Advanced Fraction Problem

[Jason76]

\(\displaystyle \dfrac{3y^{2} + xy}{x^{2} }\)

\(\displaystyle \dfrac{(3y^{2} + xy)}{(x^{2}) } * \dfrac{(x^{-2})}{(x^{-2})}\) This is an attempt to simplify by multiplying the fraction by 1 over x on the top and bottom. I can see that on the bottom the \(\displaystyle x^{2}\) and \(\displaystyle x^{-2}\) cancel out leaving one. Meanwhile, on the top, the distributive property is carried out. But I can't see what is going on, besides the distributive property being carried out. I do see that an \(\displaystyle x^{-1}\) comes out of it, that's all.

How does this get to:

\(\displaystyle 3(\dfrac{y}{x})^{2} + \dfrac{y}{x}\) ?

Actually this is a diff. eq problem, but this segment by itself is not, so i put it in this sub-forum.

 

[stapel]

\(\displaystyle \dfrac{3y^{2} + xy}{x^{2} }\)

How does this get to:

\(\displaystyle 3(\dfrac{y}{x})^{2} + \dfrac{y}{x}\) ?

Try breaking the one fraction with two terms in the numerator into two fractions over the same denominator. Factor in the first fraction. Simplify in the second.

 

[Deleted member 4993]

\(\displaystyle \dfrac{3y^2 \ + \ xy}{x^2}\)

\(\displaystyle = \ \dfrac{3y^2}{x^2} \ + \ \dfrac{xy}{x^2}\)

\(\displaystyle = \ 3\left [\dfrac{y}{x}\right ]^2 \ + \ \dfrac{y}{x}\)

Just algebra.....

 

[Jason76]

\(\displaystyle \dfrac{3y^2 \ + \ xy}{x^2}\)

\(\displaystyle = \ \dfrac{3y^2}{x^2} \ + \ \dfrac{xy}{x^2}\)

\(\displaystyle = \ 3\left [\dfrac{y}{x}\right ]^2 \ + \ \dfrac{y}{x}\)

Just algebra.....

Looking at it one more time I get the idea. The \(\displaystyle x^{-2}\) also written as \(\displaystyle \dfrac{1}{x^{2}}\) cancels out in the bottom to make 1. In the top, the distributive property gives \(\displaystyle x^{-2}\) and \(\displaystyle y^{2}\) on the top left and \(\displaystyle x^{-1}y^{2}\) on the top right. When this is expressed in fractional form, then the expression on the top left becomes true: \(\displaystyle x^{-2}y^{2}\) can be rewritten as \(\displaystyle \dfrac{y^{2}}{x^{2}}\) This is finally rewritten as \(\displaystyle (\dfrac{y}{x})^{2}\) due to algebra rules.

Finally, since \(\displaystyle \dfrac{a}{b}\) over 1 = \(\displaystyle \dfrac{a}{b}\), then we see how the final form came into being.