مجموعة حل المعادلة |س + 5| = 2 هي:
- 1 {-3, -7}
- 2 {3, 7}
- 3 {-3, 7}
- 4 {3, -7}
This unit teaches students how to solve modulus (absolute value) equations and inequalities both graphically and algebraically. It begins by explaining that the solution set of an equation f(x)=g(x) is the set of x-coordinates of the intersection points of the two functions' graphs. Students learn to solve equations like |ax+b|=c by graphing the modulus function and a constant function, then finding intersection points, and also by using the piecewise definition of the modulus function. The unit also covers properties of absolute value, such as |ab|=|a||b| and |a+b|≤|a|+|b|, and then moves to solving inequalities like |x-a|<b or |x-a|>b, both graphically (by comparing curves) and algebraically (using rules like if |x|<a then -a<x<a). A real-life application involving temperature deviation is included. By the end, students should be able to solve modulus equations and inequalities using both methods and apply them to model problems.
Multiple choice questions on this lesson in General Mathematics for 2nd secondary, First Term, with practice exercises for revision and exam preparation.
مجموعة حل المعادلة |س + 5| = 2 هي:
إذا كان منحنى الدالة ص = |س| والمستقيم ص = 4 يتقاطعان في نقطتين، فإن مجموع إحداثيي س لنقطتي التقاطع يساوي:
مجموعة حل المتباينة |س - 3| > 0 هي:
إذا كانت |أ + ب| = 9 و |أ| = 5 و |ب| = 3، فإن العلاقة الصحيحة هي:
مجموعة حل المتباينة |2س - 1| ≤ 7 هي:
عدد حلول المعادلة |س| = -3 هو:
Generate a quiz from this unit, send it to your class, and let the answers be graded for you.
Start free