Algebraic Solution of Linear Inequalities in One Variable

The direct answer is A: x>8. Begin with 3−(x+2)/5<1. Subtract 3 from both sides: −(x+2)/5<−2. Multiply both sides by 5, which is positive: −(x+2)<−10. Now multiply by −1, or divide by −1. Because this number is negative, reverse the sign: x+2>10. Subtract 2: x>8. Option A is correct. Option B, x<8, has the inequality direction reversed incorrectly; the negative factor requires reversal in the opposite way. Option C, x>−8, has the wrong boundary and may result from mishandling the +2. Option D, x<−8, has both the wrong boundary and wrong direction. Testing x=9 gives 3−11/5=0.8<1, so it works. Testing x=8 gives 1, not a strict value below 1, so x=8 is excluded. Testing x=7 gives 1.4, which fails. Memory cue: whenever you remove a negative coefficient, reverse the inequality sign; with <, the boundary itself is not included.

Multiplying all parts of the inequality by the positive number 2 does not reverse the signs, giving \\(-6\leq x+1<8\\). Subtracting 1 from all parts yields \\(-7\leq x<7\\), so option A is correct. Option B incorrectly reverses the endpoint conditions. Exam tip: an inequality sign reverses only when multiplying or dividing by a negative number.

For a strict inequality, the boundary value is not included. Since the variable is restricted to integers, after solving the inequality we must choose the first whole number that satisfies it. This is different from simply rounding without checking the inequality.

From \(\frac{x-3}{2}>1\), multiply both sides by the positive number 2, so the sign does not change: \(x-3>2\). Adding 3 gives \(x>5\). Integers greater than 5 begin with 6, so the least integer solution is 6. The value 5 is not allowed because it makes the left side equal to 1, not greater than 1. Option A is correct.

‘At least 13’ means 13 or more, so the inequality is \(2x-5\geq 13\). Adding 5 to both sides gives \(2x\geq 18\), and dividing by the positive number 2 gives \(x\geq 9\). The inequality sign does not change because the divisor is positive. Remember that \(x=9\) is included, so \(x>9\) is not correct.

The governing method is to expand, collect like terms, and determine the truth of the resulting statement. Starting with 5(x − 1) + 2 < 5x, expansion gives 5x − 5 + 2 < 5x, or 5x − 3 < 5x. Subtracting 5x from both sides leaves −3 < 0. This is a true numerical statement and places no restriction on x. Consequently, every real number satisfies the original inequality, so the solution set is ℝ. Option B would be correct only if simplification produced a contradiction such as −3 > 0. Options C and D incorrectly restrict the answer to one value or to positive values. Therefore option A is correct.

Multiply both sides by \(2\) (positive, so inequality sign stays the same):
\[\frac{3x-1}{2}\le x+4\Rightarrow 3x-1\le 2x+8\Rightarrow 3x-2x\le 8+1\Rightarrow x\le 9.\]

The governing concept is solving a linear inequality in one variable while preserving the inequality direction. First expand the left side: 4(x − 2) + 3 = 4x − 8 + 3 = 4x − 5. The inequality therefore becomes 4x − 5 ≤ 2x + 9. Subtract 2x from both sides to get 2x − 5 ≤ 9, then add 5 to obtain 2x ≤ 14. Dividing by the positive number 2 gives x ≤ 7; the sign remains unchanged because division by a positive number does not reverse an inequality. Thus option A is correct. Option B reverses the direction without justification, option C incorrectly removes equality, and option D comes from an arithmetic error.

This is an algebraic linear inequality in one variable. First subtract 1.2 from both sides: 0.4x ≤ 2.4. Since 0.4 is positive, dividing by it does not reverse the inequality sign. Therefore x ≤ 2.4 ÷ 0.4 = 6. The endpoint 6 is included because the original sign is ≤, so the answer is option C. Option A reverses the direction incorrectly, option B excludes the valid endpoint and uses the wrong direction, and option D results from an arithmetic error. Equivalently, using fractions, 0.4 = 2/5 and 2.4 = 12/5, giving (2/5)x ≤ 12/5 and hence x ≤ 6. The solution set is (-∞, 6].

The governing rule is that equivalent operations may be performed on both sides of a linear inequality, while the sign reverses only when multiplying or dividing by a negative number. Expand the left side: 2(3x − 1) − 5 = 6x − 2 − 5 = 6x − 7. Thus 6x − 7 ≥ 4x + 7. Subtract 4x from both sides to obtain 2x − 7 ≥ 7, then add 7: 2x ≥ 14. Division by the positive number 2 gives x ≥ 7, so option C is correct. The value 7 is included because the original relation is ≥. Option A reverses the direction, option B has an incorrect boundary, and option D wrongly excludes 7.

