Question

Resolva os sistemas a seguir:

a) 4(x+1)=2y+5
x+y=4

b) 3x+2=y+3
x+y=4

c) x/2+y=7
3x+y=x+13



preciso pra agoraa...

Answer 1

$a)\\ \\ 4(x+1)=2y+5\\ \\ 4x+4=2y+5\\ \\ 4x-2y=5-4\\ \\ 4x-2y=1~..........~equacao~I\\ \\ x+y=4~.............~equacao~II\\ \\ Metodo~da~substituicao:~na~equacao~II\\ \\ x=4-y\\ \\ substitua~o~valor~de~x~(x=4-y)~na~equacao~I\\ \\ 4x-2y=1\\ \\ 4(4-y)-2y=1\\ \\ 16-4y-2y=1\\ \\ -6y=1-16\\ \\ -6y=-15~.(-1)\\ \\ 6y=15\\ \\ y=\frac{15}{6} =\frac{5}{2} \\ \\ y=\frac{5}{2} \\ \\ substitua~(y=\frac{5}{2} )~em~uma~das~equacoes:\\ \\ x+y=4~.....~equacao~II\\ \\ x+\frac{5}{2} =4\\ \\ x=4-\frac{5}{2}$

$mmc=2\\ \\ x=\frac{8-5}{2} =\frac{3}{2} \\ \\ x=\frac{3}{2} \\ \\ S=(x=\frac{3}{2},~~~y=\frac{5}{2} )$

$b)\\ \\ 3x+2=y+3\\ \\ 3x-y=3-2\\ \\ 3x-y=1~...........~equacao~I\\ \\ x+y=4~.............equacao~II\\ \\ o~metodo~mais~adequado~parece~ser~o~metodo~da~adicao:\\ \\ some~as~duas~equacoes:\\ \\ ~~~~~3x-y=1\\ +~~x+y=4\\ -------\\ ~~~~2x+0=5\\ \\ 2x=5\\ \\ x=\frac{5}{2} \\ \\ substitua~x=\frac{5}{2} ~em~uma~das~equacoes:\\ \\ x+y=4~...........~equacao~II\\ \\ \frac{5}{2} +y=4\\ \\ y=4-\frac{5}{2} \\ \\ mmc=2\\ \\ y=\frac{8-5}{2} =\frac{3}{2} \\ \\ y=\frac{3}{2} \\ \\ S=(x=\frac{5}{2} ,~~~y=\frac{3}{2} )$

$c)\\ \\ \frac{x}{2} +y=7\\ \\ mmc=2\\ \\ \frac{x+2y=14}{2} \\ \\ elimine~o~denominador\\ \\ x+2y=14~..........~equacao~I\\ \\ 3x+y=x+13\\ \\ 3x-x+y=13\\ \\ 2x+y=13~..........~equacao~II\\ \\ metodo~da~substituicao:\\ \\ na~equacao~I\\ \\ x+2y=14\\ \\ x=14-2y\\ \\ substitua~x=14-2y~na~equacao~II\\ \\ 2x+y=13\\ \\ 2(14-2y)+y=13\\ \\ 28-4y+y=13\\ \\ -3y=13-28\\ \\ -3y=-15~.(-1)\\ \\ 3y=15\\ \\ y=\frac{15}{3} \\ \\ y=5\\ \\ substitua~(y=5)~em~uma~das~equacoes:$

$x+2y=14~..........~equacao~I\\ \\ x+2.5=14\\ \\ x+10=14\\ \\ x=14-10\\ \\ x=4\\ \\ S=(x=4,~~y=5)$