Let's solve each equation step by step:
(a) \(6x + 4x - 6 + 24 + 9x\)
- Combine like terms: \[ 6x + 4x + 9x - 6 + 24 = (6x + 4x + 9x) + (-6 + 24) \] \[ = 19x + 18 \]
Since this expression is not set equal to anything, we interpret \(6x + 4x - 6 + 24 + 9x = 0\), which simplifies to: \[ 19x + 18 = 0 \]
- Solve for \(x\): \[ 19x = -18 \] \[ x = -\frac{18}{19} \]
Since we found a single solution, the answer has one solution and is classified as neither an identity nor a contradiction.
(b) \(25 - 4x + 15 - 3x + 10 - x\)
- Combine like terms: \[ 25 + 15 + 10 - 4x - 3x - x = (25 + 15 + 10) + (-4x - 3x - x) \] \[ = 50 - 8x \]
Again, interpret expression as equaling zero (\(50 - 8x = 0\)): \[ 50 - 8x = 0 \]
- Solve for \(x\): \[ -8x = -50 \] \[ x = \frac{50}{8} = \frac{25}{4} \]
This gives us one solution and is classified as neither an identity nor a contradiction.
(c) \(4x + 8 = 2x + 7 + 2x - 20\)
- Simplify the right-hand side: \[ 2x + 7 + 2x - 20 = (2x + 2x) + (7 - 20) \] \[ = 4x - 13 \]
So the equation becomes: \[ 4x + 8 = 4x - 13 \]
- Subtract \(4x\) from both sides: \[ 8 = -13 \]
This is a contradiction because \(8\) cannot equal \(-13\). Thus, the equation has no solutions and is classified as a contradiction.
Summary:
- (a) One solution: \(x = -\frac{18}{19}\) (Neither)
- (b) One solution: \(x = \frac{25}{4}\) (Neither)
- (c) No solution (Contradiction)