8. \( \frac{3}{4}(4 x-12)+x=\frac{2}{3}(9 x-5)+1 \) 9. \( (x-2)(x+1)=0 \) 10. \( x^{2}-4 x=0 \) 11. \( x^{2}-16=0 \)

Solve the quadratic equation by following steps: - step0: Solve using square roots: \(x^{2}-16=0\) - step1: Add or subtract both sides: \(x^{2}=16\) - step2: Simplify the expression: \(x=\pm \sqrt{16}\) - step3: Simplify the expression: \(x=\pm 4\) - step4: Separate into possible cases: \(\begin{align}&x=4\\&x=-4\end{align}\) - step5: Rewrite: \(x_{1}=-4,x_{2}=4\) Solve the equation \( (x-2)(x+1)=0 \). Solve the quadratic equation by following steps: - step0: Solve by factoring: \(\left(x-2\right)\left(x+1\right)=0\) - step1: Separate into possible cases: \(\begin{align}&x-2=0\\&x+1=0\end{align}\) - step2: Solve the equation: \(\begin{align}&x=2\\&x=-1\end{align}\) - step3: Rewrite: \(x_{1}=-1,x_{2}=2\) Solve the equation \( x^{2}-4 x=0 \). Solve the quadratic equation by following steps: - step0: Solve by factoring: \(x^{2}-4x=0\) - step1: Factor the expression: \(x\left(x-4\right)=0\) - step2: Separate into possible cases: \(\begin{align}&x-4=0\\&x=0\end{align}\) - step3: Solve the equation: \(\begin{align}&x=4\\&x=0\end{align}\) - step4: Rewrite: \(x_{1}=0,x_{2}=4\) Solve the equation \( \frac{3}{4}(4 x-12)+x=\frac{2}{3}(9 x-5)+1 \). Solve the equation by following steps: - step0: Solve for \(x\): \(\frac{3}{4}\left(4x-12\right)+x=\frac{2}{3}\left(9x-5\right)+1\) - step1: Simplify: \(4x-9=6x-\frac{7}{3}\) - step2: Multiply both sides of the equation by LCM: \(\left(4x-9\right)\times 3=\left(6x-\frac{7}{3}\right)\times 3\) - step3: Calculate: \(12x-27=\left(6x-\frac{7}{3}\right)\times 3\) - step4: Calculate: \(12x-27=18x-7\) - step5: Move the expression to the left side: \(12x-18x=-7+27\) - step6: Add and subtract: \(-6x=-7+27\) - step7: Add and subtract: \(-6x=20\) - step8: Change the signs: \(6x=-20\) - step9: Divide both sides: \(\frac{6x}{6}=\frac{-20}{6}\) - step10: Divide the numbers: \(x=-\frac{10}{3}\) Here are the solutions for each of the equations provided: ### 8. Solve \( \frac{3}{4}(4 x-12)+x=\frac{2}{3}(9 x-5)+1 \) 1. Distribute the terms: \[ \frac{3}{4} \cdot 4x - \frac{3}{4} \cdot 12 + x = \frac{2}{3} \cdot 9x - \frac{2}{3} \cdot 5 + 1 \] This simplifies to: \[ 3x - 9 + x = 6x - \frac{10}{3} + 1 \] 2. Combine like terms: \[ 4x - 9 = 6x - \frac{10}{3} + 1 \] 3. Rearranging gives: \[ 4x - 6x = -\frac{10}{3} + 1 + 9 \] \[ -2x = -\frac{10}{3} + \frac{3}{3} + \frac{27}{3} \] \[ -2x = -\frac{10}{3} + \frac{30}{3} = \frac{20}{3} \] 4. Solving for \( x \): \[ x = -\frac{10}{3} \] ### 9. Solve \( (x-2)(x+1)=0 \) Setting each factor to zero: 1. \( x - 2 = 0 \) gives \( x_1 = 2 \) 2. \( x + 1 = 0 \) gives \( x_2 = -1 \) Thus, the solutions are: \[ x_1 = 2, \quad x_2 = -1 \] ### 10. Solve \( x^{2}-4 x=0 \) Factoring out \( x \): \[ x(x - 4) = 0 \] Setting each factor to zero: 1. \( x = 0 \) 2. \( x - 4 = 0 \) gives \( x_2 = 4 \) Thus, the solutions are: \[ x_1 = 0, \quad x_2 = 4 \] ### 11. Solve \( x^{2}-16=0 \) Factoring as a difference of squares: \[ (x - 4)(x + 4) = 0 \] Setting each factor to zero: 1. \( x - 4 = 0 \) gives \( x_1 = 4 \) 2. \( x + 4 = 0 \) gives \( x_2 = -4 \) Thus, the solutions are: \[ x_1 = 4, \quad x_2 = -4 \] ### Summary of Solutions: - **8:** \( x = -\frac{10}{3} \) - **9:** \( x_1 = 2, \quad x_2 = -1 \) - **10:** \( x_1 = 0, \quad x_2 = 4 \) - **11:** \( x_1 = 4, \quad x_2 = -4 \)