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Split the compound inequality into two separate inequalities.

Chemistry

ScanSolve AI Answer

Answer

b
3, 4, 5

Step-by-step solution

4 steps

Step 1: Split the compound inequality into two separate inequalities.
The given inequality is 12(5x−4)≤x+512<3x+112\frac{1}{2}(5x-4) \le x + 5\frac{1}{2} < 3x + 1\frac{1}{2}.
This can be written as:

Part a

12(5x−4)≤x+512\frac{1}{2}(5x-4) \le x + 5\frac{1}{2}

Part b3, 4, 5

x+512<3x+112x + 5\frac{1}{2} < 3x + 1\frac{1}{2}

  1. Solve the first inequality.

    12(5x−4)≤x+512\frac{1}{2}(5x-4) \le x + 5\frac{1}{2}

    Convert the mixed number to an improper fraction: 512=1125\frac{1}{2} = \frac{11}{2}.

    12(5x−4)≤x+112\frac{1}{2}(5x-4) \le x + \frac{11}{2}

    Multiply the entire inequality by 2 to eliminate fractions:

    2×12(5x−4)≤2×x+2×1122 \times \frac{1}{2}(5x-4) \le 2 \times x + 2 \times \frac{11}{2} 5x−4≤2x+115x-4 \le 2x + 11

    Subtract 2x2x from both sides:

    5x−2x−4≤115x - 2x - 4 \le 11 3x−4≤113x - 4 \le 11

    Add 4 to both sides:

    3x≤11+43x \le 11 + 4 3x≤153x \le 15

    Divide by 3:

    x≤5x \le 5
  2. Solve the second inequality.

    x+512<3x+112x + 5\frac{1}{2} < 3x + 1\frac{1}{2}

    Convert mixed numbers to improper fractions: 512=1125\frac{1}{2} = \frac{11}{2} and 112=321\frac{1}{2} = \frac{3}{2}.

    x+112<3x+32x + \frac{11}{2} < 3x + \frac{3}{2}

    Multiply the entire inequality by 2 to eliminate fractions:

    2×x+2×112<2×3x+2×322 \times x + 2 \times \frac{11}{2} < 2 \times 3x + 2 \times \frac{3}{2} 2x+11<6x+32x + 11 < 6x + 3

    Subtract 2x2x from both sides:

    11<6x−2x+311 < 6x - 2x + 3 11<4x+311 < 4x + 3

    Subtract 3 from both sides:

    11−3<4x11 - 3 < 4x 8<4x8 < 4x

    Divide by 4:

    84<x\frac{8}{4} < x 2<x2 < x
  3. Combine the solutions.

    From Step 2, we have x≤5x \le 5.
    From Step 3, we have x>2x > 2.
    Combining these two conditions gives 2<x≤52 < x \le 5.

  4. Find the integral values of xx.

    The integral values of xx that are greater than 2 and less than or equal to 5 are 3, 4, and 5.

    The integral values are 3,4,5\boxed{3, 4, 5}.

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