To determine which equations have infinitely many solutions, we need to simplify each equation and see if they lead to an identity (e.g., something that simplifies to a true statement for all values of \( x \)) or if they become a contradiction (e.g., something that leads to a false statement).
Let's evaluate each option:
A. \( 6(x - 9) = 6x - 54 + x \)
Simplifying the left side: \[ 6(x - 9) = 6x - 54 \] The right side simplifies to: \[ 6x - 54 + x = 7x - 54 \]
So the equation becomes: \[ 6x - 54 = 7x - 54 \] Subtract \( 6x \) from both sides: \[ -54 = x - 54 \] Adding \( 54 \) to both sides gives: \[ 0 = x \] This equation has a unique solution (not infinitely many).
B. \( 5(x - 8) + 10 = 5(x + 2) \)
Simplify the left side: \[ 5(x - 8) + 10 = 5x - 40 + 10 = 5x - 30 \]
The right side is: \[ 5(x + 2) = 5x + 10 \]
Thus, the equation simplifies to: \[ 5x - 30 = 5x + 10 \] Now, subtract \( 5x \) from both sides: \[ -30 = 10 \] This is a contradiction, which means this equation has no solutions (not infinitely many).
C. \( 20 - 8x - 6 = 2(7 - 4x) \)
On the left: \[ 20 - 8x - 6 = 14 - 8x \]
On the right: \[ 2(7 - 4x) = 14 - 8x \]
Thus, we have: \[ 14 - 8x = 14 - 8x \] This is an identity, so this equation has infinitely many solutions.
D. \( 4(x + 4) = 2(2x + 5) + 6 \)
Left side: \[ 4(x + 4) = 4x + 16 \]
Right side: \[ 2(2x + 5) + 6 = 4x + 10 + 6 = 4x + 16 \]
So the equation simplifies to: \[ 4x + 16 = 4x + 16 \] This is another identity, so this equation also has infinitely many solutions.
E. \( 2x + 9 = 18 + 2x \)
Subtracting \( 2x \) from both sides: \[ 9 = 18 \] This is a contradiction, which means this equation has no solutions.
From the analysis, the two equations that have infinitely many solutions are:
C. \( 20 - 8x - 6 = 2(7 - 4x) \)
D. \( 4(x + 4) = 2(2x + 5) + 6 \)