Question

Resolva os sistemas de equações abaixo utilizando o método da adição.
com as contas,por favor!​

Answer 1

Resposta:

$a)\\ \\~~~~~~~2x+y=5\\+~~~3x-y=10\\~~~~~------\\~~~~~~~5x+0=15\\ \\5x=15\\ \\x=\frac{15}{5}\\ \\x=3\\ \\substitua~(x=3)~em~uma~das~equacoes\\ \\2x+y=5\\ \\2.3+y=5\\ \\6+y=5\\ \\y=5-6\\ \\y=-1\\ \\S:~~\left \{ {{x=3,~~y=-1} \}} \right.$

$b)\\ \\~~~~~~~~5x-y=7\\+~~~~2x+y=7\\~~~~~~~~-----\\~~~~~~~~7x+0=14\\ \\7x=14\\ \\x=\frac{14}{7}\\ \\x=2\\ \\substitua~(x=2)~em~uma~das~equacoes\\ \\5x-y=7\\ \\5.2-y=7\\ \\10-y=7\\ \\-y=7-10\\ \\-y=-3~~.(-1)~~multiplique~por~(-1)~para~inverter~os~sinais\\ \\y=3\\ \\S:~~\left \{ {{x=2;~~y=3} \}} \right.$

$c)\\ \\4x-y=4\\5x-y=1\\ \\multiplique~uma~das~equacoes~por~(-1)~para~isolar~a~incognita~"x"\\ \\4x-y=4~.(-1)=-4x+y=-4\\ \\monte~o~sistema\\ \\~~~~~~-4x+y=-4\\+~~~~~~5x-y=~~1\\~~~~~~~~~------\\~~~~~~~~~~~~x+0=-3\\ \\x=-3\\ \\substitua~(x=-3)~em~uma~das~equacoes\\ \\4x-y=4\\ \\4.(-3)-y=4\\ \\-12-y=4\\ \\-y=4+12\\ \\-y=16~~.(-1)~multiplique~por~(-1)~para~inverter~os~sinais\\ \\y=-16\\ \\S:~~\left \{ {{x=-3;~~y=-16} \} \right.$

$d)\\ \\x+2y=6\\x-y=2\\ \\multiplique~uma~das~equacoes~por~(-1)~para~isolar~uma~incognita\\ \\x+2y=6~.(-1)=-x-2y=-6\\ \\monte~o~sistema\\ \\~~~~~~~-x-2y=-6\\ +~~~~~~~x-~y=~~2\\~~~~~~~~-------\\~~~~~~~~~~~0-3y=-4\\ \\-3y=-4~.(-1)~multiplique~por~(-1)~para~inverter~os~sinais\\ \\3y=4\\ \\y=\frac{4}{3}\\ \\substitua~(y=\frac{4}{3})~em~uma~das~equacoes\\  \\x-y=2\\ \\x-\frac{4}{3}=2\\ \\mmc=3\\ \\\frac{3x-4=6}{3}~~~~~despreze~o~denominador~3\\ \\3x-4=6\\ \\3x=6+4\\ \\3x=10\\ \\x=\frac{10}{3}$

$S:~~\left \{ {{x=\frac{10}{3};~y=\frac{4}{3}  } \}} \right.$

$e)\\ \\x-y=0\\2x+3y=5\\ \\multiplique~a~primeira~equacao~por~(+3)~para~isolar~uma~incognita\\ \\x-y=0~.(+3)=~~3x-3y=0\\ \\monte~o~sistema\\ \\~~~~~~~~3x-3y=0\\+~~~~2x+3y=5\\~~~~~~~------\\~~~~~~~~5x+~0=~5\\ \\5x=5\\ \\x=\frac{5}{5}\\ \\x=1\\ \\substitua~(x=1)~em~uma~das~equacoes\\ \\2x+3y=5\\ \\2.1+3y=5\\ \\2+3y=5\\ \\3y=5-2\\ \\3y=3\\ \\y=1\\ \\S:~~\left \{ {{x=1;~~y=1;~~x=y } \}} \right.$

$f)\\ \\2x+6y=10\\4x-2y=-1\\ \\Podemos~multiplicar~a~segunda~equacao~por~(+3)~para~isolar~uma~incognita\\ \\4x-2y=-1~.(+3)=12x-6y=-3\\ \\monte~o~sistema\\ \\~~~~~~~~2x+6y=~10\\+~~~12x-6y=-3\\~~~~~--------\\~~~~~~~14x+~~0=~~7\\ \\14x=7\\ \\x=\frac{7}{14}\\ \\x=\frac{1}{2}\\ \\substitua~(x=\frac{1}{2})~em~uma~das~equacoes\\ \\2x+6y=10\\ \\2.(\frac{1}{2})+6y=10\\ \\\frac{2}{2}+6y=10\\ \\1+6y=10\\ \\6y=10-1\\ \\6y=9\\ \\y=\frac{9}{6}~~~~~~simplifique\\ \\y=\frac{3}{2}$

$S:\left \{ {x=\frac{1}{2};~~y=\frac{3}{2}  } \}} \right.$

$g)\\ \\x+2y=4\\2x-y=0\\ \\multiplique~asegunda~equacao~por~(+2)~para~isolar~uma~incognita\\ \\2x-y=0~.(+2)=4x-2y=0\\ \\monte~o~sistema\\ \\~~~~~~~~x+2y=4\\+~~~4x-2y=0\\~~~~~~------\\~~~~~~~5x+~0=4\\ \\5x=4\\ \\x=\frac{4}{5}\\ \\substitua~(x=\frac{4}{5})~em~uma~das~equacoes\\ \\x+2y=4\\ \\\frac{4}{5}+2y=4\\ \\mmc=5\\ \\\frac{4+10y=20}{5}~~~~despreze~o~denominador\\ \\4+10y=20\\ \\10y=20-4\\ \\10y=16\\ \\y=\frac{16}{10}~~~~~~~simplifique\\ \\y=\frac{8}{5}$

$S:~~\left \{ {{x=\frac{4}{5};~~y=\frac{8}{5}} \}} \right.$