Question

resolver os sistemas por metodo de adição

3x-4y=10
5x-3y=2

2x/3 +y = 31/3
x/4-y/2=-1/2

x+y/7=2/13(2x-y)
x+y=14

2(x-5)=7(3-y)
x-y/3-x+y/4 =-37/12​

Answer 1

Resposta:

$a)~multiplique~a~primeira~equacao~por~(-5)~e~a~segunda~por~(3)\\ \\3x-4y=10~.(-5)\\5x-3y=2~.(3)\\ \\~~~~~~-15x+20y=-50\\+~~~~~~15x-9y=~~~6\\~~~~~~----------\\~~~~~~~~~~~~0+~11y=~~46\\ \\11y=46\\ \\y=\frac{46}{11}\\ \\substitua~na~segunda~equacao\\ \\ 15x-9y=6\\ \\15x-9(\frac{46}{11})=6\\ \\15x-\frac{414}{11}=6\\ \\mmc=11\\ \\\frac{165x-414=66}{11}\\ \\despreze~o~denominador\\ \\165x-414=66\\ \\165x=66+414\\ \\165x=480\\ \\x=\frac{480}{165}\\ \\x=\frac{16}{33}$

$S~~\left \{ {{y=\frac{46}{11};~~x=\frac{16}{33}} \}} \right.$

$b)~multiplique~a~segunda~equacao~por~2\\ \\ \frac{2x}{3}+y=\frac{31}{3}\\ \\\frac{x}{4}-\frac{y}{2}=-\frac{1}{2} ~~.(2)=\frac{x}{2} -y=-1\\ \\~~~~~~\frac{2x}{3}+y=\frac{31}{3}\\ \\+~~~\frac{x}{2}-y=-1\\~~~~~~-------\\(\frac{2x}{3}-\frac{x}{2})+0=(\frac{31}{3}-1)\\  \\\frac{2x}{3}-\frac{x}{2}=\frac{31}{3}-1\\ \\mmc=6\\ \\\frac{4x-3x=62-6}{6}\\ \\4x-3x=62-6\\ \\x=56\\ \\substitua~(x=56)\\ \\\frac{2x}{3}+y=\frac{31}{3}\\ \\\frac{2.(56)}{3}+y=\frac{31}{3}\\ \\\frac{112}{3}+y=\frac{31}{3}\\ \\ mmc=3$

$\frac{112+3y=31}{3}\\ \\3y=31-112\\ \\3y=-81\\ \\y=-\frac{81}{3}\\ \\y=-27\\ \\S~~\left \{ {{y=-27;~~x=56} \}} \right.$

$c)~~organize~a~primeira~equacao\\ \\x+\frac{y}{7}=\frac{4x-2y}{13}\\ \\13(x+y)=7(4x-2y)\\ \\13x+13y=28x-14y\\ \\28x-13x=14y-13y\\ \\15x=27y\\ \\~~~~~~15x-27y=0\\+~~~~~~x+y=14~~.(-15)\\ \\~~~~~~~~15x-27y=~~~~~~0\\+~-15x-15y=-210\\~~~~~~~~---------\\0-42y=-210\\ \\y=\frac{-210}{-42}\\ \\y=+5\\ \\x+y=14\\ \\x+5=14\\ \\x=14-5\\ \\x=9\\ \\S~~\left \{ {{y=5;~~x=9} \}} \right.$

Desculpe-me, não houve tempo para resolver a última, mas a idéia é a mesma.

Espero ter ajudado um pouco

Bons estudos!