QUESTION IMAGE
Question
3 the product of a number and 1.5 is less than the absolute value of the difference between 20 and 5. what are all the possible values of the number? show your work.
solution
4 consider the inequalities $-\frac{1}{4}a > 3$ and $b - 12 > -3$. what values, if any, make both inequalities true? show your work.
solution
5 solve the inequality $x + 2\frac{1}{2} > -\frac{1}{2}$. graph the solution set on a number line. show your work.
Problem 3
Step1: Define the number and absolute value
Let the number be \( x \). The absolute value of the difference between 20 and 5 is \( |20 - 5| \).
\( |20 - 5| = 15 \)
Step2: Set up the inequality
The product of the number and 1.5 is \( 1.5x \), and this is less than 15. So the inequality is \( 1.5x < 15 \).
Step3: Solve the inequality
Divide both sides by 1.5: \( x < \frac{15}{1.5} \)
\( x < 10 \)
Step1: Solve the first inequality \( -\frac{1}{4}a > 3 \)
Multiply both sides by -4. When multiplying or dividing an inequality by a negative number, the direction of the inequality sign changes.
\( a < 3\times(-4) \)
\( a < -12 \)
Step2: Solve the second inequality \( b - 12 > -3 \)
Add 12 to both sides: \( b > -3 + 12 \)
\( b > 9 \)
Step3: Analyze the solution sets
The solution for \( a \) is \( a < -12 \) and for \( b \) is \( b > 9 \). There is no overlap between the set of numbers less than -12 and the set of numbers greater than 9.
Step1: Subtract \( 2\frac{1}{2} \) from both sides
The inequality is \( x + 2\frac{1}{2} > -\frac{1}{2} \). Subtract \( 2\frac{1}{2} \) (which is \( \frac{5}{2} \)) from both sides:
\( x > -\frac{1}{2} - 2\frac{1}{2} \)
Step2: Simplify the right - hand side
\( -\frac{1}{2}-\frac{5}{2}=-\frac{6}{2}=-3 \)
So \( x > - 3 \)
Step3: Graph the solution on a number line
- Draw a number line.
- Place an open circle at -3 (since \( x > - 3 \), -3 is not included in the solution set).
- Draw an arrow to the right of -3 to represent all numbers greater than -3.
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All possible values of the number are \( x < 10 \)