Use the distributive law carefully on both sides. The left side becomes −2x + 8 + 5 = −2x + 13, while the right side becomes 3 − 3x. Therefore −2x + 13 < 3 − 3x. Add 3x to both sides: x + 13 < 3. Subtract 13: x < −10. No sign reversal occurs in these steps because only terms are being added or subtracted. Hence option D is correct. A quick check with x = −11 gives 35 < 36, which is true, whereas x = −10 gives equality, not a strict inequality, so −10 must be excluded. Option A has the wrong direction, option B includes the invalid endpoint, and option C has the wrong numerical boundary.

Solve each inequality separately and then take their intersection. From \(3x+2\ge -4\), subtracting 2 gives \(3x\ge -6\), so division by the positive number 3 gives \(x\ge -2\). The second condition is \(x<5\). Hence, over the real numbers, the combined interval is \([-2,5)\). Since the question restricts \(x\) to integers, the permitted values are \(-2,-1,0,1,2,3,4\). Thus option A is correct. Option B omits -2 and includes 5, option C incorrectly includes 5, and option D starts below the lower bound and misses 4. The strict sign at 5 excludes that endpoint.

This is a linear inequality with a positive decimal coefficient. Add 1.5 to both sides: 0.25x ≥ 3.5. Since 0.25 is positive, division preserves the ≥ sign. Thus x ≥ 3.5 ÷ 0.25 = 14. The same calculation can be made without decimals: 0.25 = 1/4 and 3.5 = 7/2, so x/4 ≥ 7/2; multiplying by 4 gives x ≥ 14. Therefore option A is correct, and equality is included because the original sign is ≥. Option B reverses the direction incorrectly, option C excludes the valid endpoint 14, and option D comes from an incorrect division. The solution set is [14, ∞).

For a common solution, solve each inequality and take their intersection. From 4x − 1 > 7, add 1 to obtain 4x > 8, then divide by positive 4: x > 2. From 2x + 3 ≤ 17, subtract 3 to get 2x ≤ 14, then divide by positive 2: x ≤ 7. A number must satisfy both conditions, so it must be greater than 2 and at most 7. Therefore the intersection is 2 < x ≤ 7, which is option C. The lower endpoint 2 is excluded because the first inequality is strict, while 7 is included because the second inequality is non-strict. Option B incorrectly includes 2, option D incorrectly excludes 7, and option A describes a disjoint condition rather than the common intersection.

The least common denominator of 4 and 6 is 12. Multiplying the entire inequality by 12 is valid because 12 is positive, so the inequality sign remains unchanged: 12(x/4) + 12(x/6) ≤ 12·5. This gives 3x + 2x ≤ 60, hence 5x ≤ 60. Dividing by positive 5 yields x ≤ 12. Therefore option A is correct. Equality is allowed because the original sign is ≤; indeed, x = 12 gives 12/4 + 12/6 = 3 + 2 = 5. Option B reverses the direction, option C wrongly excludes x = 12, and option D comes from adding the denominators or otherwise mishandling the fractional coefficients. The solution set is (−∞, 12].

The phrase “at least 40 MB” translates into the inequality 120 − 8x ≥ 40. Subtract 120 from both sides: −8x ≥ −80. Dividing by −8 requires reversing the inequality sign, because division by a negative number reverses order; therefore x ≤ 10. Option B is correct. A practical check confirms it: if x = 10, the remaining data is 120 − 80 = 40 MB, so equality is allowed. If x = 8, 56 MB remains, also satisfying the condition. Option A reverses the direction and would permit too much usage, option C wrongly excludes x = 10, and option D results from an incorrect calculation. In a real context x may also represent a nonnegative quantity, but among the given choices the required inequality is x ≤ 10.

First expand both sides carefully. The left side is \(9-2(4x-3)=9-8x+6=15-8x\), while the right side is \(5(x+2)=5x+10\). Thus \(15-8x<5x+10\). Subtracting 10 gives \(5-8x<5x\), and adding \(8x\) gives \(5<13x\). Since 13 is positive, division preserves the inequality sign, yielding \(x>5/13\). Therefore option A is correct. The other options either reverse the direction, replace the strict inequality by a non-strict one, or state the opposite solution set. The endpoint \(5/13\) itself is excluded because the original relation is strictly less than.

The governing rule is that an inequality with rational expressions can be solved by multiplying by a positive common denominator; because the multiplier is positive, the inequality sign does not change. The least common denominator is 10. Thus 10 times the left side is 2(2x+3)-5(x-4)=4x+6-5x+20=26-x, while the right side becomes 7-x. Therefore 26-x\le7-x, and adding x to both sides gives 26\le7, which is false. The variable cancels completely, so no real value of x can satisfy the statement. Hence option C is correct. Option A would require an identity, option B or D would require a genuine interval, but the resulting contradiction gives an empty solution set.

