\[ 3x = 17 – 5 \] \[ 3x = 12 \] \[ x = \frac{12}{3} \] \[ \boxed{x = 4} \]

\[ 2x – 6 = 5x + 1 \] \[ 2x – 5x = 1 + 6 \] \[ -3x = 7 \] \[ x = -\frac{7}{3} \] \[ \boxed{x \approx -2,33} \]

\[ \frac{x}{2} – \frac{2x}{5} = -1 – 3 \] \[ \frac{5x – 4x}{10} = -4 \] \[ \frac{x}{10} = -4 \] \[ \boxed{x = -40} \]

\[ 4x \leq 12 \] \[ x \leq 3 \]

Solution : \(\boxed{x \in ]-\infty; 3]}\)

Soit \( x \) = âge de Karim

\[ x + (x + 5) = 31 \] \[ 2x + 5 = 31 \] \[ 2x = 26 \] \[ \boxed{x = 13 \text{ ans}} \]

Ahmed a donc \( 13 + 5 = 18 \) ans.

Racines : \( x = 3 \) et \( x = -3 \)

Intervalle	Signe
]−∞; −3]	+
[−3; 3]	−
[3; +∞[	+

Solution : \(\boxed{x \in ]-\infty; -3] \cup [3; +\infty[}\)

\[ 2(2x – 1) = 3(x + 4) \] \[ 4x – 2 = 3x + 12 \] \[ 4x – 3x = 12 + 2 \] \[ \boxed{x = 14} \]

Valeurs interdites : \( x = -1 \)

Racine : \( x = 2 \)

Intervalle	Signe
]−∞; −1[	+
]−1; 2[	−
]2; +∞[	+

Solution : \(\boxed{x \in ]-1; 2[}\)

Soit \( x \) = largeur, alors \( 2x \) = longueur

\[ 2(x + 2x) = 30 \] \[ 2(3x) = 30 \] \[ 6x = 30 \] \[ \boxed{x = 5 \text{ cm}} \]

Dimensions : largeur = 5 cm, longueur = 10 cm

Par substitution :

\[ x = y + 1 \] \[ 2(y + 1) + 3y = 12 \] \[ 2y + 2 + 3y = 12 \] \[ 5y = 10 \] \[ \boxed{y = 2} \] \[ x = 2 + 1 \] \[ \boxed{x = 3} \]

Solution : \(\boxed{(3; 2)}\)