Matemática, perguntado por ludyfelloni, 1 ano atrás

Resolva os sistema:

a) 2x + y = 5
3x - y = 5

b) -2x + 3y = 1
4x - 5y = 0

c) x = -2y
x - 3y = 17,5

d) 2x + 5y = 14
x=y

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

f) x + 2y = 5
2x + 3y =7

Soluções para a tarefa

Respondido por LuanaSC8
71
a)~~\begin{cases}2x+y=5\\ 3x-y=5\end{cases}\\\\\\\\ 2x+y=5\to~~2x=-y+5\to~~ x= \dfrac{-y+5}{2} \\\\\\ 3x-y=5\to~~ 3x=y+5\to~~ x= \dfrac{y+5}{3} \\\\\\\\ \dfrac{-y+5}{2}=\dfrac{y+5}{3}\to~~ multiplique~~cruzado:\\\\ -3y+15=2y+10\to~~ -3y-2y=10-15\to ~~-5y=-5~(-1)\to\\\\ 5y=5\to~~y= \dfrac{5}{5} \to ~~ \boxed{y=1}\\\\\\ x= \dfrac{-y+5}{2}\to ~~x= \dfrac{-1+5}{2}\to ~~x= \dfrac{4}{2}\to~~ \boxed{x=2}\\\\\\ \Large\boxed{\boxed{S=\{ 2~;~1\}}}





b)~~\begin{cases}-2x+3y=1\\ 4x-5y=0\end{cases} \\\\\\ -2x+3y=1\to~~-2x=-3x+1~(-1)\to~~ 2x=3y-1~~x= \dfrac{3y-1}{2} \\\\ 4x-5y=0\to~~4x=5y\to~~x= \dfrac{5y}{4} \\\\\\\\ \dfrac{3y-1}{2}=\dfrac{5y}{4} \to~~multiplique~~cruzado:\\\\ 12y-4=10y\to~~12y-10y=4\to~~2y=4\to~~y=\dfrac{4}{2}\to~~\boxed{y=2 }


x= \dfrac{5y}{4}\to~~ x= \dfrac{5.2}{4}\to~~ x= \dfrac{ 10}{4}\to~~ \boxed{x= \dfrac{5}{2}}\\\\\\ \Large\boxed{\boxed{S= \left\{ \frac{5}{2}~;~2\right\} }}




c)~~\begin{cases}x=-2y\\ x-3y=17,5\end{cases}\\\\\\\\ x-3y=17,5\to~~-2y-3y=17,5\to~~ -5y=17,5\to\\\\ y=- \dfrac{17,5}{5} \to~~ \boxed{y=-3,5}\\\\\\x=-2y\to~~ x=-2.(-3,5)\to~~ \boxed{x=7}\\\\\\\\ \Large\boxed{\boxed{ S=\{7~;~-3,5\}}}





d)~~\begin{cases}2x + 5y = 14\\ x=y\end{cases}\\\\\\ 2x + 5y = 14\to~~ 2y + 5y = 14\to~~7y=14\to~~y= \dfrac{14}{7} \to~~\boxed{y=2} \\\\\\ x=y\to~~ \boxed{x=2} \\\\\\\\ \Large\boxed{\boxed{ S=\{2~;~2\}}}





e)~~ \begin{cases}2x=3y\\ y=2x+4\end{cases}\\\\\\ 2x=3y\to~~ 2x=3(2x+4)\to~~2x=6x+12\to~~2x-6x=12\to\\\\ -4x=12~(-1)\to~~ 4x=-12\to ~~ x=- \dfrac{12}{4} \to ~~ \boxed{x=-3}\\\\\\ y=2x+4\to~~ y=2.(-3)+4\to~~y=-6+4\to ~~ \boxed{y=-2}\\\\\\\\ \Large\boxed{\boxed{ S=\{-3~;~-2\}}}





f)~~ \begin{cases}x+2y=5\\ 2x+3y=7\end{cases}\\\\\\ x+2y=5\to~~x=-2y+5\\\\\\ 2x+3y=7\to ~~ 2.(-2y+5)+3y=7 \to ~~ -4y+10+3y=7\to \\\\ -4y+3y=7-10\to~~ -y=-3~(-1)\to ~~ \boxed{y=3}\\\\\\x=-2y+5\to~~ x=-2.3+5\to~~ x=-6+5\to ~~ \boxed{ x=-1}\\\\\\\\  \Large\boxed{\boxed{ S=\{-1~;~3\}}}
Respondido por Usuário anônimo
104
a) Método da Adição: 

2x + y = 5
3x - y = 5 (+)
5x = 10
x = 10/5
x = 2

2x + y = 5
2.2 + y = 5
4 + y = 5
y = 5 - 4
y = 1
----------------------------------------
b)  Método da Adição:

- 2x + 3y = 1   (2)
  4x - 5y = 0

- 4x + 6y = 2
  4x - 5y = 0  (+)
y = 2

4x - 5y = 0
4x - 5.2 = 0
4x - 10 = 0
4x = 10 (:2)
2x = 5
x = 5/2

prova real:

4. 5  - 5.2 = 10 - 10 = 0
    2

- 2.5 + 3.2 = - 5 + 6 =  1
     2
------------------------------------------
c) Método da substituição:

x = - 2y
x - 3y = 17,5

Substituir "I" em "II":

   x - 3y = 17,5
- 2y - 3y = 17,5
- 5y = 17,5
5y = - 17,5
y = - 17,5/5
y = - 3,5

x = - 2y
x = - 2.(-3,5)
x = 2.3,5
x = 7
--------------------------------------------
Substituição:

d)
2x + 5y = 14    (I)
x = y              (II)

2x + 5y = 14
2y + 5y = 14
7y = 14
y = 14/7
y = 2

x = y
x = 2
--------------------------------
e)

Substituição:

2x = 3y
y = 2x +4

2x = 3y

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

2x = 3y
2x = 3.(-2)
2x = - 6
x = - 6/2
x = - 3
-----------------------------
f) Método da Adição

x + 2y = 5     ( - 2)
2x + 3y = 7

- 2x - 4y = - 10
2x + 3y = 7

- y = - 3 ( - 1)
y = 3

x + 2y = 5
x + 2.3 = 5
x + 6 = 5
x = 5 - 6
x = - 1
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