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4x + 3 /leq 5x - 3 or 1 - 2x < -3x + 3

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To solve the inequalities \(4x + 3 \leq 5x - 3\) and \(1 - 2x < -3x + 3\), we can solve them one at a time.

Inequality 1: \(4x + 3 \leq 5x - 3\)

  1. Subtract \(4x\) from both sides: \[ 3 \leq 5x - 4x - 3 \] This simplifies to: \[ 3 \leq x - 3 \]

  2. Add \(3\) to both sides: \[ 3 + 3 \leq x \] Thus, we have: \[ 6 \leq x \quad \text{or} \quad x \geq 6 \]

Inequality 2: \(1 - 2x < -3x + 3\)

  1. Add \(2x\) to both sides: \[ 1 < -3x + 2x + 3 \] This simplifies to: \[ 1 < -x + 3 \]

  2. Subtract \(3\) from both sides: \[ 1 - 3 < -x \] Which simplifies to: \[ -2 < -x \]

  3. Multiply both sides by \(-1\) (remember, this reverses the inequality): \[ 2 > x \quad \text{or} \quad x < 2 \]

Combining the Solutions

We have two inequalities from our solutions:

  1. \(x \geq 6\)
  2. \(x < 2\)

Since there are no values of \(x\) that satisfy both inequalities simultaneously, the solution set is:

\[ \text{No solution} \]

In conclusion, there is no value of \(x\) that satisfies either of the inequalities.