Complex Fractions: [4 + (x/y^2)]/[3 - (1/y^2)], etc

[Richay]

How would I solve a problem like this

And one like this

I've tried the method on this site http://www.purplemath.com/modules/compfrac.htm and I understand it a bit but I need it broken down.. In a way that's easier to understand.

 

[skeeter]

there is no "solving" here ... simplification, maybe.

for the first complex fraction, multiply the numerator and denominator by y²

for the second complex fraction, multiply the numerator and denominator by (x+y) ... there will be some more simplification (combining like terms) for this one after you do the multiplication.

 

[Richay]

there is no "solving" here ... simplification, maybe.

for the first complex fraction, multiply the numerator and denominator by y²

for the second complex fraction, multiply the numerator and denominator by (x+y) ... there will be some more simplification (combining like terms) for this one after you do the multiplication.

For the first one. I don't really get it. If I multiply by the denominator then it's y^2 x y^2? I'm not too sure how to do that. Two y's would make a 1?

 

[stapel]

You don't multiply the denominator by the denominator; you multiply the fraction by the denominator.

As the referenced lesson demonstrates, you find the common denominator of all the fractions contained within the "complex" fraction (which, in your first example, is y²), and multiply the numerator of the complex fraction and the denominator of the complex fraction by this term.

. . . . .\(\displaystyle \L \left(\frac{4\,+\,\frac{x}{y^2}}{3\,-\,\frac{1}{y^2}}\right)\,\left(\frac{\frac{y^2}{1}}{\frac{y^2}{1}}\right)\)

. . . . .\(\displaystyle \L \frac{\left(4\,+\,\frac{x}{y^2}\right)\left(\frac{y^2}{1}\right)}{\left(3\,-\,\frac{1}{y^2}\right)\left(\frac{y^2}{1}\right)}\)

Multiply through onto each of the terms:

. . . . .\(\displaystyle \L \frac{4(y^2)\,+\,\left(\frac{x}{y^2}\right)\left(\frac{y^2}{1}\right)}{3(y^2)\,-\,\left(\frac{1}{y^2}\right)\left(\frac{y^2}{1}\right)}\)

Then simplify.

Eliz.