Question 7. Find the value of \( m \) for which the pair of simultaneous equations \( 3 x+m y=5 \) and \( (m+2) x+5 y=m \) have: a. infinitely many solutions - Workspace

To determine the value of \( m \) for which the given system of equations has infinitely many solutions, we'll analyze the conditions under which two linear equations represent the same line. When two equations are multiples of each other, they depict the same line, leading to infinitely many solutions. **Given Equations:** 1. \( 3x + my = 5 \)     (Equation 1) 2. \( (m + 2)x + 5y = m \)  (Equation 2) **Condition for Infinitely Many Solutions:** For the system to have infinitely many solutions, the ratios of the corresponding coefficients of \( x \), \( y \), and the constants must be equal. Mathematically, this means: \[ \frac{3}{m + 2} = \frac{m}{5} = \frac{5}{m} \] **Step-by-Step Solution:** 1. **Set Up the Ratios:** \[ \frac{3}{m + 2} = \frac{m}{5} \] \[ \frac{m}{5} = \frac{5}{m} \] 2. **Solve the Second Ratio:** \[ \frac{m}{5} = \frac{5}{m} \] Cross-multiplying gives: \[ m^2 = 25 \] \[ m = \pm 5 \] 3. **Check Both Possible Values of \( m \):** - **Case 1: \( m = 5 \)** Substitute \( m = 5 \) into the first ratio: \[ \frac{3}{5 + 2} = \frac{3}{7} \quad \text{and} \quad \frac{5}{5} = 1 \] \[ \frac{3}{7} \neq 1 \quad \text{(Not equal)} \] *Conclusion:* \( m = 5 \) does **not** satisfy the condition for infinitely many solutions. - **Case 2: \( m = -5 \)** Substitute \( m = -5 \) into the first ratio: \[ \frac{3}{-5 + 2} = \frac{3}{-3} = -1 \quad \text{and} \quad \frac{-5}{5} = -1 \quad \text{and} \quad \frac{5}{-5} = -1 \] All ratios are equal to \(-1\). *Conclusion:* \( m = -5 \) satisfies the condition for infinitely many solutions. **Final Answer:** The value of \( m \) that leads to infinitely many solutions is **\( m = -5 \)**.