Matemática, perguntado por gabriella0009, 10 meses atrás

RESOLVA PRA MIM PFVR

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Soluções para a tarefa

Respondido por B0Aventura
0

método da adição:

 a)\\ \\ ~~~~~~~x+y=9\\ ~+~~x-y=5\\ ~~~~~------\\ ~~~~~2x+0=14\\ \\ 2x=14\\ \\ x=\frac{14}{2} \\ \\ x=7\\ \\ substitua~(x=7)~em~uma~das~equacoes:\\ \\ x+y=9\\ \\ 7+y=9\\ \\ y=9-7\\ \\ y=2~~~~~~~~~~S=(x=7;~~y=2)

 b)\\ \\ ~~~~~~4x-y=8\\ ~+~~~x+y=7\\ --------\\ ~~~~~~5x+0=15\\ \\ 5x=15\\ \\ x=\frac{15}{5} \\ \\ x=3\\ \\ substitua~(x=3)~em~uma~das~equacoes:\\ \\ x+y=7\\ \\ 3+y=7\\ \\ y=7-3\\ \\ y=4~~~~~~~~~~~~~S=(x=3;~~y=4)

 c)\\ \\ x-3y=5\\ 2x+4y=0\\ \\ multiplique~a~equacao~1~por (-2)\\ \\ x-3y=5~.~(-2)\\ \\ ~~~~~~-2x+3y=-5\\ ~+~~~~~2x+4y=~~0\\ ~~~----------\\ ~~~~~~~~~~~~0+7y=-5\\ \\ ~7y=-5\\ \\ y=-\frac{7}{5} \\ \\ substitua~(y=-\frac{7}{5} )~em~uma~das~equacoes:\\ \\ 2x+4y=0\\ \\ 2x+4(-\frac{7}{5} )=0\\ \\ 2x-\frac{28}{5} =0\\ \\ 2x=\frac{28}{5} \\ \\ x=\frac{28}{5} ~.~\frac{1}{2} \\ \\ x=\frac{28}{10} =\frac{14}{5} ~~~~~~~~~~~~~S=(x=\frac{7}{5} ;~~y=\frac{14}{5} )

 d)\\ \\ 2x+3y=~~2\\ 4x-9y=-1\\ \\ multiplique~a~equacao~1~por~3:\\ \\ 2x+3y=2~.~(3)\\ \\ ~~~~~~~~6x+9y=~~6\\ ~+~~~4x-9y=-1\\ ~~~~~--------\\ ~~~~~~~~10x+0=~~5\\ \\ 10x=5\\ \\ x=\frac{5}{10} =\frac{1}{2} \\ \\ substitua~(x=\frac{1}{2} )~em~uma~das~equacoes:\\ \\ 2x+3y=2\\ \\ 2~.\frac{~1}{2} +3y=2\\ \\ 1+3y=2\\ \\ 3y=2-1\\ \\ 3y=1\\ \\ y=\frac{1}{3} ~~~~~~~~~~~~~S=(x=\frac{1}{2} ;~~y=\frac{1}{3} )

 e)\\ \\ 3x+2y=5\\ 4x+y=5\\ \\ multiplique~a~equacao~2~por~(-2)\\ \\ 4x+y=5~.~(-2)=\\\\ -8x-2y=-10 \\ \\ ~~~~~~~~~~~~~~3x+2y=~~~~5\\ ~~+~~~~~-8x-2y=-10\\ ~~~~~~~~~~---------\\ ~~~~~~~~~~~~-5x+0=~~-5\\ \\ -5x=-5~.~(-1)\\ \\ 5x=5\\ \\ x=\frac{5}{5} \\ \\ x=1\\ \\ substitua~(x=1)~em~uma~das~equacoes:\\ \\ 3x+2y=5\\ \\ 3.1+2y=5\\ \\ 2y=5-3\\ \\ 2y=2\\ \\ y=\frac{2}{2} =1~~~~~~~~~~~~S=(x=y;~~x=1;~~y=1)

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