Pure Mathematics · 2nd secondary · First Term

Functions of a Real Variable and Drawing Curves — Solving absolute value equations and inequalities

About this lesson

This unit covers solving absolute value equations and inequalities, both graphically and algebraically. It begins by explaining how to solve equations of the form |ax - b| = c, |ax + b| = cx + d, and |ax + b| = |cx + d|, using graphical methods (plotting the modulus function and a linear or constant function, then finding intersection points) and algebraic methods (using the piecewise definition of the modulus). It also covers properties of absolute value, such as |ab| = |a||b| and |a + b| ≤ |a| + |b|, and includes life applications, like calculating the area of a piece of land between two curves. The unit then moves to solving absolute value inequalities graphically (by comparing curves) and algebraically (using rules like if |x| < a then -a < x < a, and if |x| > a then x > a or x < -a). Exercises are provided for practice. The text also includes an introduction to a separate unit on logarithms, listing objectives and key terms, but the main content of this unit is absolute value equations and inequalities.

Main topics in this lesson

  • Solving absolute value equations graphically and algebraically
  • Solving absolute value inequalities graphically and algebraically
  • Properties of absolute value
  • Applications of absolute value equations (e.g., area calculation)
  • Introduction to logarithms (unit objectives and key terms)

Key terms and vocabulary

  • Absolute value
  • Equation
  • Inequality
  • Graphical solution
  • Algebraic solution
  • Modulus function
  • Intersection points
  • Solution set
  • Interval
  • Exponential function
  • Logarithmic function
  • Inverse function
  • Rational exponents
  • Logarithm laws

Questions and answers on Functions of a Real Variable and Drawing Curves — Solving absolute value equations and inequalities - practice and revision

Multiple choice questions on this lesson in Pure Mathematics for 2nd secondary, First Term, with practice exercises for revision and exam preparation.

1

إذا كان |س - 1| = 5 فإن مجموع حلي المعادلة يساوي

  • 1 2
  • 2 -2
  • 3 6
  • 4 4
2

حل المعادلة |2س + 3| = س + 6 هو

  • 1 س = 3 أو س = -3
  • 2 س = 3 أو س = 9
  • 3 س = -3 أو س = 9
  • 4 س = 1 أو س = -1
3

مجموعة حل المتباينة |س| ≤ 7 هي

  • 1 -7 ≤ س ≤ 7
  • 2 س ≤ 7
  • 3 س ≥ -7
  • 4 س ≤ -7 أو س ≥ 7
4

إذا كانت المسافة بين نقطة س والنقطة 5 تساوي 3 وحدات فإن قيمة س هي

  • 1 8 أو 2
  • 2 8 أو -2
  • 3 -8 أو 2
  • 4 5 أو 3
5

عدد حلول المعادلة |س| = 4 هو

  • 1 حلان
  • 2 حل واحد
  • 3 ثلاثة حلول
  • 4 لا يوجد حل
6

مجموعة حل المتباينة |3س + 2| < 8 هي

  • 1 -10/3 < س < 2
  • 2 -2 < س < 10/3
  • 3 س < 2 أو س > -10/3
  • 4 -8 < س < 8

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