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 \[ 2 x-1=\square x-1 \] (3) Infinitely many solutions \[ \frac{2}{5} x+3=\frac{2}{5} x+3 \] No solution \[ \frac{1}{3} x+2=\frac{\square}{\square} x-3 \] (4) One solution \[ 2 \sqrt{\square}-2)=6 x+5 \] (6) No solution (5) One solution \( \square \) \( x+4= \) \( \square \) \[ ] x-8 \] \[ \frac{2}{3} x+\square=\frac{\square}{\square} x+7 \]

To get an equation with one solution for \( 2x - 1 = \square x - 1 \), you can choose \(\square = 2\). This way both sides equal \(2x - 1\), meaning the equation has one unique solution when \(x\) is solved. For infinitely many solutions in \(\frac{2}{5}x + 3 = \frac{2}{5}x + 3\), the left side exactly matches the right side because they are identical expressions. This means any value for \(x\) will satisfy the equation, leading to endless solutions—voilà, it's a math party! To create an equation with no solution with \(\frac{1}{3}x + 2 = \frac{\square}{\square} x - 3\), you can fill in the boxes with contradictory terms. For instance, set \(\square = 0\) for the second box. This gives you \(\frac{1}{3}x + 2 = -3\), which is impossible because the left side will never equal a constant negative value from the right. To have another equation with one solution like \(2\sqrt{\square} - 2 = 6x + 5\), choose \(\square = 25\). This sets \(2 \times 5 - 2 = 6x + 5\) which simplifies to \(8 = 6x + 5\), ultimately leading to one clear solution when solved. For a no-solution scenario with your last equation, if you want \(\square x + 4 = \square\) to yield no solutions, you could set both squares to \(0\). This transforms the left side to \(4 = -8\), which is clearly a contradiction, confirming that no value for \(x\) can satisfy this equation!