Question

Va rog... Ofer coronița primului care răspunde

Answer 1

Am ayadat poza cu rezolvarea

Answer 2

$\displaystyle\\ 7d) \frac{x}{2}=\frac{y}{9}=>\frac{x}{y}=\frac{2}{9}\\\\ \frac{10x-2y}{5x+y}~\text{Simplificam fortat cu y.}\\\\ \frac{10x-2y}{5x+y}=\frac{10\cdot\dfrac{x}{y}-2\cdot\dfrac{y}{y}}{5\cdot\dfrac{x}{y}+\dfrac{y}{y}}=\frac{10\cdot\dfrac{x}{y}-2}{5\cdot\dfrac{x}{y}+1}=\frac{10\cdot\dfrac{2}{9}-2}{5\cdot\dfrac{2}{9}+1}=\\\\\\ =\frac{\dfrac{20}{9}-2}{\dfrac{10}{9}+1}=\frac{\dfrac{20-2\cdot9}{9}}{\dfrac{10+9}{9}}=\frac{\dfrac{2}{9}}{\dfrac{19}{9}}=\frac{2}{9}\cdot\frac{9}{19}= \boxed{\frac{2}{19}}$

8a)

Fractia o simplificam fortat cu y.

$\displaystyle\\ \frac{x+y}{2x+y}=\frac{4}{5}\\\\\\ \frac{\dfrac{x}{y}+\dfrac{y}{y}}{2\cdot\dfrac{x}{y}+\dfrac{y}{y}}=\frac{4}{5}\\\\\\ \frac{\dfrac{x}{y}+1}{2\cdot\dfrac{x}{y}+1}=\frac{4}{5}\\\\\\ 5\Big(\frac{x}{y}+1\Big)=4\Big(2\cdot\frac{x}{y}+1\Big)\\\\ 5\cdot\frac{x}{y}+5=4\cdot2\cdot\frac{x}{y}+4\\\\ 8\cdot\frac{x}{y} - 5\cdot\frac{x}{y} =5-4\\\\ 3\cdot\frac{x}{y}=1\\\\ \boxed{\frac{x}{y}=\frac{1}{3}}$

8b)

Fractia o simplificam fortat cu y.

$\displaystyle\\ \frac{3x+4y}{2x+y}=\frac{7}{3}\\\\\\ \frac{3\cdot\dfrac{x}{y}+4\cdot\dfrac{y}{y}}{2\cdot\dfrac{x}{y}+\dfrac{y}{y}}=\frac{7}{3}\\\\\\ \frac{3\cdot\dfrac{x}{y}+4}{2\cdot\dfrac{x}{y}+1}=\frac{7}{3}\\\\\\ 7\Big(2\cdot\frac{x}{y}+1\Big)=3\Big(3\cdot\frac{x}{y}+4\Big)\\\\ 7\cdot2\cdot\frac{x}{y}+7=3\cdot3\cdot\frac{x}{y}+12\\\\ 7\cdot2\cdot\frac{x}{y}+7=3\cdot3\cdot\frac{x}{y}+12\\\\ 14\cdot\frac{x}{y}-9\cdot\frac{x}{y}=12-7\\\\ 5\cdot\frac{x}{y}=5\\\\ \frac{x}{y}=\frac{5}{5}=\boxed{1}$