Solve for \( x \), then represent your solution on a number line and in interval notation where possible: \( \begin{array}{ll}\text { 1. }-2 \leq 2 x-4<2 ; x \in \mathbb{N} & \text { 2. }-2<-x+3<10 ; x \in \mathbb{N}_{0} \\ \text { 3. }-4 \leq 8-3 x<11 ; x \in \mathbb{Z} & \text { 4. } 2+3 x>4 x-7 ; x \in \mathbb{R} \\ \text { 5. } 6-5 x \geq 4 x-9 ; x \in \mathbb{R} & \text { 6. } \frac{2}{3} x+2>2 ; x \in \mathbb{R} \\ \text { 7. Solve for } m: \frac{3}{2}(2 m+3)-\frac{4 m}{3}<2-\frac{8 m-3}{6} ; m \in \mathbb{R}\end{array} \)

Let’s solve these inequalities one by one. 1. For \( -2 \leq 2x - 4 < 2 \): - From \( -2 \leq 2x - 4 \): \[ 2x \geq 2 \rightarrow x \geq 1 \] - From \( 2x - 4 < 2 \): \[ 2x < 6 \rightarrow x < 3 \] - Combining both conditions: \( 1 \leq x < 3 \), so \( x = 1, 2 \). 2. For \( -2 < -x + 3 < 10 \): - From \( -2 < -x + 3 \): \[ -5 < -x \rightarrow x < 5 \] - From \( -x + 3 < 10 \): \[ -x < 7 \rightarrow x > -7 \] - As we need \( x \in \mathbb{N}_{0} \), we consider \( x = 0, 1, 2, 3, 4 \). 3. For \( -4 \leq 8 - 3x < 11 \): - From \( -4 \leq 8 - 3x \): \[ -12 \leq -3x \rightarrow x \leq 4 \] - From \( 8 - 3x < 11 \): \[ -3x < 3 \rightarrow x > -1 \] - So \( -1 < x \leq 4 \), leading to \( x = 0, 1, 2, 3, 4 \). 4. For \( 2 + 3x > 4x - 7 \): \[ 9 > x \rightarrow x < 9 \] - Therefore, \( x \in (-\infty, 9) \). 5. For \( 6 - 5x \geq 4x - 9 \): \[ 15 \geq 9x \rightarrow x \leq \frac{5}{3} \] - Thus, \( x \in (-\infty, \frac{5}{3}] \). 6. For \( \frac{2}{3}x + 2 > 2 \): \[ \frac{2}{3}x > 0 \rightarrow x > 0 \] - So, \( x \in (0, \infty) \). 7. For \( \frac{3}{2}(2m + 3) - \frac{4m}{3} < 2 - \frac{8m - 3}{6} \): - Simplifying this gives a linear expression leading to \( m < \frac{1}{4} \). - Thus, \( m \in (-\infty, \frac{1}{4}) \). To represent each solution on a number line, we can have points for the specific integers involved and shade the valid regions: 1. Points: 1, 2 (shaded area in between). 2. Points: 0, 1, 2, 3, 4 (shaded area). 3. Points: 0, 1, 2, 3, 4 (shaded area). 4. Shade everything to the left of 9. 5. Shade everything to the left of or on \(\frac{5}{3}\). 6. Shade to the right of 0. 7. Shade to the left of \(\frac{1}{4}\). In interval notation, the solutions can be stated as follows: 1. \([1, 3)\) 2. \([0, 5)\) 3. \((0, 4]\) 4. \((-\infty, 9)\) 5. \((-\infty, \frac{5}{3}]\) 6. \((0, \infty)\) 7. \((-\infty, \frac{1}{4})\)