Question 6
A student simplifies the inequality [tex]$7 t-6\ \textgreater \ -5$[/tex] to [tex]$7 t\ \textgreater \ 1$[/tex]. Which step did the student perform?
A. Multiplied both sides of the inequality by 7
B. Divided both sides of the inequality by 7
C. Added 6 to each side of the inequality
D. Subtracted 7 from each side of the inequality
Question 7
A student showed their work for solving the inequality [tex]$-6(2 c-1)+30 \leq c-3$[/tex] for [tex]$c$[/tex].
| | [tex]$-6(2 c-1)+30 \leq c-3$[/tex] | |
| :--------- | :--------------------------------- | :------------------- |
| Step One | [tex]$-12 c+6 + 30 \leq c-3$[/tex] | Distribute. |
| Step Two | [tex]$-12 c+36 \leq c-3$[/tex] | Combine like terms. |
| Step Three | [tex]$-13 c+36 \leq -3$[/tex] | Subtract c on both sides. |
| Step Four | [tex]$-13 c \leq-39$[/tex] | Subtract 36 on both sides. |
| Step Five | [tex]$c \leq 3$[/tex] | Divide by -13 on both sides. |
A student made an error in Step Five. What is the correct inequality for Step Five?
A. [tex]$c \leq 3$[/tex]
B. [tex]$c \leq 507$[/tex]
C. [tex]$c \geq 507$[/tex]
D. [tex]$c \geq 3$[/tex]
Answer (1)
Question 6: Adding 6 to both sides of -5"> 7 t − 6 > − 5 results in 1"> 7 t > 1 .
Question 7: Distribute, combine like terms, and isolate the variable in − 6 ( 2 c − 1 ) + 30 ≤ c − 3 .
Dividing both sides of − 13 c ≤ − 39 by -13 requires flipping the inequality sign.
The correct inequality for Step Five is c g e 3 .
Explanation
Problem Analysis The problem consists of two multiple-choice questions related to solving linear inequalities. The first question asks us to identify the operation performed in simplifying an inequality, and the second asks us to identify and correct an error in the steps of solving an inequality.
Solving Question 6 For Question 6, we need to determine what operation transforms -5"> 7 t − 6 > − 5 into 1"> 7 t > 1 . We can see that adding 6 to both sides of the original inequality achieves this:
-5 + 6"> 7 t − 6 + 6 > − 5 + 6 simplifies to 1"> 7 t > 1 .
Solving Question 7 For Question 7, we need to find the error in the student's work and correct it. Let's go through each step:
Original Inequality: − 6 ( 2 c − 1 ) + 30 ≤ c − 3
Step One: − 12 c + 6 + 30 ≤ c − 3 (Distribute)
Step Two: − 12 c + 36 ≤ c − 3 (Combine like terms)
Step Three: − 13 c + 36 ≤ − 3 (Subtract c from both sides)
Step Four: − 13 c ≤ − 39 (Subtract 36 from both sides)
Step Five: c g e 3 (Divide by -13)
The error is in Step Five. When dividing both sides of an inequality by a negative number, you must flip the inequality sign. So, dividing − 13 c ≤ − 39 by -13 should result in c g e 3 , not c ≤ 3 .
Final Answer Therefore, the correct answer for Question 6 is 'Added 6 to each side of the inequality', and the correct inequality for Step Five in Question 7 is c g e 3 .
Examples
Understanding how to solve inequalities is crucial in many real-world scenarios. For example, imagine you're managing a budget and need to ensure your expenses don't exceed your income. Inequalities help you model and solve such constraints, ensuring you stay within your financial limits. Similarly, in engineering, inequalities are used to design systems that operate within safe and efficient ranges. Whether it's optimizing resource allocation or ensuring product quality, the ability to manipulate and solve inequalities is a valuable skill.
