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Algebraic Inequalities Practice Questions

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Updated on Oct 3, 2011

To review these concepts, go to Algebraic Inequalities Study Guide.

Algebraic Inequalities Practice Questions

Problems

Practice 1

Solve each inequality for x. Show the solution on a number line with values from –10 to 10.

  1. 6x + 11 < 29
  2. 8 –7x > 50
  3. 3x – 6 < 0
  4. –4x + 7 ≤ 19

Practice 2

Solve each compound inequality for x. Show the solution on a number line with values from –10 to 10.

  1. 0 < x – 4 ≤ 4
  2. 49 > –7x > –21
  3. –13≤ 3x–7< 20

Practice 3

Solve each inequality for y in terms of x.

  1. y – 4 > x
  2. 8y –2x ≤ 24
  3. 15x – 5y > 100

Solutions

Practice 1

1. Subtract 11 from both sides of the inequality and divide by 6:
      6x + 11 < 29
      6x < 18
      x < 3
Because x is less than 3, 3 is not part of the solution set. Draw a hollow circle around 3 and shade to the left, where values are less than 3:
Algebraic Inequalities
2. Multiply both sides of the inequality by 4:
      x≥ –8
Because x is greater than or equal to –8, –8 is part of the solution set. Draw a closed circle around –8 and shade to the right, where values are greater than –8:
Algebraic Inequalities
3. Subtract 8 from both sides of the inequality and divide by –7. Because we are dividing by a negative number, change the inequality symbol from greater than to less than:
      8 – 7x > 50
      –7x > 42
      x < –6
Because x is less than –6, –6 is not part of the solution set. Draw a hollow circle around –6 and shade to the left, where values are less than –6:
Algebraic Inequalities
4. Add 6 to both sides of the inequality and divide by 3:
      3x – 6 < 0
      3x < 6
      x < 2
Because x is less than 2, 2 is not part of the solution set. Draw a hollow circle around 2 and shade to the left, where values are less than 2:
Algebraic Inequalities
5. Subtract 7 from both sides of the inequality and divide by –4. Because we are dividing by a negative number, change the inequality symbol from less than or equal to greater than or equal:
      –4x + 7 ≤ 19
      –4x ≤ 12
      x ≥ –3
Because x is greater than or equal to –3, –3 is part of the solution set. Draw a closed circle around –3 and shade to the right, where values are greater than –3:
Algebraic Inequalities

 

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