Inequalities, fractions and solution sets
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| intersection/ˌɪntəˈsekʃn/ | 交集 | jiāo jí |
A value must satisfy two conditions at once. How can we keep the allowed interval without losing a boundary?
- A value must satisfy two conditions at once. How can we keep the allowed interval without losing a boundary?
- This lesson studies intersection 交集: The values common to two solution sets, corresponding to AND.
Choose the mathematical structure
- Adding the same expression preserves an inequality; multiplying or dividing by a negative number reverses it. Clear constant denominators using a positive common multiple. Factor a quadratic and test the sign on the intervals separated by its roots. AND takes the intersection of sets; OR takes their union. Distinguish included and excluded endpoints.
- State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.
Which description correctly defines intersection?
The values common to two solution sets, corresponding to AND.
Work through a checked case
- Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.
For (2x−1)/3−(x+2)/2≥1/6, multiply by 6: 4x−2−3x−6≥1, so x≥9. For (x−2)(x−5)<0, the product is negative between its roots: 2<x<5. Combining with x≥3 gives 3≤x<5, or [3,5). Taking OR with x≤0 gives (−∞,0]∪[3,5). A denominator depending on x needs a separate sign/domain check: (x−1)/(x+2)≥0 has x<−2 or x≥1. The zero numerator at 1 is allowed; x=−2 is undefined.
Inequalities, fractions and solution sets
Adding the same expression preserves an inequality; multiplying or dividing by a negative number reverses it
Match solution-set and boundary decisions to their conditions.
Find the lower bound in (2x−1)/3−(x+2)/2≥1/6.
Multiply by positive 6: 4x−2−3x−6≥1, hence x≥9.
Test a tempting shortcut
- Do not multiply by an expression of unknown sign as though it were positive. For −2x<6, dividing by −2 gives x>−3. A root can be included only for a non-strict inequality and only where the expression is defined. Infinity is never a real endpoint to include.
- When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.
AND always means combine every value from both sets. This claim is false. Explain which definition or assumption it violates.
How many integers satisfy 2<x<5 AND x≥3?
The combined interval is [3,5), containing integers 3 and 4.
AND always means combine every value from both sets.
Do not multiply by an expression of unknown sign as though it were positive. For −2x<6, dividing by −2 gives x>−3. A root can be included only for a non-strict inequality and only where the expression is defined. Infinity is never a real endpoint to include.
Interpret a new situation
- Use a sign table or test value on every interval, then write the whole set. The variable-denominator example extends the domain-checking method; B5’s explicit core is linear/quadratic inequalities with brackets and fractions, and combined sets. Check x=4 and x=6 in the quadratic case to confirm which side is allowed.
- A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.
Find the larger boundary root of (x−2)(x−5)=0.
The product is zero at 2 and 5, so the larger root is 5.
Match each part of a complete solution to its purpose.
An assumption justifies the model; a check tests the result; interpretation connects it to the question.
Use this in your course
- 7357 · A-level · B. Match the target tier and specification before assigning extensions.
- Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.
The values common to two solution sets, corresponding to AND. Choose the relationship, show the method, check its assumptions and interpret the result.