Question

Me Ajudem pfvr pessoal !!!!!!
MUITOOO OBRIGADAAA

Answer 1

MÉTODO DA SUBSTITUIÇÃO:

$3x + y = 11 \\ x + 5y = 7$

x=7-5y

3(7-5y)+y=11

21-15+y=11

y=11-21+15

y=-10+15

y=5

x=7-5.5

x=7-25

x=-18

(-18,5)

$4x - y = 2 \\ x + y = 1$

x=1-y

4(1-y)-y=2

4-4y-y=2

-4y-y=2-4

-5y=-2

y=

$\frac{ - 2}{ - 5}$

y=-0,4

x=1-(-0,4)

x=1+0,4

x=1,4

(1,4,-0,4)

(LETRA C eu pulei pq falta um sinal)

$10x - 5y = 11\\ 2x - 4y = 18$

♡

Answer 2

Resposta:

$3x+~y=11~...~eq~I\\ ~~x+5y=~7~...~eq~II\\ \\ na~eq~II\\ \\ x=7-5y\\ \\ substitua~(x=7-5y)~na~eq~I\\ \\3.(7-5y)+y=11\\ \\ 21-15y+y=11\\ \\ -14y=11-21~.(-1)\\ \\ 14y=-11+21\\ \\14y=20\\ \\ y=\frac{20}{14}=\frac{10}{7}\\ \\ substitua~(y=\frac{10}{7})~na~~eq~II\\  \\ x+5y=7\\ \\ x+5.(\frac{10}{7})=7\\  \\ x+\frac{50}{7}=7\\  \\ x=7+\frac{50}{7} \\ \\ mmc=7\\ \\ \frac{7x=49+50}{7}\\ \\ despreze~o~denominador\\ \\ 7x=49+50\\ \\ 7x=99\\ \\ x=\frac{99}{7} \\ \\ substitua~(x=\frac{99}{7})~na~eq~I$

$3x+y=11\\ \\ 3.(\frac{99}{7})+y=11\\  \\ 297+y=11\\ \\ y=\frac{297}{11} \\ \\ S~(x=\frac{99}{7};~~y=\frac{297}{11})$

$B)~~adicao\\ \\ ~~~~~~~~4x-y=2~...~eq~I\\ +~~~~~~x+y=1~...~eq~II\\ ~~~~~----------\\ 5x+0=3\\ \\ 5x=3\\ \\ x=\frac{3}{5} \\ \\ substitua~(x=\frac{3}{5})~em~uma~das~equacoes\\  \\ x+y=1\\ \\ \frac{3}{5} +y=1\\ \\ y=1-\frac{3}{5} \\ \\ mmc=5\\ \\\frac{5y=5-3}{5}~despreze~o~denominador\\ \\ 5y=5-3\\ \\ 5y=2\\ \\ y=\frac{2}{5} \\ \\ S~~(x=\frac{3}{5};~~y=\frac{2}{5})$