Algebra Exercises - Equations and Inequalities
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1) If a < b, then -a > -b Why?
2) What role of operations that applies when you are solving an equation does not apply when you are solving an inequality?
---Give examples to explain your answers.
Section 1.6: Exercises 100 and 104
Section 2.1: Exercises 92 and 94
Section 2.2: Exercises 78, 86, and 90
Section 2.3: Exercises 14, 22, 32, 54, and 68
Section 2.4: Exercises 32, 54, 72, and 82
Section 2.6: Exercise 8
Please see attached for the details of the problems.
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This posting contains the solution and step-by-step procedures to solve each of the given problems.
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Solutions:
1.) For this problem, it is best to explain with the aid of a number line. Suppose a < b, then b is much farther to zero and a to the right since b is much greater.
If using the same numbers we take the negative, we can see that b is still farther to a and zero but to the left.
Any number getting farther to zero to the left will have lesser value. So when we take the negative of a and b, the conditions will be reversed: -a > -b. For example, if a = 2 and b = 5 then: 2 < 5. But if we get the negative of each, -a = -2 and -b = -5, then: -2 > -5.
2.) There are certain operations in equality that you cannot simply apply in inequalities:
- When multiplying or dividing negative numbers on both sides, the sense of inequality must be reversed. A common mistake is the solver will forget to reverse the sense of inequality when multiplying or dividing a negative number. For example, -x > -10. When we multiply -1 to both sides, the inequality must be x < 10, and NOT x > 10.
- One cannot simply relate two inequalities having a different sense. Meaning, if a < b and b > c, one cannot ...
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