prologos





PÁGINA 141


E cuaciones sencillas

1 Resuelve mentalmente.

a) x + 4 = 5
b) x – 3 = 6
c) 7 + x = 10
d) 7 – x = 5
e) 11 = x + 5
f ) 2 = x – 9
g) 5 = 2 + x
h) 9 = 15 – x
i) 2 – x = 9
a) x = 1
b) x = 9
c) x = 3
d) x = 2
e) x = 6
f ) x = 11
g) x = 3
h) x = 6
i) x = –7


2 Resuelve.
a) 2x + x = 5 b) 7x – 3x = 10 – 7
c) x – 9x = 9 – 7 d) 5x – x = 3 – 5
e) 6= 12x – 2x f ) 2 – 8 = x + 2x
g) 5x – 13x = 6 – 10 h) 2x + 4 + 5x = 18
i) 11x + 17 – 6x = 2 j) 9 = 12x – 6 – 7x
k) 2x – 5 + 3x + 1 = 3x – 2 l) x + 7 = 12x – 3 – 8x + 1 m) 6x – 1 + x = 4 – 5x + 3 n) x + 2x + 3x – 5 = 4x – 9 ñ) 5x + 4 – 6x = 7 – x – 3 o) 4x +2 + 7x = 10x + 3 + x

a) x = 5
3

b) x = 3
4

c) x = – 1
4

d) x = – 1
2

e) x = 3
5

f ) x = –2 g) x = 1
2h) x = 2
i) x = –3 j) x = 3 k) x = 1 l) x = 3
m) x = 2
3

n) x = –2
ñ) Es una identidad. Tiene infinitas soluciones. o) Incompatible. Sin solución.

3 Quita paréntesis y resuelve.
a) 6(x + 1) – 4x = 5x – 9 b) 18x – 13 = 8 – 4(3x – 1)
c) 3x + 5(2x – 1) = 8 – 3(4 – 5x) d) 5 – (4x + 6) = 3x + (7 – 4x) e) x – 7(2x + 1) = 2(6 – 5x) – 13 f ) 11 – 5(3x + 2) + 7x = 1 – 8x g) 13x – 5(x + 2) =4(2x – 1) + 7
a) 6x + 6 – 4x = 5x – 9 8 15 = 3x 8 x = 5

b) 18x – 13 = 8 – 12x + 4 8 30x = 25 8 x = 5
6



c) 3x + 10x – 5 = 8 – 12 + 15x 8 –1 = 2x 8 x = – 1
2

d) 5 – 4x – 6 = 3x + 7 – 4x 8 –8 = 3x 8 x = – 8
3
e) x – 14x – 7 = 12 – 10x – 13 8 –6 = 3x 8 x = –2

f ) 11 – 15x – 10 + 7x = 1 – 8x 8 1 – 8x = 1 – 8x 8
8 Identidad. Infinitas soluciones.

g) 13x – 5x – 10 = 8x – 4 +7 8 8x – 10 = 8x + 3 8
8 Incompatible. No tiene solución.


E cuaciones de primer grado con denominadores

4 Quita denominadores y resuelve.
a) x + 1 = x
b) 5x + 1 = 5 + x
3 3 3 6
c) 3x – 1 = x – 7x – 1
d) x + 4 – x = 1 – 7x
5 4 10 5
3 15
6 10
e) 7x – 1 – x = x + 5x + 1 f ) x + 1 – x = x – 2 + 5
4 8 8
2 6 3 6 3 6

a) 3x + 1 = x 8x = – 1
2

c) 12x – 5 = 20x – 14x – 4 8 x = 1
6
b) 10x + 6 = 5 + 6x 8 x = – 1
4

d) 10x + 8 – 30x = 5 – 21x 8 x = –3
e) 14x – 8 – x = 8x + 5x + 8 8 0x = 16 8 Sin solución.
f) 3x +1 – 2x = x – 4 + 5 8 x +1 = x +1 8 Identidad. Tiene infinitas soluciones.

5 Elimina los paréntesis y los denominadores y resuelve. a) 2x – 5 = 1 (x – 3) b) 5 (2x – 1) – x = x
2 2 6 6
5 ( 5 ) 36
c) x – 1 = 2 x – 4
d) x – 1 = 1 (2x – 5)


a) 4x – 5 = x – 3 8 x = 2
3

b) 5(2x – 1) – 6x = x 8 10x – 5 – 6x = x 8 x = 5
3

c) x
5
– 1 = 2x – 8
5
8 x – 5 = 10x – 8 8 x = 1
3

d) x – 1 = x – 5
8 6x – 2 = 2x – 5 8 x = – 3
3 3 6 4



6 Resuelve las ecuaciones siguientes:
5 2 ( 5 ) 3 3
a) 1 (2 + 5x) = 1
x – 1
b) 2(x – 3) – 1 = x – 1 (x – 1)


c) 1 – 3x= 3 – 1 (x – 2) d) x – 3x = 1 (2x – 1) + x
8 4 2
(4 10 ) 2 ( 2 )
4 3 6


7 3 7
e) 5 x – 1
= 1 3x – 1
f ) 1 – 3 (x + 1) = 2x – 1

a) 2 + x = x
– 1
8 4 + 10x = 5x – 1 8 x = –1
5 2 10

b) 2x – 6 – 1 = x – x + 1

8 6x – 18 – 1 = 3x – x + 1 8 x = 5
3 3 3

c) 1 – 3x = 3 – x

+ 1 8 8 – 3x = 6 – 4x + 8 8 x = 6
8 4 2

d) x – 3x = 2x – 1 + x

8 12x– 9x = 8x – 4 + 2x 8 x = 4
4 3 3 6 7

e) 5x – 1 = 3x – 1

8 5x – 2 = 6x – 1 8 x = –1
4 2 2 4

f ) 21 – 9(x + 1) = 14x – 3 8 21 – 9x – 9 = 14x – 3 8 x = 15
23

7 Elimina denominadores y resuelve.
a) x – x – 3 = 1 b) 1 – x + 1= 2x – 1
5

c) 1 – 1 – x = x + 1
3 3

d) 3x – 1 = 3x + 2
3 2 2 4

e) 3x – 1 – 1 = 2x – 2 f ) x + 2 – 3x = x + 1
2 5 2

g) 2x +x – 3 = x – 3

h) 3x

– 1 = x – x + 1
2 4 5 2

i) x – x + 2= x

j) x – 5 + x – 2 = x – 2
5 15 3 3 5

k) x + 3 – x – 6 = 1 l)

1 – x – x – 1 = 3x – 1
5 7 3 12 4

a) 5x – (x – 3) = 5 8 5x – x + 3 = 5 8 x = 1
2

b) 3 – (x + 1) = 6x – 1 8 3 – x – 1 = 6x – 1 8 x = 3
7

c) 6 – 2(1 – x) = 6x + 3 8 6 – 2 + 2x = 6x + 3 8 x = 1
4
d) 6x – 4 = 3x + 2...