Question

5- (Valor da questão: 14,0) Resolva as equações do 1o grau com 1 incógnita.

a) −6x = −4

b) 3n − 17 = 7

c) 4y + 12 = y − 10

d) 5 − 7x = 2x − 5

e) 5x − 3 + x = 3x − 2 − 1

f) 4 − 2x = x + 31

g) 7a − (2a − 1) = 4 + (a − 6)

h) 5 − 3(x − 2) = 8 − (4 − 2x)

i) −2(3x − 4 − x) = 26 − 7x

j) 1

2

(x − 3) =

1

2

+ (x + 3)


k) 2

3

(x − 2) = 3(x − 2)

l) 3 (

2x

3

− 1) = x + 4

m) 7 − 6x = 1 + 3(5 + 2x)

n) 2(4 − 6x) = −4(1 − x)

eu dou melhor resposta para qm acertar
tbm coração e 5 estrelas

Answer 1

a) −6x = −4

$\frac{-6x}{-6}=\frac{-4}{-6}\\x=\frac{2}{3}$

b) 3n − 17 = 7

$3n-17+17=7+17\\3n=24\\\frac{3n}{3}=\frac{24}{3}\\n=8$

c) 4y + 12 = y − 10

$4y+12-12=y-10-12\\4y=y-22\\4y-y=y-22-y\\3y=-22\\\frac{3y}{3}=\frac{-22}{3}\\y=-\frac{22}{3}$

d) 5 − 7x = 2x − 5

$5-7x-5=2x-5-5\\-7x=2x-10\\-7x-2x=2x-10-2x\\-9x=-10\\\frac{-9x}{-9}=\frac{-10}{-9}\\x=\frac{10}{9}$

e) 5x − 3 + x = 3x − 2 − 1

$5x+x-3=3x-2-1\\6x-3=3x-2-1\\6x-3=3x-3\\6x-3+3=3x-3+3\\6x=3x\\3x=0\\\frac{3x}{3}=\frac{0}{3}\\x=0$

f) 4 − 2x = x + 31

$4-2x-4=x+31-4\\-2x=x+27\\-2x-x=x+27-x\\-3x=27\\\frac{-3x}{-3}=\frac{27}{-3}\\x=-9$

g) 7a − (2a − 1) = 4 + (a − 6)

$7a-\left(2a-1\right)=4+a-6\\7a-\left(2a-1\right)=a+4-6\\7a-\left(2a-1\right)=a-2\\5a+1=a-2\\5a+1-1=a-2-1\\5a=a-3\\5a-a=a-3-a\\4a=-3\\\frac{4a}{4}=\frac{-3}{4}\\a=-\frac{3}{4}$

h) 5 − 3(x − 2) = 8 − (4 − 2x)

$-3x+11=2x+4\\\\-3x+11-11=2x+4-11\\-3x=2x-7\\-3x-2x=2x-7-2x\\-5x=-7\\\frac{-5x}{-5}=\frac{-7}{-5}\\x=\frac{7}{5}$

i) −2(3x − 4 − x) = 26 − 7x

$-4x+8=26-7x\\-4x+8-8=26-7x-8\\-4x=-7x+18\\-4x+7x=-7x+18+7x\\3x=18\\\frac{3x}{3}=\frac{18}{3}\\x=6$

j) 12(x - 3) = 12 +(x + 3)

$12\left(x-3\right)=12+x+3\\12\left(x-3\right)=x+12+3\\12\left(x-3\right)=x+15\\12x-36=x+15\\12x-36+36=x+15+36\\12x=x+51\\11x=51\\\frac{11x}{11}=\frac{51}{11}\\x=\frac{51}{11}$

k) 23(x - 2) = 3(x - 2)

$23x-46=3x-6\\23x-46+46=3x-6+46\\23x=3x+40\\23x-3x=3x+40-3x\\20x=40\\\frac{20x}{20}=\frac{40}{20}\\x=2$

l) 3(2x3 - 1) = x + 4

$3\left(\left(2x^3-1\right)=x+4\right)\\=3\left(=x\left(2x^3-1\right)+4\right)\\3\left(2x^3-1\right)=x+3\cdot \:4\\6x^3-3=x+12$

m) 7 − 6x = 1 + 3(5 + 2x)

$7-6x=6x+16\\7-6x-7=6x+16-7\\-6x=6x+9\\-6x-6x=6x+9-6x\\-12x=9\\\frac{-12x}{-12}=\frac{9}{-12}\\x=-\frac{3}{4}$

n)2(4 − 6x) = −4(1 − x)

$8-12x=-4+4x\\8-12x-8=-4+4x-8\\-12x=4x-12\\-12x-4x=4x-12-4x\\-16x=-12\\\frac{-16x}{-16}=\frac{-12}{-16}\\x=\frac{3}{4}$