Question

Resolva as equações e determine a quais conjuntos numéricos (N , Z , Q , I ou R) pertence a solução de cada uma delas.
a) 8b - 3 = 5b + 6
b) 2y + 9 = 10 - y
c) z- 7 = 1 + 2z
d) 5 + 4a = a - 3
e) 9c + 6/5 = 8 - c
f) 5(2x - 1) = 3(x - 4)

Answer 1

a)8b-5b=6+3
3b=9
b=9/3
b=3 pertence a N
b)2y+y=10-9
3y=1
y=1/3
y=0,333 pertence a Q
c)z-2z=1+7
-z=8 . (-1)
z=-8/1
z=-8 pertence a I
d)4a-a=-3-5
3a=-8
a=-8/3
a=2.666 pertence a Q
e)9c-c=8-6/5
8c=8-1,2
c=6,8/8
c=0.85 pertence a R
f)10x-2=3x-12
10x-3x=-12+2
7x=-10
x=-10/7
x=1.42857143 pertence a Z
Espero ter ajudado.

Answer 2

$a) ~~8b - 3 = 5b + 6\to~8b-5b=6+3\to~ 3b=9\to~b= \dfrac{9}{3} \to~\boxed{b=3}\\\\ \large\boxed{\mathbb{N}} \\\\\\\\ b)~~ 2y + 9 = 10 - y\to~2y+y=10-9\to~3y=1\to~\boxed{y= \frac{1}{3} }\\\\ \large\boxed{\mathbb{Q}} \\\\\\\\c) ~~z- 7 = 1 + 2z\to~z-2z=1+7\to~-z = 8~(-1)\to\boxed{z=-8}\\\\\\ \large\boxed{\mathbb{Z}}$



$d)~~ 5 + 4a = a - 3\to~4a-a=-3-5\to~3a=-8\to~\boxed{a=- \frac{8}{3}} \\\\\large\boxed{\mathbb{Q}} \\\\\\\\e) ~~9c + \dfrac{6}{5} = 8 - c\to~45c+6=40-5c\to~45c+5c=40-6\to\\\\ 50c=36\to~c= \dfrac{36}{50} \to~ \boxed{c= \dfrac{18}{25} }\\\\ \large\boxed{\mathbb{Q}}\\\\\\\\ f) ~~5(2x - 1) = 3(x - 4) \to~10x-5=3x-12\to\\\\10x-3x=-12+5\to~7x=-7\to~x=- \dfrac{7}{7} \to~\boxed{x=-1}\\\\ \large\boxed{\mathbb{Z}}$