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Pearson Algebra 1 Common Core, 2011

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Pearson Algebra 1 Common Core, 2011 View details

4. Solving Multi-Step Inequalities

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Exercise 12 Page 190

Begin by isolating p on the left-hand side.

p≤5

Practice makes perfect

Let's isolate p to solve the inequality.

6-3p≥-9

LHS-6≥RHS-6

-3p≥-15

Divide by -3 and flip inequality sign

p≤5

The solution p≤5 means that any number less than or equal to 5 will satisfy the inequality.

Checking Our Answer

Checking Our Answer

Let's check this by choosing an arbitrary value for p and plugging it into the original inequality. Let's choose 1 for p. Since 1 is less than 5, we expect the result to be true.

6-3p≥-9

p= 1

6 -3* 1 ? ≥ -9

Multiply

6-3\stackrel{?}\geq\text{-}9

Subtract term

3≥-9

This result is true, so our inequality is correct for this test value.

Subchapter links

Lesson Check p.189

Practice and Problem-Solving Exercises p.190-192

Standardized Test Prep p.192

Mixed Review p.192