Solve the following inequalities and graph the solution. − 6 ≤ 0.5 ( x − 4 ) < − 3 -6\le0.5\left(x-4\right)<-3

Welcome back. In this problem, we'd like to solve the three part inequality negative six is less than or equal to 0.5 times x minus four, which is less than negative three, and we'd like to graph the solution. As a reminder, for three part inequalities, the goal is to get our variable all by itself in the center. Also, any operation we do to one side of the inequality, we have to do to all three sides of the inequality, left, center, and right. Looking at our inequality, we do have a decimal, and just like with equations, it's a good idea to clear that decimal first. That decimal has one decimal place, so we can clear it by multiplying by 10 to the first power, which is just 10. When we multiply the center by 10, we will clear the decimal. But if we multiply the center by 10, we have to multiply all sides of the inequality by 10. So that will get us on the left, 10 times negative six, which is less than or equal to, in the center, 10 times 0.5 times x minus four, which is less than, on the right, 10 times negative three. Let's simplify. On the left, we have 10 times negative six, which is negative 60. That's going to be less than or equal to in the center, we have 10 times 0.5 times x minus four. And when we do 10 times 0.5, we just move the decimal point one place to the right, so we get five times x minus four in the center. And on the right, we do 10 times negative three to get negative 30. Now looking at our inequality, we can see that we have some parentheses in the center, so we should simplify the center by distributing that five. When we distribute the five, we get five times x, which is five x, and we have five times negative four, which is minus 20. Now all we did was rewrite the center so that it's easier to work with. We didn't actually perform any new operation to the center. So that means that the left side remains the same, negative 60, and the right side also remains the same at negative 30. Now looking at the center of our inequality, we want to get the variable term five x by itself, which means we have to get rid of the minus 20. To get rid of the minus 20, we have to add 20 to the center of the inequality, which means we also have to add 20 to the left and to the right. On the left, we have negative 60 plus 20, which is negative 40. That's less than or equal to the center where we have minus 20 and plus 20 canceling, so we just have five x left, and that's less than on the right where we have negative 30 plus 20, which is negative 10. Now looking at the center, we need to get rid of this coefficient of five to get the x all by itself. So we need to divide the center of the inequality by five, which means we also need to divide the left side of the inequality by five and the right side of the inequality by five. On the left side of the inequality, we have negative 40 divided by five, which is negative eight. And that's less than or equal to in the center where the fives cancel and we're left with just x, which is less than on the right, negative 10 divided by five, which is negative two. We know that we're all set solving the inequality because our variable x is all alone in the center, so our solution is negative eight is less than or equal to x is less than negative two, but we're not done with the problem because we still need to graph the solution. We can see that x is between negative eight and negative two, so on the graph, we should draw our line between negative eight and negative two. But now we need to decide whether to add parentheses or brackets at negative eight and negative two. Looking first at the left side of the inequality where we have negative eight is less than or equal to x, we can see that x could be negative eight, so we wanna make sure that negative eight is included as a solution. To make sure that negative eight is included, we have to add a bracket at negative eight on our graph. Looking at the right side of our inequality, x is less than negative two, we can see that x can't be negative two, so we don't want negative two to be included as a solution. Because we don't want negative two to be included, we have to do a parenthesis at negative two. So our graph has a bracket at negative eight, we draw our line all the way to negative two, and then we have a parenthesis at negative two, to include negative eight, but to not include negative two as a solution. Great work.

5. Algebra Review

Linear Inequalities in One Variable

Quantitative Reasoning

5. Algebra Review

Linear Inequalities in One Variable

Struggling with Quantitative Reasoning?

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Multiple Choice

Solve the following inequalities and graph the solution.
$- 6 \leq 0.5 (x - 4) < - 3$

Number line from -8 to 8 with a red bracket indicating the interval from -4 inclusive to -1 exclusive.

Number line from -8 to 8 with a red bracket highlighting the interval between 1 and 4, including 1 but excluding 4.

Number line from -8 to 8 with a red bracket highlighting the interval from 2 to 8, including 2 but excluding 8.

Number line graph showing the solution to a compound inequality from -6 to -2, including -6 but excluding -2.

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Struggling with Quantitative Reasoning?

Solve the following inequalities and graph the solution.−6≤0.5(x−4)<−3-6\(\le\)0.5\(\left\)(x-4\(\right\))<-3

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Solve the following inequalities and graph the solution.
$- 6 \leq 0.5 (x - 4) < - 3$