Question

directions: drag each tile to the correct box. solve the given equations. place the name of each equation in order by the least number of solutions to the greatest number of solutions. equation a: $4x + 12 = 2(2x + 3) + 6$; equation b: $5(x - 1) = 3(x + 7)$; equation c: $2x - 4 = 2(x + 18)$; tiles: equation b, equation a, equation c; boxes: , ,

Explanation:

Step1: Solve Equation A

Simplify \(4x + 12 = 2(2x + 3)+6\).
Expand RHS: \(4x + 12 = 4x + 6 + 6\)
Simplify RHS: \(4x + 12 = 4x + 12\)
Subtract \(4x\) and 12: \(0 = 0\).
This is an identity, so infinitely many solutions.

Step2: Solve Equation B

Simplify \(5(x - 1)=3(x + 7)\).
Expand: \(5x - 5 = 3x + 21\)
Subtract \(3x\) and add 5: \(2x = 26\)
Divide: \(x = 13\).
One solution.

Step3: Solve Equation C

Simplify \(2x - 4 = 2(x + 18)\).
Expand RHS: \(2x - 4 = 2x + 36\)
Subtract \(2x\): \(-4 = 36\), which is false.
No solutions.

Step4: Order by solutions

Least (0 solutions): Equation C
Next (1 solution): Equation B
Most (infinite): Equation A

Wait, correction: Wait, Equation C has 0 solutions, Equation B has 1, Equation A has infinite. So order from least to greatest: Equation C, Equation B, Equation A. Wait, initial step had a mix. Let recheck:

Equation C: \(2x -4 = 2x +36\) → \( -4=36\) (no solution, 0 solutions)
Equation B: \(5x -5 = 3x +21\) → \(2x=26\) → \(x=13\) (1 solution)
Equation A: \(4x +12 =4x +12\) → \(0=0\) (infinite solutions)

So number of solutions:
Equation C: 0
Equation B: 1
Equation A: infinite

Thus order from least to greatest: Equation C, Equation B, Equation A.

Answer:

Equation C, Equation B, Equation A