MATHEMATIQUES – EQUATIONS A UNE INCONNUE

x – 5 = 5 – x =>

x – 5 – 5 + x = 0
2x – 10 = 0
2x = 10
x = \(\frac{10}{2}\) = 5

4x • 2 + 4 = 6x – 8 =>

8x + 4 = 6x – 8
8x – 6x + 4 + 8 = 0
2x + 12 = 0
2x = -12
x = \(\frac{-12}{2}\) = 2

la réponse est -5 = -5 donc les solutions sont infinies.

10x + 5 = 10x + 2 • 7 =>

10x + 5 = 10x + 14
10x – 10x + 5 – 14 = 0
– 9 = 0 => résultat impossible

La multiplication s’effectue en premier.

4(x – 2) + x = 6 + 6x =>

4x – 8 + x = 6 + 6x
4x + x – 6x – 8 – 6 = 0
– x -14 = 0
-x = 14 (équivaut à -1x)
x = -14

5x – 3 = 3x + 5 =>

5x – 3x -3 – 5 = 0
2x – 8 = 0
2x = 8
x = \(\frac{8}{2}\) = 4

Résultat -20 = -20 donc résultat est infini.

4x • 3 – 2(x – 1) = 5x – 2 + x =>

12x – 2x + 2 = 5x – 2 + x
12x – 2x – 5x – x + 2 + 2 = 0
4x + 4 = 0
4x = -4
x = \(\frac{-4}{4}\) = -1

\(3x^2\) + 4x – 2 = 3x(x – 3) + 1 =>

\(3x^2\) + 4x – 2 = \(3x^2\) – 9x + 1
\(3x^2\) – \(3x^2\) + 4x + 9x – 2 – 1 = 0
13x – 3 = 0
13x = 3
x = \(\frac{3}{13}\)

7 – (8 – 2x) = 7x + (2x – 1) • 3 =>

7 – 8 + 2x = 7x + 6x – 3
7 – 8 – 3 + 2x – 7x – 6x = 0
2 – 11x = 0
-11 x = -2
x = \(\frac{-2}{-11}\)
Les  « – » s’annulent et deviennent des « + »
x = \(\frac{2}{11}\)