Many Solutions Write a number in each box so that the equation has the indicated number of solutions. (1) One solution \[ 2 x-1=4 x-1 \] 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 \[ 2(2 \sqrt{2}-2)=6 x+5 \] (5) One solution \( \square \) \( x+4= \) \( \square \) \( x-8 \) (6) No solution \[ \frac{2}{3} x+7=\frac{2}{3} x+7 \]

Certainly! Let's address each equation individually to ensure they have the indicated number of solutions. For equations that require filling in numbers (denoted by boxes), I'll provide appropriate values that meet the specified conditions. --- ### 1. **One Solution** \[ 2x - 1 = 4x - 1 \] **Verification:** \[ 2x - 1 = 4x - 1 \] Subtract \(2x\) from both sides: \[ -1 = 2x - 1 \] Add \(1\) to both sides: \[ 0 = 2x \] Divide by \(2\): \[ x = 0 \] **Conclusion:** There is **one solution**, \(x = 0\), as required. --- ### 2. **No Solution** \[ \frac{1}{3}x + 2 = \frac{1}{3}x - 3 \] **Verification:** Subtract \(\frac{1}{3}x\) from both sides: \[ 2 = -3 \] This is a contradiction. **Conclusion:** There is **no solution**, as the equation leads to an impossible statement. --- ### 3. **Infinitely Many Solutions** \[ \frac{2}{5}x + 3 = \frac{2}{5}x + 3 \] **Verification:** Subtract \(\frac{2}{5}x\) from both sides: \[ 3 = 3 \] This is always true for any \(x\). **Conclusion:** There are **infinitely many solutions**, as the equation holds for all real numbers \(x\). --- ### 4. **One Solution** \[ 2(2\sqrt{2} - 2) = 6x + 5 \] **Solving for \(x\):** First, simplify the left side: \[ 2 \times 2\sqrt{2} = 4\sqrt{2} \] \[ 2 \times (-2) = -4 \] \[ 4\sqrt{2} - 4 = 6x + 5 \] Subtract \(5\) from both sides: \[ 4\sqrt{2} - 9 = 6x \] Divide by \(6\): \[ x = \frac{4\sqrt{2} - 9}{6} \] **Conclusion:** There is **one solution**, \(x = \frac{4\sqrt{2} - 9}{6}\), as required. --- ### 5. **One Solution** \[ \boxed{\phantom{a}}\,x + 4 = \boxed{\phantom{a}}\,x - 8 \] **Goal:** Fill in the blanks with numbers such that the equation has exactly one solution. **Strategy:** For the equation \(ax + 4 = bx - 8\) to have one solution, \(a\) and \(b\) must be different (\(a \neq b\)). **Example:** Let’s choose \(a = 3\) and \(b = 5\): \[ 3x + 4 = 5x - 8 \] **Solving for \(x\):** Subtract \(3x\) from both sides: \[ 4 = 2x - 8 \] Add \(8\) to both sides: \[ 12 = 2x \] Divide by \(2\): \[ x = 6 \] **Conclusion:** With \(a = 3\) and \(b = 5\), the equation has **one solution**, \(x = 6\). Therefore, one possible way to fill the boxes is: \[ 3x + 4 = 5x - 8 \] *Note:* Any pair of different numbers for \(a\) and \(b\) will ensure exactly one solution. --- ### 6. **No Solution** \[ \frac{2}{3}x + 7 = \frac{2}{3}x + 7 \] **Issue:** As written, this equation simplifies to \(7 = 7\), which is always true for any \(x\). Therefore, it has **infinitely many solutions**, not "no solution" as intended. **Correction Needed:** To achieve **no solution**, the constants on both sides must differ, making the equation a contradiction. **Corrected Example:** \[ \frac{2}{3}x + 7 = \frac{2}{3}x + 5 \] **Verification:** Subtract \(\frac{2}{3}x\) from both sides: \[ 7 = 5 \] This is a contradiction. **Conclusion:** By modifying the equation to \(\frac{2}{3}x + 7 = \frac{2}{3}x + 5\), it now has **no solution**. --- **Summary:** - **Equations 1-4:** Already correctly set with the desired number of solutions. - **Equation 5:** Example filled as \(3x + 4 = 5x - 8\) to ensure one solution. - **Equation 6:** Original equation had infinitely many solutions; corrected to \(\frac{2}{3}x + 7 = \frac{2}{3}x + 5\) for no solution. Feel free to adjust the coefficients in Equation 5 as long as they are different to maintain exactly one solution.