philipssonicare wrote:
GMATGuruNY great explanation
Could someone please decipher what MathRevolution has done?
Surely this is not a sub-600 question
philipssonicare they divided the numerator and denominator with\(b^2\) .
It is valid since\(b \neq 0\) .
Following is a more legible solution.
\( \begin{alignat}{2}
&&\cfrac{a^2+ab}{a^2+b^2} \\
&= & \frac{\left(\frac{a^2 + ab}{b^2}\right)}{\left(\frac{a^2 + b^2}{b^2}\right)} \\
&= & \frac{\left(\frac{a^2}{b^2}\right)+\left(\frac{ab}{b^2}\right)}{\left(\frac{a^2}{b^2}\right)+\left(\frac{b^2}{b^2}\right)}\)
...









