6. Resolución de ecuaciones de primer grado con paréntesis

a) 2 (x + 3) = 1

b) 3 (2 x + 5) = 4 x – 9

c) 5 = – 4 (3 x + 7)

d) -2 (x + 1) = -3 x – 9

e) 7 (3 x – 8 ) = 4 (x – 5 )

f) – 5 (2 x – 9 ) = – (2 x + 5)

g) 4 (2 – x ) – ( x – 8 ) = -4

h) 16 x – 8 (2 x – 1 ) = 7 – x

i) 2 (x – 1) – 6 (x + 1) = 3 x – 21

j) 4 (x + 2) – 18 = 28 – 2 (x + 1)

k) x – 3 (x – 1 ) = 15 + 4 (2 – x )

l) 3 – (1 – 6 x) = 2 + 4 x