For an ‘or’ statement, the solution is the union of the solution sets of the two inequalities, not their intersection. First, 4x+7\le2x+15 gives 2x\le8 and hence x\le4. Second, 5-3x<2 gives -3x<-3; dividing by the negative number -3 reverses the sign, so x>1. Thus the combined set is (-\infty,4]\cup(1,\infty). Every real number belongs to this union: numbers at most 4 satisfy the first inequality, while numbers greater than 1 satisfy the second. For example, a number above 4 satisfies the second, and a number at most 1 satisfies the first. Therefore option C, all real x, is correct. Option D is only the intersection and is inappropriate for ‘or’.

The governing concept is preservation of an inequality under multiplication by a positive number. The denominators 5 and 10 have positive common denominator 10, so multiplying both sides by 10 does not reverse the sign. We obtain 2(7-2x)\le3x+1, or 14-4x\le3x+1. Subtracting 1 gives 13-4x\le3x; adding 4x gives 13\le7x. Finally, division by positive 7 yields x\ge13/7. Therefore option A is correct, and the endpoint 13/7 is included because the original relation is ‘less than or equal to’. Option B reverses the direction, option C incorrectly interchanges the numerator and denominator of the boundary value, and option D has an incorrect sign.

The governing idea is to solve the compound inequality first and then count only the integer points in the resulting interval. Since 3 is positive, multiplying 1\le(5x-2)/3<8 by 3 preserves both inequality signs: 3\le5x-2<24. Adding 2 throughout gives 5\le5x<26. Dividing by positive 5 yields 1\le x<26/5, or 1\le x<5.2. Because x is restricted to the integers, the possible values are x=1,2,3,4,5, exactly five values. The upper endpoint 26/5 is not an integer and is excluded anyway, while x=1 is included because the left inequality is non-strict. Hence option B is correct. Options A, C, and D result from miscounting an endpoint or ignoring the integer restriction.

The correct answer is D: all real numbers. Multiply the entire inequality by 8, which is positive, so the inequality sign does not change. We get 4x − 2(x − 3) > 2x + 5. Expand the left side: 4x − 2x + 6 = 2x + 6. Therefore 2x + 6 > 2x + 5. Subtracting 2x from both sides gives 6 > 5, which is always true. Hence every real value of x satisfies the inequality. Option A includes only numbers below 1, so it misses valid values such as x=2. Option B includes only numbers above 1 and misses x=0. Option C is similarly too restrictive. Option D is correct. Memory cue: multiplying by a positive number preserves the sign; a true statement independent of x means all real numbers are solutions.

The governing concept for a combined ‘and’ statement is intersection: a number must satisfy both inequalities simultaneously. Solve the first inequality: 5x-2<3x+10 gives 2x<12, hence x<6. Solve the second: x+4\ge2 gives x\ge-2. The common portion of x<6 and x\ge-2 is the interval [-2,6), written -2\le x<6. The left endpoint is included because the second inequality permits equality at -2. The right endpoint is excluded because the first inequality is strict and does not permit x=6. Therefore option A is correct. Option C omits the lower restriction, option D omits the upper restriction, and option B incorrectly reverses which endpoints are included.

The relevant rule is that multiplication by a positive common denominator preserves the inequality direction. The denominators 3 and 2 have positive common denominator 6. Multiplying by 6 gives 24-2(2x+1)\ge3(1-x). Expanding both sides produces 24-4x-2\ge3-3x, so 22-4x\ge3-3x. Subtracting 3 gives 19-4x\ge-3x; adding 4x gives 19\ge x. Therefore x\le19, so option A is correct. The equality value x=19 is included because the original relation is non-strict. Options B and D reverse the direction, while option C introduces an erroneous factor of one-half during denominator clearing. A quick check at x=0 confirms the result: 4-1/3 is at least 1/2, so 0 belongs to the solution.

Start with (5-2x)/3 ≥ x-4. Since 3 is positive, multiplying both sides by 3 does not reverse the inequality: 5-2x ≥ 3x-12. Add 12 to both sides and add 2x to both sides to obtain 17 ≥ 5x. Dividing by the positive number 5 preserves the direction, giving x≤17/5. Therefore option A is correct. The endpoint is included because the original sign is ≥, not >. A quick check at x=0 gives 5/3≥-4, which is true, and 0≤17/5 is also true. At x=4, the original inequality gives -1≥0, false, while 4≤17/5 is false. Option B reverses the solution direction, option C incorrectly excludes the equality endpoint, and option D has the wrong sign. The essential rule is to reverse an inequality only when multiplying or dividing by a negative number.

The governing concept is the algebraic solution of a linear inequality in one variable. The denominators 2 and 5 are positive, so multiply the entire inequality by their least common multiple, 10; the inequality sign does not change. This gives 5(x+4) − 2(x−3) < 40, or 5x+20−2x+6 < 40. Hence 3x+26 < 40, so 3x < 14 and therefore x < 14/3. Thus option A is correct. Option B reverses the direction without justification, option C introduces an incorrect negative value, and option D results from faulty fraction manipulation. The endpoint 14/3 is excluded because the original sign is strict.