Writing an Equation with No, One, or Infinitely Many Solutions Write a number in each box so that the equation has the indicated number of solutions. (1) One solution \[ 2 x-1=4 x-1 \] (3) Infinitely many solutions \[ \frac{2}{5} x+3=\frac{2}{5} x+3 \] (5) One solution \[ 8 x+4=10 x-8 \] 2. No solution \[ \frac{1}{3} x+2=\frac{1}{3} x-3 \] (4) One solution \[ 2 \sqrt{2} x-2)=6 x+5 \] (6) No solution

Let's solve each of the equations step by step to ensure they meet the specified conditions of having one solution, infinitely many solutions, or no solution. ### (1) One solution The equation is: \[ 2x - 1 = 4x - 1 \] **Step 1:** Rearranging the equation: \[ 2x - 4x = -1 + 1 \] \[ -2x = 0 \] \[ x = 0 \] This equation has one solution, \( x = 0 \). ### (2) No solution The equation is: \[ \frac{1}{3} x + 2 = \frac{1}{3} x - 3 \] **Step 1:** Rearranging the equation: \[ \frac{1}{3} x - \frac{1}{3} x = -3 - 2 \] \[ 0 = -5 \] This is a contradiction, indicating that there is no solution. ### (3) Infinitely many solutions The equation is: \[ \frac{2}{5} x + 3 = \frac{2}{5} x + 3 \] **Step 1:** Rearranging the equation: \[ \frac{2}{5} x - \frac{2}{5} x = 3 - 3 \] \[ 0 = 0 \] This is always true, indicating that there are infinitely many solutions. ### (4) One solution The equation is: \[ 2\sqrt{2} x - 2 = 6x + 5 \] **Step 1:** Rearranging the equation: \[ 2\sqrt{2} x - 6x = 5 + 2 \] \[ (2\sqrt{2} - 6)x = 7 \] **Step 2:** Solving for \( x \): \[ x = \frac{7}{2\sqrt{2} - 6} \] This equation has one solution. ### (5) One solution The equation is: \[ 8x + 4 = 10x - 8 \] **Step 1:** Rearranging the equation: \[ 8x - 10x = -8 - 4 \] \[ -2x = -12 \] \[ x = 6 \] This equation has one solution, \( x = 6 \). ### (6) No solution To create an equation with no solution, we can use a similar structure to the previous no-solution example. For instance: \[ 2x + 1 = 2x - 5 \] **Step 1:** Rearranging the equation: \[ 2x - 2x = -5 - 1 \] \[ 0 = -6 \] This is a contradiction, indicating that there is no solution. ### Summary of Equations 1. One solution: \( 2x - 1 = 4x - 1 \) 2. No solution: \( \frac{1}{3} x + 2 = \frac{1}{3} x - 3 \) 3. Infinitely many solutions: \( \frac{2}{5} x + 3 = \frac{2}{5} x + 3 \) 4. One solution: \( 2\sqrt{2} x - 2 = 6x + 5 \) 5. One solution: \( 8x + 4 = 10x - 8 \) 6. No solution: \( 2x + 1 = 2x - 5 \)