### Assignments:

Unfinished Assignment Study Questions for Lesson 6

### Lesson Objectives:

- Solve linear inequalities.
- Solve compound inequalities.
- Solve an applied problem using an inequality.

An inequality is a sentence with <, >, <=, or >=.

Multiplication principle for inequalities. When both sides of an inequality are multiplied by a negative number, the inequality sign must be reversed.

Solve the linear inequality, and graph the solution set.

3x-2 > x+6

So let's start by adding 2 to both sides of our inequality. We'll get 3x > x+8. Now subtract both sides by x, so we can have all of our x values on one side of our inequality. This will give us 2x > 8. Now divide both sides by 2 and we get x > 4. The solution in set notation is {x| x > 4}. Then in interval notation, we have (4, oo), since parentheses indicate that we don't have an equal sign in our inequality, or that we're not including the value 4.

We can graph the solution set by drawing a number line, labeling 4, then drawing our open parenthesis to the right, and then shading to the right.

Solve the linear inequality and graph the solution set.

x+5 < 4x-5

So let's subtract both sides by 5, and we get x < 4x-10. Then subtract 4x on both sides, and we get -3x < -10. Now we divide both sides by -3. So we get x > 10/3. Notice that we flipped the sign here, because we divided by a negative number.

So the solution to our linear inequality in set notation is {x| x > 10/3}. And then in interval notation, we have (10/3, oo), since our inequality does not include 10/3.

Now the graph to our solution set is a number line with 10/3 in open parenthesis to the right, shaded to the right.

Find the domain of the function which has a square root.

h(x) = sqrt(x-8)

So the domain will include all values of x for which x-8 >= 0. Since you can only take the square root of a positive number or zero. So x-8 >= 0. If you add 8 to both sides, this gives you x >= 8. So the domain in set notation is {x | x >= 8}, and then in interval notation, we have [8, oo). Notice the bracket, because our inequality includes the equal signs, or includes the value 8.

Find the domain of the function which has a square root.

g(x) = x/sqrt(4+x)

So our x value in the numerator can be any real number, but in the denominator, it's restricted. 4+x >= 0, since you can only take the square root of a positive number. And then since the radical is in the denominator, the sqrt(4+x) != 0, since zero in the denominator would give us an undefined number.

So the domain must follow both of these requirements. So we have to solve the inequality 4+x > 0. First, we'll subtract 4 from both sides, and this will give us x > -4.

In set notation, our domain is {x| x > -4}. And then in interval notation, our domain is (-4, oo), with parentheses, because our domain does not include -4.

Solve the compound inequality involving a conjunction and write in interval notation. Then graph the solution set.

-1 <= x+2 < 5

So let's subtract 2 on all three sides. We'll get -3 <= x < 3. In interval notation, this is written [-3, 3). The bracket indicates that we include -3, and the parenthesis indicates that we don't include 3.

Now let's graph the solution set by drawing a number line, labeling -3 and 3, putting a bracket over -3, and a parenthesis over 3, and shading in between.

Solve the compound inequality involving a disjunction and write in interval notation. Then graph the solution set.

2x <= -6 or x-1 > 1

So for the first portion of our compound inequality, let's divide both sides by 2. So we get x <= -3. Then for the second part, add 1 to both sides. So we get x > 2.

This can be written in interval notation as (-oo, -3] (bracket since we're including -3) union (2, oo).

Now let's graph the solution set. Draw a number line, label -3 and 2, then draw an open parenthesis to the right at 2 and a bracket opening to the left at -3. Now shade to the right of 2 and shade to the left of -3.

Dave's Painters charges $300 plus$50 an hour to paint a home. Joe's Painters charges \$80 an hour. For what lengths of time does it cost less to hire Joe's Painters?

So if we let x = "the amount of time in hours", then the amount it costs to hire Dave's Painters is 300 + 50x. And the amount it costs to hire Joe's Painters is 80x. Then since we want to know what length of time it costs less to hire Joe's Painters, set 80x < 300 + 50x.

Now solve for x. So if we subtract 50x on both sides, we get 30x < 300. And then divide both sides by 30, we can cancel the zero, and then 3 goes into 30, 10 times. So x < 10.

The cost is less to hire Joe's Painters for less than 10 hours.