Résoudre des équations du premier degré à une inconnue

A) \( 4x - 5 = 2x + 3 \)
\( 4x - 5 + (-2x) = 2x + 3 + (-2x) \)
\( 2x - 5 = 3 \)
\( 2x - 5 + 5 = 3 + 5 \)
\( 2x = 8 \)
\( \frac{1}{2} \times 2x = \frac{1}{2} \times 8 \)
\( x = 4 \). L'ensemble solution est \( S = \{4\} \).

B) \( 7 - 3x = 1 - 5x \)
\( 7 - 3x + 5x = 1 - 5x + 5x \)
\( 7 + 2x = 1 \)
\( 7 + 2x + (-7) = 1 + (-7) \)
\( 2x = -6 \)
\( \frac{1}{2} \times 2x = \frac{1}{2} \times (-6) \)
\( x = -3 \). L'ensemble solution est \( S = \{-3\} \).

C) \( 5x + 1 = 5x - 8 \)
\( 5x - 5x = -8 - 1 \)
\( 0x = -9 \). C'est impossible.
L'équation n'a pas de solution. \( S = \emptyset \)

D) \( 2(x + 3) = 2x + 6 \)
\( 2x + 6 = 2x + 6 \)
\( 0x = 0 \). C'est toujours vrai.
Tous les nombres réels sont solution. \( S = \mathbb{R} \)

E) \( 10x - 4 = 6x \)
\( 10x - 4 + (-6x) = 6x + (-6x) \)
\( 4x - 4 = 0 \)
\( 4x - 4 + 4 = 0 + 4 \)
\( 4x = 4 \)
\( \frac{1}{4} \times 4x = \frac{1}{4} \times 4 \)
\( x = 1 \). L'ensemble solution est \( S = \{1\} \).

A) \( 3(x - 2) = 5(x + 4) \)
\( 3x - 6 = 5x + 20 \)
\( -2x = 26 \)
\( (-\frac{1}{2}) \times (-2x) = (-\frac{1}{2}) \times 26 \)
\( x = -13 \). L'ensemble solution est \( S = \{-13\} \).

B) \( 2(x + 1) - (x - 3) = 4 \)
\( 2x + 2 - x + 3 = 4 \)
\( x + 5 = 4 \)
\( x = -1 \). L'ensemble solution est \( S = \{-1\} \).

C) \( 7x - 4(x - 2) = 2(x + 3) \)
\( 7x - 4x + 8 = 2x + 6 \)
\( 3x + 8 = 2x + 6 \)
\( x = -2 \). L'ensemble solution est \( S = \{-2\} \).

D) \( 5(3x - 1) = 3(5x - 2) \)
\( 15x - 5 = 15x - 6 \)
\( -5 = -6 \). C'est impossible.
L'équation n'a pas de solution. \( S = \emptyset \)

E) \( x - (2x + 5) = 3(x - 1) \)
\( x - 2x - 5 = 3x - 3 \)
\( -x - 5 = 3x - 3 \)
\( -4x = 2 \)
\( (-\frac{1}{4}) \times (-4x) = (-\frac{1}{4}) \times 2 \)
\( x = -1/2 \). L'ensemble solution est \( S = \{-1/2\} \).

A) \( \frac{x}{2} + \frac{1}{3} = \frac{x}{3} + 1 \)
\( \frac{3x}{6} + \frac{2}{6} = \frac{2x}{6} + \frac{6}{6} \)
\( 3x + 2 = 2x + 6 \)
\( x = 4 \). L'ensemble solution est \( S = \{4\} \).

B) \( \frac{2x}{3} - 5 = \frac{x}{4} \)
\( \frac{8x}{12} - \frac{60}{12} = \frac{3x}{12} \)
\( 8x - 60 = 3x \)
\( 5x = 60 \)
\( \frac{1}{5} \times 5x = \frac{1}{5} \times 60 \)
\( x = 12 \). L'ensemble solution est \( S = \{12\} \).

C) \( \frac{x + 1}{2} = \frac{2x - 1}{3} \)
\( \frac{3(x+1)}{6} = \frac{2(2x-1)}{6} \)
\( 3(x + 1) = 2(2x - 1) \)
\( 3x + 3 = 4x - 2 \)
\( -x = -5 \)
\( x = 5 \). L'ensemble solution est \( S = \{5\} \).

