Which equations have infinitely many solutions? Select two that apply.

A. 6 ( x - 9 ) = 6x - 54 + x

B. 5 ( x - 8 ) + 10 = 5 ( x = 2 )

C. 20 - 8x - 6 = 2 ( 7 - 4x )

D. 4 ( x + 4 ) =2 ( 2x + 5 ) + 6

E.

1 answer

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.