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Matemáticas Académicas
 Marta Martín Sierra
Resuelve las siguientes ecuaciones:
004 7(x – 1) + 2(x – 1) – 3(x – 1) – x = – 5 (x – 1) – 1
RESOLUCIÓN:
7x – 7 + 2x – 2 – 3x + 3 – x = – 5x + 5 – 1
7x + 2x – 3x – x + 5x = 5 – 1 + 7 + 2 – 3
10x = 10
x = 10/10
x = 1
008 1 – 2(1 – 2x) – 3x + 5 = x + 3 + 2 + 1 – (x – 2) + x – 2
RESOLUCIÓN:
1 – 2 + 4x – 3x + 5 = x + 3 + 2 + 1 – x + 2 + x – 2
+ 4x – 3x – x + x – x = 3 + 2 + 1 + 2 – 2 – 5 + 2 – 1
0x = 2
0 = 2
Pero como 0 ≠ 2 . Incoherencia; no existe ningún valor de "x" que verifique la igualdad
del enunciado. Incompatible.
009 2 + 7x – 1 – 3x + 5 = 8x + 6 + 2(x – 1) + 2
RESOLUCIÓN:
2 + 7x – 1 – 3x + 5 = 8x + 6 + 2x – 2 + 2
7x – 3x – 8x – 2x = 6 – 2 + 2 – 2 + 1 – 5
– 6x = 0
6x = 0
x = 0
011 2 (x + 5) – x = 3 (5 – x) + 3 – 2(4 – 2x)
RESOLUCIÓN:
2x + 10 – x = 15 – 3x + 3 – 8 + 4x
2x – x + 3x – 4x = 15 + 3 – 8 – 10
0x = 0
0 = 0
Se verifica para cualquier valor de x. Se trata de una identidad
012 6x – (x + 2) = 4x – 1 – (3x + 1)
RESOLUCIÓN:
6x – x – 2 = 4x – 1 – 3x – 1
6x – x – 4x + 3x = 2 – 1 – 1
4x = 0
x = 0/4
x = 0
018 – 3(x – 3) + 2(– x – 1) = – 3(– x – 1) – 2
RESOLUCIÓN:
– 3x + 9 – 2x – 2 = 3x + 3 – 2
– 3x– 2x – 3x = 3 – 2 – 9 + 2
– 8x = – 6
8x = 6
Ecuaciones polinómicas de grado 1
Sencillas, con paréntesis y con denominadores
x = 6/8 = 3/4
x = 3/4
008
3
x
– 2 =
5
x
– 1
RESOLUCIÓN:
mcm: 15
5x – 30 = 3x – 15
5x – 3x = 30 – 15
2x = 15
x = 15/2
009 x +
6
1
=
3
2x
–
2
1
RESOLUCIÓN:
mcm: 6
6x + 1 = 4x – 3
6x – 4x = – 1 – 3
2x = – 4
x = – 4/2
x = – 2
014
3
4x
–
9
5x
= 2 +
3
x
RESOLUCIÓN:
mcm: 9
12x – 5x = 18 + 3x
12x – 5x – 3x = 18
4x = 18
x = 18/4
x = 9/2
015
7
x
– 1 = 7 – x
RESOLUCIÓN:
mcm: 7
x – 7 = 49 – 7x
x + 7x = 49 + 7
8x = 56
x = 7
018
4
3x
– 1 = 3x +
2
1
RESOLUCIÓN:
mcm : 4
3x – 4 = 12x + 2
3x – 12x = 2 + 4
– 9x = 6
9x = – 6
x = – 6/9
x = – 2/3

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Ecuaciones grado1 blog

  • 1. Matemáticas Académicas  Marta Martín Sierra Resuelve las siguientes ecuaciones: 004 7(x – 1) + 2(x – 1) – 3(x – 1) – x = – 5 (x – 1) – 1 RESOLUCIÓN: 7x – 7 + 2x – 2 – 3x + 3 – x = – 5x + 5 – 1 7x + 2x – 3x – x + 5x = 5 – 1 + 7 + 2 – 3 10x = 10 x = 10/10 x = 1 008 1 – 2(1 – 2x) – 3x + 5 = x + 3 + 2 + 1 – (x – 2) + x – 2 RESOLUCIÓN: 1 – 2 + 4x – 3x + 5 = x + 3 + 2 + 1 – x + 2 + x – 2 + 4x – 3x – x + x – x = 3 + 2 + 1 + 2 – 2 – 5 + 2 – 1 0x = 2 0 = 2 Pero como 0 ≠ 2 . Incoherencia; no existe ningún valor de "x" que verifique la igualdad del enunciado. Incompatible. 009 2 + 7x – 1 – 3x + 5 = 8x + 6 + 2(x – 1) + 2 RESOLUCIÓN: 2 + 7x – 1 – 3x + 5 = 8x + 6 + 2x – 2 + 2 7x – 3x – 8x – 2x = 6 – 2 + 2 – 2 + 1 – 5 – 6x = 0 6x = 0 x = 0 011 2 (x + 5) – x = 3 (5 – x) + 3 – 2(4 – 2x) RESOLUCIÓN: 2x + 10 – x = 15 – 3x + 3 – 8 + 4x 2x – x + 3x – 4x = 15 + 3 – 8 – 10 0x = 0 0 = 0 Se verifica para cualquier valor de x. Se trata de una identidad 012 6x – (x + 2) = 4x – 1 – (3x + 1) RESOLUCIÓN: 6x – x – 2 = 4x – 1 – 3x – 1 6x – x – 4x + 3x = 2 – 1 – 1 4x = 0 x = 0/4 x = 0 018 – 3(x – 3) + 2(– x – 1) = – 3(– x – 1) – 2 RESOLUCIÓN: – 3x + 9 – 2x – 2 = 3x + 3 – 2 – 3x– 2x – 3x = 3 – 2 – 9 + 2 – 8x = – 6 8x = 6
  • 2. Ecuaciones polinómicas de grado 1 Sencillas, con paréntesis y con denominadores x = 6/8 = 3/4 x = 3/4 008 3 x – 2 = 5 x – 1 RESOLUCIÓN: mcm: 15 5x – 30 = 3x – 15 5x – 3x = 30 – 15 2x = 15 x = 15/2 009 x + 6 1 = 3 2x – 2 1 RESOLUCIÓN: mcm: 6 6x + 1 = 4x – 3 6x – 4x = – 1 – 3 2x = – 4 x = – 4/2 x = – 2 014 3 4x – 9 5x = 2 + 3 x RESOLUCIÓN: mcm: 9 12x – 5x = 18 + 3x 12x – 5x – 3x = 18 4x = 18 x = 18/4 x = 9/2 015 7 x – 1 = 7 – x RESOLUCIÓN: mcm: 7 x – 7 = 49 – 7x x + 7x = 49 + 7 8x = 56 x = 7 018 4 3x – 1 = 3x + 2 1 RESOLUCIÓN: mcm : 4 3x – 4 = 12x + 2 3x – 12x = 2 + 4 – 9x = 6 9x = – 6 x = – 6/9 x = – 2/3