratics?
Thank you.
Eliz.
Hello, needshelp!
Factor: .\(\displaystyle \displaystyle{\frac{(2x\,-\,y)(x\,-\,2y)}{(2x\,-\,y)(x\,-\,y)}}\)
Cancel: .\(\displaystyle \displaystyle{\frac{x\,-\,2y}{x\,-\,y} }\) . . . providing \(\displaystyle 2x - y\ \neq\ 0\)
We have: .\(\displaystyle \displaystyle{\frac{2x^2\,-\,5xy\,+\,2y^2}{2x^2\,-\,3xy\,+\,y^2}}\)
Factor: .\(\displaystyle \displaystyle{\frac{(2x\,-\,y)(x\,-\,2y)}{(2x\,-\,y)(x\,-\,y)}}\)
Cancel: .\(\displaystyle \displaystyle{\frac{x\,-\,2y}{x\,-\,y} }\) . . . providing \(\displaystyle 2x - y\ \neq\ 0\)
