Inequalities with fractions and brackets

Clear fractions by multiplying both sides by the LCM of the denominators. The LCM is positive, so the sign stays.

Worked example: (2x − 1)/3 ≥ (x + 2)/4. Multiply by 12: 4(2x − 1) ≥ 3(x + 2), so 8x − 4 ≥ 3x + 6, 5x ≥ 10 and x ≥ 2. Check at x = 2: both sides equal 1, and ≥ allows equality.

Worked example: x/2 − (x − 1)/3 < 2. Multiply by 6: 3x − 2(x − 1) < 12, so 3x − 2x + 2 < 12 and x < 10. At x = 10 the left side is 5 − 3 = 2, which is not less than 2, so 10 is excluded.

Open brackets carefully: −2(x − 1) = −2x + 2. Dropping the second sign is the usual slip.

Worked example with brackets: 3(x − 2) ≤ 5(x + 2). Then 3x − 6 ≤ 5x + 10, so −16 ≤ 2x and x ≥ −8. Check at x = −8: 3(−10) = −30 and 5(−6) = −30, equal.

If the variable cancels, look at what is left. For 2(x + 3) > 2x + 1 the statement becomes 6 > 1, true for every real x. For 2x + 1 > 2(x + 3) it becomes 1 > 6, true for none.