To determine which equations have infinitely many solutions, we need to simplify each equation and see if they collapse to a statement that's always true (like \(0 = 0\)) or if they result in a contradiction (like \(5 = 6\)).
Let's evaluate each one:
A. \(6 ( x - 9 ) = 6x - 54 + x\)
Simplifying the left side: \[ 6(x - 9) = 6x - 54 \] The right side becomes: \[ 6x - 54 + x = 7x - 54 \]
So the equation simplifies to: \[ 6x - 54 = 7x - 54 \] Subtract \(6x\) from both sides: \[ -54 = x - 54 \] Add 54 to both sides: \[ 0 = x \] This has a single solution, \(x = 0\).
B. \(5 ( x - 8 ) + 10 = 5 ( x - 2 )\)
Left side: \[ 5(x - 8) + 10 = 5x - 40 + 10 = 5x - 30 \] Right side: \[ 5(x - 2) = 5x - 10 \]
Setting them equal: \[ 5x - 30 = 5x - 10 \] Subtract \(5x\) from both sides: \[ -30 = -10 \] This is a contradiction, so no solutions.
C. \(20 - 8x - 6 = 2 ( 7 - 4x )\)
Simplifying the left side: \[ 20 - 8x - 6 = 14 - 8x \]
Right side: \[ 2(7 - 4x) = 14 - 8x \] The equation becomes: \[ 14 - 8x = 14 - 8x \] This is always true, so there are infinitely many solutions.
D. \(4 ( x + 4 ) = 2 ( 2x + 5 ) + 6\)
Left side: \[ 4(x + 4) = 4x + 16 \] Right side: \[ 2(2x + 5) + 6 = 4x + 10 + 6 = 4x + 16 \] The equation becomes: \[ 4x + 16 = 4x + 16 \] This is true for all values of \(x\), so there are infinitely many solutions.
Conclusion: The equations with infinitely many solutions are:
- C. \(20 - 8x - 6 = 2 ( 7 - 4x )\)
- D. \(4 ( x + 4 ) = 2 ( 2x + 5 ) + 6\)
So, the correct choices are C and D.
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