Question

ex 14
va rog
e urgent
dau 50 p. ​

Answer 1

$\bold{a)} \ \left \{ {{x+\frac{y}{2}=10} \atop {\frac{x}{2}+y=11}} \right. \Leftrightarrow \left \{ {{x=\frac{20}{2}-\frac{y}{2}} \atop {\frac{x}{2}+y=11}} \right. \Leftrightarrow \left \{ {{x=\frac{20-y}{2}} \atop {\frac{20-y}{2}\cdot \frac{1}{2}+y=11}} \right. \Leftrightarrow \left \{ {{x=\frac{20-y}{2}} \atop {\frac{20-y}{4}+\frac{4y}{4}=11}} \right. \Leftrightarrow$

$\Leftrightarrow \left \{ {{x=\frac{20-y}{2}} \atop {20-3y=11 \cdot 4}} \right. \Leftrightarrow \left \{ {x=\frac{20-y}{2}} \atop {3y=20-44}} \right. \Leftrightarrow \left \{ {{x=\frac{20-y}{2}} \atop {y=-\frac{24}{3}}} \right. \Leftrightarrow \left \{ {{x=\frac{20+8}{2}} \atop {y=-8}} \right. \Leftrightarrow \\ \Leftrightarrow \boxed{\left \{ {{x=14} \atop {y=-8}} \right. }$

$\bold{b)} \ \left \{ {{2\cdot (x+\sqrt{2})+3\cdot (y-\sqrt2)=10\sqrt2} \atop {\frac{x}{y}=0,(3)=\frac{1}{3}}} \right. \Leftrightarrow \left \{ {{y=3x} \atop {2x+2\sqrt2+3\cdot 3x-3\sqrt2=10\sqrt2 }} \right. \Leftrightarrow \\ \\ \Leftrightarrow \left \{ {{y=3x} \atop {11x-\sqrt2=10\sqrt2}} \right. \Leftrightarrow \left \{ {{y=3x} \atop {11x=11\sqrt2}} \right. \Leftrightarrow \left \{ {{y=3x} \atop {x=\sqrt2}} \right. \Leftrightarrow \boxed{\left \{ {{x=\sqrt2} \atop {y=3\sqrt2}} \right. }$