- Simplify the given inequality through distribution, addition, and subtraction as shown in Steps 1-3.
- Isolate the variable \(x\) by dividing by \(-2\), remembering to reverse the inequality sign because \(-2\) is negative.
- Conclude that the final step results in the inequality \(\boxed{x < 3}\).
Explanation
- Analysis of the Inequality and Given Steps
The objective is to solve the linear inequality \(-2(5 - 4x) < 6x - 4\) and determine the final simplified form for \(x\). The problem provides a sequence of simplification steps:
Step 1: Distributing \(-2\) into \((5 - 4x)\) yields \(-10 + 8x < 6x - 4\).
Step 2: Subtracting \(8x\) from both sides gives \(-10 < -2x - 4\).
Step 3: Adding 4 to both sides results in \(-6 < -2x\).
We need to find the final Step 4 by solving for \(x\).
- Solving for x and Reversing the Inequality Sign
We begin with the result from Step 3:
\[-6 < -2x\]
To isolate \(x\), we divide both sides of the inequality by \(-2\). A fundamental property of inequalities is that when multiplying or dividing both sides by a negative number, the direction of the inequality sign must be reversed.
Dividing both sides by \(-2\):
\[\frac{-6}{-2} > \frac{-2x}{-2}\]
Simplifying the fractions:
\[3 > x\]
This is equivalent to the expression:
\[x < 3\]
- Conclusion
Comparing our result \(x < 3\) to the provided options:
○ \(x < -3\)
○ \(x > -3\)
○ \(x < 3\)
○ \(x > 3\)
Our result matches the option \(x < 3\). Thus, the final step in solving the inequality \(-2(5 - 4x) < 6x - 4\) leads to \(x < 3\).
Examples
Inequalities like this are foundational in real-world scenarios, such as budgeting. Suppose you are comparing the cost of two different service providers. Provider A charges a fixed fee of \$10 minus a \$8 variable discount (based on usage x), while Provider B charges \$6 times your usage x minus \$4. If you want to find the usage range where Provider A's total cost is strictly less than Provider B's, you set up the inequality \(-2(5 - 4x) < 6x - 4\). Solving this tells you that for any usage \(x < 3\), Provider A is more cost-effective.