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 \( \square \) \( x+4= \) \( \square \) \[ ] x-8 \] \[ \frac{2}{3} x+\square=\frac{\square}{\square} x+7 \]

Sure! Let's fill in the blanks for each equation to achieve the indicated number of solutions: none, one, or infinitely many. Here's how you can complete each equation: --- ### (1) One Solution \[ 2x - 1 = \boxed{3}x - 1 \] **Explanation:** - **Equation:** \(2x - 1 = 3x - 1\) - **Solution:** Subtract \(2x\) from both sides: \[ -1 = x - 1 \implies x = 0 \] - **Result:** Exactly one solution (\(x = 0\)). --- ### (2) No Solution \[ \frac{1}{3}x + 2 = \frac{\boxed{1}}{\boxed{3}}x - 3 \] **Explanation:** - **Equation:** \(\frac{1}{3}x + 2 = \frac{1}{3}x - 3\) - **Solution:** Subtract \(\frac{1}{3}x\) from both sides: \[ 2 = -3 \] - **Result:** This is a contradiction, so there is **no solution**. --- ### (3) Infinitely Many Solutions \[ \frac{2}{5}x + 3 = \frac{2}{5}x + 3 \] **Explanation:** - **Equation:** \(\frac{2}{5}x + 3 = \frac{2}{5}x + 3\) - **Solution:** Subtract \(\frac{2}{5}x\) from both sides: \[ 3 = 3 \] - **Result:** This is always true for any \(x\), so there are **infinitely many solutions**. --- ### (4) One Solution \[ 2 \sqrt{\boxed{4}} - 2 = 6x + 5 \] **Explanation:** - **Equation:** \(2 \sqrt{4} - 2 = 6x + 5\) - **Simplify:** \(2 \times 2 - 2 = 6x + 5 \implies 4 - 2 = 6x + 5 \implies 2 = 6x + 5\) - **Solution:** \[ 6x = 2 - 5 \implies 6x = -3 \implies x = -\frac{1}{2} \] - **Result:** Exactly one solution (\(x = -\frac{1}{2}\)). --- ### (5) One Solution \[ \boxed{2}x + 4 = \boxed{3}(x - 8) \] **Explanation:** - **Equation:** \(2x + 4 = 3(x - 8)\) - **Simplify:** \(2x + 4 = 3x - 24\) - **Solution:** \[ 4 + 24 = 3x - 2x \implies 28 = x \] - **Result:** Exactly one solution (\(x = 28\)). --- ### (6) No Solution \[ \frac{2}{3}x + \boxed{5} = \frac{\boxed{2}}{\boxed{3}}x + 7 \] **Explanation:** - **Equation:** \(\frac{2}{3}x + 5 = \frac{2}{3}x + 7\) - **Solution:** Subtract \(\frac{2}{3}x\) from both sides: \[ 5 = 7 \] - **Result:** This is a contradiction, so there is **no solution**. --- ### Summary of Filled Blanks | Problem | Filled Blanks | |---------|-----------------------------------------| | (1) | \(3\) | | (2) | Numerator: \(1\), Denominator: \(3\) | | (4) | \(4\) (under the square root) | | (5) | Left coefficient: \(2\), Right coefficient: \(3\) | | (6) | Left constant: \(5\), Right fractions: Numerator \(2\), Denominator \(3\) | --- By carefully choosing the numbers to fill in the blanks, we've ensured that each equation has the desired number of solutions: one, none, or infinitely many.