Question

rezolvati urmatoarele intervale

Answer 1

[tex]2.a) \left \{ {{x+y=10} \atop {x-y=2}} \right. \\ -------------\\ 2x|=12=\ \textgreater \ x=12:2=\ \textgreater \ x=6\\ x+y=10\ \textless \ =\ \textgreater \ 6+y=10=\ \textgreater \ y=10-6=\ \textgreater \ y=4\\\\ b) \left \{ {{-2x+3y=4} \atop {2x-5y=2}} \right. \\ ---------\\ |-2y=6=\ \textgreater \ y=6:(-2)=\ \textgreater \ y=-3\\ -2x+3y=4\ \textless \ =\ \textgreater \ -2x-9=4=\ \textgreater \ -2x=13=\ \textgreater \ x=-6,5\\\\ c) \left \{ {{-x-3y=7|*1\ \textless \ =\ \textgreater \ -x-3y=7} \atop {2x+y=8|*3\ \textless \ =\ \textgreater \ 6x+3y=24}} \right. \\ 5x|31=\ \textgreater \ x=31:5=\ \textgreater \ x=6,2\\ -x-3y=7\ \textless \ =\ \textgreater \ -6,2-3y=7=\ \textgreater \ -3y=7+6,2=\ \textgreater \ -3y=13,2\\ =\ \textgreater \ y=13,2:(-3)=\ \textgreater \ y=-4,4\\\\ [/tex]. d) si e) sunt la fel....iti fac f-ul inca care este cu radicali ..... [tex]f) \left \{ {{ \sqrt{2x} + \sqrt{3y}= \sqrt{8} |* \sqrt{5} =\ \textgreater \ \sqrt{10x} + \sqrt{15y}= \sqrt{40} } \atop { \sqrt{3x} + \sqrt{5y} = \sqrt{12} |*(- \sqrt{3} )=\ \textgreater \ \sqrt{9x}- \sqrt{15y} = -\sqrt{36} }} \right.\\ \sqrt{x} |= -\sqrt{4} =\ \textgreater \ x=2\\ \sqrt{8} + \sqrt{3y} = \sqrt{8} =\ \textgreater \ \sqrt{3y} =0=\ \textgreater \ y=0[/tex]