Algebraic Solution of Linear Inequalities in One Variable

The direct answer is option A: x≥8. Start with (2x-1)/3≥5. Since 3 is positive, multiply both sides by 3 without changing the sign: 2x-1≥15. Add 1: 2x≥16. Divide by positive 2: x≥8. A is correct, and x=8 is included because the original sign is ≥; substitution gives (16-1)/3=5. B, x≤8, reverses the direction incorrectly. C, x>8, wrongly excludes x=8, even though equality works. D, x<8, has the opposite direction and also excludes larger valid values such as x=9. The denominator is positive, so no sign reversal occurs when clearing it. If the denominator had been negative, reversal would be necessary. Memory cue: positive multiplication or division keeps the inequality sign; check the boundary separately.

Multiplying both sides by 2 gives \(5-x>8\), so \(-x>3\). Multiplying both sides by \(-1\) reverses the inequality sign, yielding \(x<-3\). The option \(x\leq -3\) is incorrect because at \(x=-3\), the left-hand side is \(\frac{5-(-3)}{2}=4\), whereas it must be greater than 4. Exam tip: reverse the inequality sign whenever you multiply or divide by a negative number.

Subtracting 1 from \(2x+1<9\) gives \(2x<8\), and dividing by the positive number 2 gives \(x<4\). Since \(x\) is an integer, the greatest integer less than 4 is 3. Option 4 is incorrect because \(2(4)+1=9\), which is not less than 9. Exam tip: In a strict inequality \(<\), the boundary value is not included.

Solving the inequality gives \(3x-2\leq 13\Rightarrow 3x\leq 15\Rightarrow x\leq 5\). The natural numbers satisfying this condition are \(1,2,3,4,5\), so there are 5 possible values of \(x\). Exam tip: in this context, natural numbers are taken to start from 1; including 0 would change the count.

Subtracting \(2x\) from both sides gives \(2x+3\ge -5\). Subtracting \(3\) and then dividing by \(2\) gives \(2x\ge -8\), so \(x\ge -4\). Therefore, the solution set is \(\{x\mid x\ge -4\}\). Option B reverses the inequality incorrectly, while options C and D mishandle equality at \(-4\). In an exam, remember that the inequality sign changes only when multiplying or dividing by a negative number.

The governing concept is solving a linear inequality while reversing the inequality sign whenever both sides are divided or multiplied by a negative number. Begin with 6 - 5x ≤ 16. Subtracting 6 from both sides gives -5x ≤ 10. Dividing both sides by -5 reverses the sign because -5 is negative, so the result is x ≥ -2. Hence option A is correct. The endpoint is included because the original inequality contains ≤. Option B is the common error of failing to reverse the sign. Options C and D are also incorrect because they either use a strict inequality or point in the wrong direction. Substitution confirms the boundary: x = -2 gives 6 + 10 = 16, which satisfies the inequality.

Subtracting 5x from both sides and adding 3 gives 3x < 15. Dividing by the positive number 3 keeps the inequality sign unchanged, so x < 5. The endpoint 5 is excluded because the original inequality is strict.

This is a one-variable linear inequality involving a negative coefficient. Begin with 11 - 4x > 3 and subtract 11 from both sides. This gives -4x > -8. Dividing by -4 reverses the inequality sign because the divisor is negative, so the result is x < 2. Thus option A is correct. The endpoint x = 2 is excluded because the original relation is strict: substituting x = 2 gives 3 > 3, which is false. Option B has the direction reversed, whereas C and D incorrectly include the boundary or use the wrong direction. A quick check with x = 1 gives 7 > 3, confirming the solution.

First expand the bracket: \(5(2x-1)=10x-5\). Thus \(10x-5\leq 3x+16\), which gives \(7x\leq 21\). Dividing by the positive number 7 yields \(x\leq 3\), so option A is correct. The inequality sign does not reverse because the divisor is positive; option C is incorrect because it excludes \(x=3\), which satisfies the original inequality. Exam tip: reverse the inequality sign only when multiplying or dividing by a negative number.

Expanding the brackets gives 6x + 8 > 4x + 18. Subtracting 4x and 8 from both sides yields 2x > 10, so x > 5. Therefore, option A is correct. Option C incorrectly includes x = 5; substituting x = 5 makes both sides equal, whereas the given inequality is strict. Exam tip: For a strict inequality using > or <, the boundary value is not included.

Multiplying both sides by 5 does not reverse the inequality because 5 is positive: \(x+3<10\). Subtracting 3 from both sides gives \(x<7\), so option A is correct. Option B reverses the inequality incorrectly, while options C and D add equality, which is not included in the original strict inequality. Exam tip: Reverse the inequality sign only when multiplying or dividing by a negative number.

