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Solve the inequalities and represent on a number line: i) −4x > 12 ii) 5x - 7 vert 3x + 9 iii) 3x - 5 7

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3 parts
iii

[-1, 4)

a

10000x>8000x+400010000x > 8000x + 4000

b

x>2 (The trader must sell more than 2 bags of rice to make a profit.)x > 2 \text{ (The trader must sell more than 2 bags of rice to make a profit.)}

x≤17x \le 17

Step-by-step solution

23 steps

1. Solve the inequalities and represent on a number line

Part i· 2 steps

−4x>12-4x > 12

  1. Divide both sides by −4-4.

    Remember to reverse the inequality sign when dividing by a negative number.

    −4x−4<12−4\frac{-4x}{-4} < \frac{12}{-4} x<−3x < -3
  2. Represent on a number line.

    An open circle at −3-3 indicates that −3-3 is not included, and the arrow points to the left.

    <——————–o\text{<--------------------o} −3\quad \quad \quad \quad \quad \quad \quad \quad -3

    The solution in interval notation is (−∞,−3)(-\infty, -3).

Part ii· 4 steps

5x−7≥3x+95x - 7 \ge 3x + 9

  1. Subtract $3x$ from both sides.

    5x−3x−7≥95x - 3x - 7 \ge 9 2x−7≥92x - 7 \ge 9
  2. Add 7 to both sides.

    2x≥9+72x \ge 9 + 7 2x≥162x \ge 16
  3. Divide both sides by 2.

    2x2≥162\frac{2x}{2} \ge \frac{16}{2} x≥8x \ge 8
  4. Represent on a number line.

    A closed circle at 8 indicates that 8 is included, and the arrow points to the right.

    ∙−−−−−−−−−−−−−−−−−−−−>\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \bullet--------------------> 8\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad 8

    The solution in interval notation is [8,∞)[8, \infty).

Part iii

[-1, 4)

3x−5≤73x - 5 \le 7

  1. Add 5 to both sides.

    3x≤7+53x \le 7 + 5 3x≤123x \le 12
  2. Divide both sides by 3.

    3x3≤123\frac{3x}{3} \le \frac{12}{3} x≤4x \le 4
  3. Represent on a number line.

    A closed circle at 4 indicates that 4 is included, and the arrow points to the left.

    <——————–∙\text{<--------------------}\bullet 4\quad \quad \quad \quad \quad \quad \quad \quad 4

    The solution in interval notation is (−∞,4](-\infty, 4].

    2. Solve the compound inequality: −3≤2x−1<7-3 \le 2x - 1 < 7 and express your solution set interval notation

  4. Add 1 to all parts of the inequality.

    −3+1≤2x−1+1<7+1-3 + 1 \le 2x - 1 + 1 < 7 + 1 −2≤2x<8-2 \le 2x < 8
  5. Divide all parts by 2.

    −22≤2x2<82\frac{-2}{2} \le \frac{2x}{2} < \frac{8}{2} −1≤x<4-1 \le x < 4
  6. Express the solution in interval notation.

    [−1,4)\boxed{[-1, 4)}

    3. A trader buys xx bags of rice at N8,000 each and spends an additional N4,000 on transportation. He sells all the bags for N10,000 each.

Part a

10000x>8000x+400010000x > 8000x + 4000

Write an inequality to represent the condition that the trader makes a profit.

  1. Calculate the total cost (Cost Price).

    Cost Price (CP) = (Cost per bag ×\times number of bags) + Transportation cost

    CP=8000x+4000CP = 8000x + 4000
  2. Calculate the total revenue (Selling Price).

    Selling Price (SP) = Selling price per bag ×\times number of bags

    SP=10000xSP = 10000x
  3. For a profit to be made, the Selling Price must be greater than the Cost Price (SP>CPSP > CP).

    10000x>8000x+400010000x > 8000x + 4000

    This is the inequality representing the condition for profit.

    10000x>8000x+4000\boxed{10000x > 8000x + 4000}

Part b

x>2 (The trader must sell more than 2 bags of rice to make a profit.)x > 2 \text{ (The trader must sell more than 2 bags of rice to make a profit.)}

Solve the inequality and interpret your answer.

  1. Subtract $8000x$ from both sides of the inequality.

    10000x−8000x>400010000x - 8000x > 4000 2000x>40002000x > 4000
  2. Divide both sides by 2000.

    2000x2000>40002000\frac{2000x}{2000} > \frac{4000}{2000} x>2x > 2
  3. Interpret the answer.

    The trader makes a profit if he sells more than 2 bags of rice. Since the number of bags must be a whole number, he must sell at least 3 bags to make a profit.

    x>2 (The trader must sell more than 2 bags of rice to make a profit.)\boxed{x > 2 \text{ (The trader must sell more than 2 bags of rice to make a profit.)}}

    4. Solve this inequality: 2x−13≤x+52\frac{2x - 1}{3} \le \frac{x + 5}{2}

  4. Multiply both sides by the least common multiple (LCM) of the denominators 3 and 2, which is 6.

    6(2x−13)≤6(x+52)6 \left( \frac{2x - 1}{3} \right) \le 6 \left( \frac{x + 5}{2} \right)
  5. Simplify both sides.

    2(2x−1)≤3(x+5)2(2x - 1) \le 3(x + 5)
  6. Distribute the numbers on both sides.

    4x−2≤3x+154x - 2 \le 3x + 15
  7. Subtract $3x$ from both sides.

    4x−3x−2≤154x - 3x - 2 \le 15 x−2≤15x - 2 \le 15
  8. Add 2 to both sides.

    x≤15+2x \le 15 + 2 x≤17x \le 17

    The solution in interval notation is (−∞,17](-\infty, 17].

    x≤17\boxed{x \le 17}
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