D) \( \frac{1}{4}x - 2 = \frac{3}{2} \)
\( \frac{x}{4} - \frac{8}{4} = \frac{6}{4} \)
\( x - 8 = 6 \)
\( x = 14 \). L'ensemble solution est \( S = \{14\} \).

E) \( \frac{3}{5}x + \frac{1}{2} = \frac{1}{5}x - \frac{3}{2} \)
\( \frac{6x}{10} + \frac{5}{10} = \frac{2x}{10} - \frac{15}{10} \)
\( 6x + 5 = 2x - 15 \)
\( 4x = -20 \)
\( \frac{1}{4} \times 4x = \frac{1}{4} \times (-20) \)
\( x = -5 \). L'ensemble solution est \( S = \{-5\} \).

A) \( \frac{3(x-1)}{4} = \frac{x+2}{2} \)
\( 2 \times 3(x-1) = 4(x+2) \)
\( 6x - 6 = 4x + 8 \)
\( 2x = 14 \)
\( \frac{1}{2} \times 2x = \frac{1}{2} \times 14 \)
\( x=7 \). L'ensemble solution est \( S = \{7\} \).

B) \( 2x - \frac{1}{3}(x+1) = 0 \)
\( 2x - \frac{x}{3} - \frac{1}{3} = 0 \)
\( \frac{5x}{3} = \frac{1}{3} \)
\( \frac{3}{5} \times \frac{5x}{3} = \frac{3}{5} \times \frac{1}{3} \)
\( x=1/5 \). L'ensemble solution est \( S = \{1/5\} \).

C) \( (x+1)(x-2) = x^2 + 5 \)
\( x^2 - 2x + x - 2 = x^2 + 5 \)
\( x^2 - x - 2 = x^2 + 5 \)
\( -x = 7 \)
\( (-1) \times (-x) = (-1) \times 7 \)
\( x = -7 \). L'ensemble solution est \( S = \{-7\} \).

D) \( 5(x - \frac{1}{2}) = 2(x + \frac{3}{2}) \)
\( 5x - \frac{5}{2} = 2x + 3 \)
\( 3x = 3 + \frac{5}{2} = \frac{11}{2} \)
\( \frac{1}{3} \times 3x = \frac{1}{3} \times \frac{11}{2} \)
\( x = 11/6 \). L'ensemble solution est \( S = \{11/6\} \).

E) \( \frac{x}{2} - \frac{x}{3} + \frac{x}{4} = \frac{5}{6} \)
\( \frac{6x - 4x + 3x}{12} = \frac{5}{6} \)
\( \frac{5x}{12} = \frac{5}{6} \)
\( \frac{12}{5} \times \frac{5x}{12} = \frac{12}{5} \times \frac{5}{6} \)
\( x=2 \). L'ensemble solution est \( S = \{2\} \).

A) Soit \(n\) le premier entier.
\( n + (n+1) + (n+2) = 78 \)
\( 3n + 3 = 78 \Rightarrow 3n = 75 \)
\( \frac{1}{3} \times 3n = \frac{1}{3} \times 75 \Rightarrow n = 25 \).
Les nombres sont 25, 26 et 27.

B) Soit \(x\) le nombre d'années.
\( 40 + x = 3(12 + x) \)
\( 40 + x = 36 + 3x \Rightarrow 4 = 2x \)
\( \frac{1}{2} \times 4 = \frac{1}{2} \times 2x \Rightarrow x = 2 \).
Ce sera dans 2 ans.

C) Soit \(l\) la largeur. La longueur est \(2l\).
Périmètre : \( 2(l + 2l) = 42 \)
\( 6l = 42 \)
\( \frac{1}{6} \times 6l = \frac{1}{6} \times 42 \Rightarrow l=7 \).
La largeur est de 7 cm et la longueur de 14 cm.

D) Soit \(x\) le nombre.
\( 3(x+5) = 45 \)
\( \frac{1}{3} \times 3(x+5) = \frac{1}{3} \times 45 \)
\( x+5 = 15 \Rightarrow x = 10 \).
Le nombre est 10.

E) Soit \(a\) la part d'Arthur.
Bérénice reçoit \(2a\). Chloé reçoit \(2a+10\).
\( a + 2a + (2a+10) = 120 \)
\( 5a + 10 = 120 \Rightarrow 5a = 110 \)
\( \frac{1}{5} \times 5a = \frac{1}{5} \times 110 \Rightarrow a=22 \).
Arthur reçoit 22€, Bérénice 44€, et Chloé 54€.