Multiplying both sides by 4, which is positive, does not reverse the inequality: \(3x+2\geq2x+12\). Subtracting \(2x\) and then 2 from both sides gives \(x\geq10\), so option A is correct. Option B reverses the inequality incorrectly; the sign changes only when multiplying or dividing by a negative number. Exam tip: Always check the sign of the number used to multiply or divide an inequality.

The least common multiple of the denominators 3 and 2 is 6. Multiplying both sides by 6 gives \(2(x-1)+3(x+2)\leq 24\). Thus, \(2x-2+3x+6\leq 24\), so \(5x+4\leq 24\), which leads to \(5x\leq 20\) and hence \(x\leq 4\). Therefore, option A is correct. Remember that multiplying or dividing an inequality by a positive number does not reverse its sign.

Adding 1 to both sides gives \(\frac{2x+5}{3}>5\). Multiplying by 3 yields \(2x+5>15\), so \(2x>10\) and hence \(x>5\). Since the original inequality is strict, 5 is not included; therefore, \(x\geq 5\) is not correct. In an exam, remember that a strict inequality excludes its boundary point.

Subtracting 7 from both sides gives 2x ≥ 12; dividing by 2 gives x ≥ 6. Since equality is included, 6 is part of the solution set, so the interval is [6, ∞).

Subtracting 5 from both sides gives −2x ≥ 6. Dividing by −2 reverses the inequality, giving x ≤ −3. The endpoint −3 is included, so the interval is (−∞, −3].

Solving \\(2x-3\leq x+1\\) gives \\(x\leq 4\\), so \\(x=4\\) belongs to the solution set. Substituting \\(x=4\\) gives 5 on both sides, so \\(5\leq 5\\) is true. The symbol \\(\leq\\) includes equality, which is why \\(x<4\\) is not the correct solution. Exam tip: Include the boundary value when the inequality sign is \\(\leq\\) or \\(\geq\\).

The key concept is an identity inequality: when the variable terms on both sides are identical, they cancel without placing any restriction on the variable. Starting with 6x + 1 > 6x - 5, subtract 6x from both sides. The result is 1 > -5, which is always true. Therefore every real value of x satisfies the original inequality, so option A is correct. There is no division or sign reversal involved. Option B would be correct only if cancellation produced a false statement such as 1 > 5. Options C and D impose restrictions that do not follow from the simplified constant inequality. For example, x = 0 and x = 100 both satisfy the original relation.

Subtracting \(4x\) from both sides gives \(9<2\), which is always false. Therefore, no real value of \(x\) can satisfy the inequality, so the solution set is empty. Option B would apply only if simplification produced an always-true statement. Exam tip: when the variable terms cancel and leave a false constant inequality, the answer is no solution.

Starting with \(ax+b<c\), subtracting \(b\) from both sides gives \(ax<c-b\). Since \(a>0\), division by \(a\) does not reverse the inequality sign, so \(x<\frac{c-b}{a}\). Option B would be correct only for a negative coefficient \(a\), while C and D incorrectly include equality. Exam tip: reverse an inequality sign only when multiplying or dividing by a negative number.

‘Not more than 20’ means less than or equal to 20, so the inequality is \(x+7\leq20\). Subtracting 7 from both sides gives \(x\leq13\). Option B reverses the inequality, while options C and D incorrectly exclude or reverse the equality case at 13. Exam tip: Translate ‘not more than’ or ‘at most’ as \(\leq\).

Let the number be \(x\). The inequality is \(3x-5\geq 16\). Adding 5 to both sides gives \(3x\geq 21\), and dividing by 3 gives \(x\geq 7\). Therefore, option A is correct. Option B reverses the inequality incorrectly, while option C excludes the value \(x=7\), even though ‘at least’ includes equality. Exam tip: ‘at least’ means \(\geq\), whereas ‘greater than’ means \(>\).

If Ravi buys x notebooks, the amount left is \(50-8x\) rupees. The condition for saving at least 10 rupees is \(50-8x\ge 10\), which gives \(x\le 5\). Therefore, the greatest possible whole number of notebooks is 5. Option 6 is incorrect because buying 6 notebooks costs 48 rupees and leaves only 2 rupees. Exam tip: Form the inequality first, then choose the greatest feasible whole-number value.

Subtracting 12 from both sides of \(x+12\ge 40\) gives \(x\ge 28\). Therefore, the smallest possible value of \(x\) is 28. Option 27 is incorrect because \(27+12=39\), which is less than 40 and does not satisfy the passing condition. Exam tip: ‘At least’ means \(\ge\), so the boundary value is included.

Expanding the brackets gives 2x + 2 - 3x + 6 < 10, so -x + 8 < 10 and hence -x < 2. Multiplying both sides by -1 reverses the inequality sign, giving x > -2. Therefore, option A is correct. Exam tip: always reverse the inequality sign when multiplying or dividing by a negative number; this is why x < -2 is incorrect.