Question

PFV ME AJUDEM NESSA EQUAÇÃO!!!!

A) y+4.(y-1) =9-2.(y+3)

B)x-(x+1)=12-(3x-2)

C)2y:3 - 1:2 = y:4 - 3 : 2

Answer 1

Explicação passo-a-passo:

a)

y+4y-4=9-2y-6

5y-4=3-2y |+2y

7y-4=3 |+4

7y=7|÷7

y=1

b)

x-x-1=12-3x+2

-1=14-3x |-14

-15=-3x |÷(-3)

x=5

c)

$\frac{2y}{3} - \frac{1}{2} = \frac{y}{4} - \frac{3}{2} < + \frac{3}{2} \\ \frac{2y}{3} + \frac{2}{2} = \frac{y}{4} < - \frac{2y}{3} \\ 1 = \frac{y}{4} - \frac{2y}{ 3} \\ 1 = \frac{3y}{12} - \frac{8y}{12} \\ 1 = - \frac{5y}{12} < \times 12 \\ 12 = - 5y < \div ( - 5) \\ - 2.4 = y$

Answer 2

Explicação passo-a-passo:

Equações do primeiro grau:

A)

y + 4(y - 1) = 9 - 2(y + 3)

y + 4y - 4 = 9 - 2y - 6

5y - 4 = 3 - 2y

5y + 2y = 3 + 4

7y = 7

y = 7 ÷ 7

y = 1

B)

x - ( x + 1) = 12 - (3x - 2)

x - x - 1 = 12 - 3x + 2

-1 = 14 - 3x

3x = 14 + 1

3x = 15

x = 15 ÷ 3

x = 5

C)

$\mathsf{\dfrac{2y}{3}-\dfrac{1}{2}~=~\dfrac{y}{4}-\dfrac{3}{2} }$

$\mathsf{\dfrac{2y}{3}-\dfrac{y}{4}~=~\dfrac{1}{2}-\dfrac{3}{2} }$

$\mathsf{\dfrac{8y-3y}{12}~=~-\dfrac{2}{2} }$

$\mathfsf{5y~=~-12 }$

$\mathsf{y~=~-\dfrac{12}{5} }$

Espero ter ajudado bastante!)