body{margin:0;} noscript{z-index: 10000; position: fixed; display: block; height: 100%; width: 100%; background: #fff;} noscript .fa{font-size: 40px; display:block; color: #daa520;} noscript div{margin-top: 6%; padding-left: 20px; padding-right: 20px; text-align: center;} ! You must have JavaScript enabled to use this site.

Pearson Algebra 1 Common Core, 2011

PA

Pearson Algebra 1 Common Core, 2011 View details

4. Solving Multi-Step Inequalities

Finish the assignment

Continue to next subchapter

Exercise 9 Page 190

Begin by isolating f on the left-hand side.

f≤3

Practice makes perfect

Let's isolate f to solve the inequality.

5f+7≤22

LHS-7≤RHS-7

5f≤15

.LHS /5.≤.RHS /5.

f≤3

The solution f ≤ 3 means that any number less than or equal to 3 will satisfy the inequality.

Checking Our Answer

Checking Our Answer

Let's check our answer with a few arbitrary values.

f	5f+7≤22	Simplify	True or False?
2	5* 2+7? ≤22	17≤22	True
3	5* 3+7? ≤22	22≤22	True
4	5* 4+7? ≤22	27≰22	False

We can see that we correctly solved the inequality since the values less than or equal to 3 make the inequality true.

Subchapter links

Lesson Check p.189

Practice and Problem-Solving Exercises p.190-192

Standardized Test Prep p.192

Mixed Review p.192