Question

aplicar la propiedad distributiva y resolver
a) (x+2)5=20
B)(x-6)7=49
c)3(x+2+4)=21
d)(2x+5)4=100
e)(x+6-5)3=12
f)5(2+3+x)=10

Answer 1

Respuesta:

Explicación paso a paso:

a) (x+2)5=20

(5)(x) + (2)(5) = 20

5x + 10 = 20

5x = 20 - 10

5x = 10

x = 2

b) (x-6)7=49

(7)(x) + (-6)(7) = 49

7x - 42 = 49

7x = 49 + 42

7x = 91

x = 13

c) 3(x+2+4)=21

(3)(x) + (3)(2) + (3)(4) = 21

3x + 6 + 12 = 21

3x + 18 = 21

3x = 21 - 18

3x = 3

x = 1

d) (2x+5)4=100

(2x)(4) + (5)(4) = 100

8x + 20 = 100

8x = 100 - 20

8x = 80

x = 10

e) (x+6-5)3=12

(3)(x) + (3)(6) + (3)(-5) = 12

3x + 18 - 15 = 12

3x + 3 = 12

3x = 12 - 3

3x = 9

x = 3

f) 5(2+3+x)=10

(5)(2) + (5)(3) + (5)(x) = 10

10 + 15 + 5x = 10

25 + 5x = 10

5x = 10 - 25

5x = -15

x = -3