Solve: $\frac{1}{3}(5-3 x)<\frac{2}{5}(3-7 x)$
Answer (2)
Multiply both sides of the inequality by 15 to eliminate fractions: 5 ( 5 − 3 x ) < 6 ( 3 − 7 x ) .
Expand and simplify the inequality: 25 − 15 x < 18 − 42 x .
Isolate x by adding 42 x and subtracting 25 from both sides: 27 x < − 7 .
Divide by 27 to solve for x : x < − 27 7 .
Explanation
Problem Analysis We are given the inequality 3 1 ( 5 − 3 x ) < 5 2 ( 3 − 7 x ) . Our goal is to solve for x .
Eliminating Fractions To eliminate the fractions, we multiply both sides of the inequality by 15 (the least common multiple of 3 and 5): 15 ⋅ 3 1 ( 5 − 3 x ) < 15 ⋅ 5 2 ( 3 − 7 x ) .
Simplifying Simplifying both sides, we get: 5 ( 5 − 3 x ) < 6 ( 3 − 7 x ) .
Expanding Expanding both sides of the inequality: 25 − 15 x < 18 − 42 x .
Isolating x term Adding 42 x to both sides: 25 − 15 x + 42 x < 18 − 42 x + 42 x , which simplifies to 25 + 27 x < 18 .
Isolating x term (cont.) Subtracting 25 from both sides: 25 + 27 x − 25 < 18 − 25 , which simplifies to 27 x < − 7 .
Solving for x Dividing both sides by 27: 27 27 x < 27 − 7 , which simplifies to x < − 27 7 .
Final Answer Therefore, the solution to the inequality is x < − 27 7 .
Examples
Understanding linear inequalities is crucial in various real-life scenarios, such as budgeting and resource allocation. For instance, imagine you are managing a project with a limited budget, and you need to determine how many resources you can allocate to different tasks while staying within budget. Linear inequalities help you model these constraints and find feasible solutions. In this case, solving an inequality helps determine the maximum amount you can spend on one task given the constraints on other tasks and the overall budget. For example, if you have a total budget of $5000 and one task costs $1000, you can use an inequality to determine how much you can spend on another task, ensuring that the total expenses do not exceed the budget.
To solve the inequality 3 1 ( 5 − 3 x ) < 5 2 ( 3 − 7 x ) , we first eliminate fractions by multiplying by 15, leading to the simplified form 25 − 15 x < 18 − 42 x . By rearranging and isolating x , we find that the solution is x < − 27 7 .
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