Part III. Solve for x1.8x+1=43x+22.< 0x²-6x+9​

Answer:Let's solve for x in the given equations: 1.     First, we can clear the denominators by cross-multiplying to eliminate the fractions:         Expanding the parentheses:         Combining like terms and simplifying:     Solving for x:       Therefore, the solution for x in the first equation is    . 2.         This equation is a quadratic inequality. We can factorize the trinomial to solve the inequality:       Since the product of the same factor squared is always positive, there is no solution for this inequality. Therefore, there is no solution for x in the second equation. In conclusion, the solution for the first equation is    , while there is no solution for the second equation.

Answered by tanimahabduljalil | 2024-09-07

Answer:Let's solve each equation step-by-step:### **1. Solving for x in the equation:**\[\frac{8}{x+1} = \frac{4}{3(x+2)}\]**Steps:**1. **Cross-Multiply:**\[ 8 \cdot 3(x+2) = 4 \cdot (x+1)\]\[ 24(x+2) = 4(x+1)\]2. **Expand and Simplify:**\[ 24x + 48 = 4x + 4\]\[ 24x - 4x = 4 - 48\]\[ 20x = -44\]\[ x = \frac{-44}{20}\]\[ x = \frac{-11}{5}\]**Solution:** x = \frac{-11}{5}### **2. Solving the inequality:**\[x^2 - 6x + 9 < 0\]**Steps:**1. **Factor the Quadratic Expression:**\[ x^2 - 6x + 9 = (x - 3)^2\] So, the inequality becomes:\[ (x - 3)^2 < 0\]2. **Analyze the Inequality:** - The square of any real number is always non-negative (\geq 0). - Therefore, (x - 3)^2 < 0 has no real solutions because a square can never be negative.**Solution:** There are no real values of x that satisfy this inequality.

Answered by sukangmatamis | 2024-09-07