Solve for \( x \) : \( \begin{array}{l}\text { 2.1.1 } \quad p x+q x=a \\ 2.1 .2 \quad 2 x^{2}-5 x+2=0\end{array} \)

To solve for \( x \) in the given equations, we will address each equation separately. ### Equation 2.1.1: \( p x + q x = a \) We can factor out \( x \) from the left side: \[ (p + q)x = a \] Now, to solve for \( x \), we divide both sides by \( p + q \) (assuming \( p + q \neq 0 \)): \[ x = \frac{a}{p + q} \] ### Equation 2.1.2: \( 2x^2 - 5x + 2 = 0 \) This is a quadratic equation, and we can solve it using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, \( a = 2 \), \( b = -5 \), and \( c = 2 \). Plugging in these values: 1. Calculate the discriminant: \[ b^2 - 4ac = (-5)^2 - 4 \cdot 2 \cdot 2 = 25 - 16 = 9 \] 2. Now apply the quadratic formula: \[ x = \frac{-(-5) \pm \sqrt{9}}{2 \cdot 2} = \frac{5 \pm 3}{4} \] This gives us two possible solutions: \[ x_1 = \frac{5 + 3}{4} = \frac{8}{4} = 2 \] \[ x_2 = \frac{5 - 3}{4} = \frac{2}{4} = \frac{1}{2} \] ### Summary of Solutions - From Equation 2.1.1, we have: \[ x = \frac{a}{p + q} \] - From Equation 2.1.2, the solutions are: \[ x = 2 \quad \text{and} \quad x = \frac{1}{2} \] Thus, the solutions for \( x \) are: \[ x = \frac{a}{p + q}, \quad x = 2, \quad x = \frac{1}{2} \]