O
Szczegółowe wyjaśnienie:
zad.1
[tex]\left \{ {{4x+y=5} \atop {x-3y=13~~\mid \cdot (-4)}} \right. \\\\\left \{ {{4x+y=5} \atop {-4x+12y=-52}} \right. ~~\mid + ~~dodaje~~stronami\\\\4x+y-4x+12y=5-52\\\\13y=-47~~\mid \div 13\\\\y=- 3\dfrac{8}{13} \\\\4x+y=5~~\land ~~ y=- 3\dfrac{8}{13} ~~ \Rightarrow ~~4x - 3\dfrac{8}{13} =5\\\\4x- 3\dfrac{8}{13} =5\\\\4x=5+3\dfrac{8}{13} \\\\4x=8\dfrac{8}{13} \\\\4x=\dfrac{112}{13} ~~\mid \div 4\\\\x=\dfrac{28}{13} \\\\x=2\dfrac{2}{13}[/tex]
Odp: Rozwiązaniem jest: [tex]\left \{ {{y=-3\frac{8}{13} } \atop {x=2\frac{2}{13} }} \right.[/tex].
zad.2
[tex]\left \{ {{x+3y=7} ~~\mid \cdot (-2)\atop {2x+y=4}} \right. \\\\\left \{ {{-2x-6y=-14} \atop {2x+y=4}} \right. ~~\mid ~~+~~dodaje~~stronami\\\\-2x-6y+2x+y=-14+4\\\\-5y=-10~~\mid \div (-5)\\\\y=2\\\\x+3y=7~~\land~~y=2~~\Rightarrow ~~x+3\cdot 2 =7\\\\x+6=7\\\\x=7-6\\\\x=1\\\\[/tex]
Odp: Rozwiązaniem jest: [tex]\left \{ {{y=2} \atop {x=1}} \right.[/tex] .
