Pearson Algebra 1 Common Core, 2011
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Pearson Algebra 1 Common Core, 2011 View details
2. Solving Inequalities Using Addition or Subtraction
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Exercise 67 Page 176

Practice makes perfect
a We are asked to tell if the statement shown below is true or not.

If 56 + 58 = t does t =56 + 58 ? Since the relation used is an equality, we can change the left-hand side for the right-hand side freely. This is because equations are symmetric, if a=b then b=a. Therefore, the statement is true.

b Here, we are asked to decide if the statement shown below is true or not.

If 56 + 58 ≤ r, is r ≤ 56 + 58 ? This statement is not true in general. The first inequality states that r is greater than or equal to 56+58. In contrast, the second inequality states that r is less than or equal to 56+58. The inequalities imply opposite things and, therefore, they cannot be both true at the same time. It would only be valid for the case where r equals 56+58.

c What is different between the previous two statements? The key difference is that the first statement is an equation. It states that both sides are equal and, therefore, we can exchange side for the other.

If a=b, then b=a. âś“ However, the second statement uses an inequality. This implies a relation where one quantity is greater than the other. In those cases, we cannot exchange one side for the other without also reversing the inequality sign. If a>b, then b>a. *