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 7 Page 174

Practice makes perfect
a We are given two equivalent inequalities. We need to think what we can do to the first inequality to get it to its second and equivalent form.
x+4 ≤ 10 ⇒ ? ⇒ x ≤ 6 Notice that in the second inequality x is isolated. To achieve this, we can undo the addition in the first inequality by subtracting the same quantity from both sides.
x+4 ≤ 10
x+4-4 ≤ 10-4
x ≤ 6
b Once again, we need to think what we can do to the first inequality to get it to its second and equivalent form.
m-1 > 3 ⇒ ? ⇒ m > 4 This time, in the second inequality m is isolated. To achieve this, we can undo the subtraction in the first inequality by adding the same quantity to both sides.
m-1 > 3
m-1+1 > 3+1
m > 4
c Just like before, we need to think what we can do to the first inequality to get it to its second and equivalent form.
5 ≥ 3 + n ⇒ ? ⇒ 2 ≥ n Now we have that n is isolated in the second inequality. To achieve this, we can undo the addition by subtracting the same quantity from both sides.
5 ≥ 3 + n
5 -3 ≥ 3 + n -3
2 ≥ n
d One last time, we need to get from the first inequality to its second and equivalent form.
- 6 < y -2 ⇒ ? ⇒ - 4 < y In the second inequality, y is isolated. We can undo the subtraction in the first inequality by adding the same quantity to both sides.
- 6 < y -2
- 6 +2 < y -2 + 2
- 4 < y