So I was looking through the homepage of Youtube to see if there were any math equations that I thought that I might be able to solve, when I happened to come across this math equation by Learncommunolizer. The equation in question was$$\text{Solve }\text{ }\frac{\sqrt{27}+\sqrt{75}}{\sqrt{48}}$$which I thought that I might be able to solve. Here is my attempt at solving the equation:$\color{white}{\require{cancel}{.}}$ $$\frac{\sqrt{27}+\sqrt{75}}{\sqrt{48}}$$$$=\frac{\sqrt{3\cdot3\cdot3}+\sqrt{5\cdot5\cdot3}}{\sqrt{2\cdot2\cdot2\cdot2\cdot3}}$$$$=\frac{3\sqrt3+5\sqrt3}{4\sqrt{3}}$$$$=\frac{3\sqrt3}{4\sqrt3}+\frac{5\sqrt3}{4\sqrt3}$$$$=\frac{3\cancel{\sqrt3}}{4\cancel{\sqrt3}}+\frac{5\cancel{\sqrt3}}{4\cancel{\sqrt3}}$$$$=\frac{3}{4}+\frac{5}{4}$$$$=\text{ }\frac{8}{4}$$$$=2$$My question
Is my solution correct, or what could I do to attain the correct solution, or attain it more easily?
You could also proceed as follows :
$\dfrac{\sqrt{27}\!+\!\sqrt{75}}{\sqrt{48}}= \dfrac{\sqrt3\!\left(\sqrt{27}\!+\!\sqrt{75}\right)}{\sqrt3\!\cdot\!\sqrt{48}}=\dfrac{\sqrt{81}\!+\!\sqrt{225}}{\sqrt{144}}=\\[12pt]=\dfrac{9\!+\!15}{12}=\dfrac{24}{12}=2\;.$