Question

$\frac{x+4y}{12} + \frac{2x+5y}{12}$
și$\frac{6 c^{2} }{3c-2} - 2c-5$

Answer 1

[tex] \frac{x+4y}{12}+ \frac{2x+5y}{12}= \frac{3c+9y}{12}^{(3}= \frac{x+3y}{4} \\ \\ \frac{6c^2}{3c-2} -^{3c-2)}2c-^{3c-2)}5= \frac{6c^2}{3c-2} - \frac{6c^2-4c}{3c-2}- \frac{15c-10}{3c-2}= \frac{-11c-10}{3c-2} [/tex]

Answer 2

             
$\frac{x+4y}{12} + \frac{2x+5y}{12} = \frac{x+4y+2x+5y}{12} = \frac{3x+9y}{12}=\frac{3(x+3y)}{3*4} =\frac{(x+3y)}{4}$


$\frac{6 c^{2} }{3c-2} - 2c-5 = \frac{6 c^{2} }{3c-2} - \frac{(2c+5)(3c-2)}{3c-2}= \\ \\= \frac{6 c^{2} }{3c-2} - \frac{(6c^{2}+15c - 4c-10)}{3c-2}= \\ \\= \frac{6 c^{2} }{3c-2} - \frac{(6c^{2}+11c -10)}{3c-2} = \\ \\= \frac{6 c^{2} -(6c^{2}+11c -10) }{3c-2}= \\ \\ = \frac{6 c^{2} -6c^{2}-11c +10}{3c-2}= \boxed{ \frac{-11c +10}{3c-2}}$