Algebraic Solution of Linear Inequalities in One Variable

Solving \(2x+8>0\) gives \(2x>-8\), and hence \(x>-4\). Therefore, its solution set is \((-4,\infty)\); \(-4\) is excluded because the inequality is strict. Option B gives \(x<-4\), while options C and D give \(x>4\) and \(x<4\), respectively. Exam tip: an open endpoint means that the boundary value is not included in the solution set.

Subtract 2x from both sides to get 2x + 1 < 9; then subtract 1 and divide by 2 to obtain x < 4. Thus, the interval is (−∞, 4), with 4 excluded.

Multiplying both sides by 4 gives 2 − x > 4, so −x > 2. Dividing by −1 reverses the inequality sign, giving x < −2. Therefore, option A is correct.

Expanding the bracket gives \\(6x-2<5x+6\\). Subtracting \\(5x\\) from both sides gives \\(x-2<6\\), so \\(x<8\\). In option B, the inequality sign has been reversed incorrectly; the sign reverses only when multiplying or dividing by a negative number. In an exam, expand the bracket first and collect the variable terms on one side.

For \(-5\leq x<2\), the integer solutions are \(-5,-4,-3,-2,-1,0,1\). The endpoint \(-5\) is included because of \(\leq\), whereas 2 is excluded because of \(<\). Hence, there are 7 solutions. Exam tip: for consecutive included integers, subtract the first value from the last value and add 1.

Adding 1 to both sides of \(-4x-1<7\) gives \(-4x<8\). Dividing both sides by \(-4\) reverses the inequality sign, so \(x>-2\). Option B results from forgetting this reversal. Exam tip: always reverse an inequality sign when multiplying or dividing by a negative number.

Expanding the brackets gives \(8-6x-2\leq 0\), or \(6-6x\leq 0\). Thus, \(-6x\leq -6\). Dividing both sides by \(-6\) reverses the inequality sign, so \(x\geq 1\). Exam tip: whenever an inequality is multiplied or divided by a negative number, reverse the inequality sign.

The governing method is clearing positive denominators before collecting the variable terms. The denominators 2 and 3 have least common multiple 6. Multiplying the whole inequality by 6 is safe because 6 is positive, so the direction does not change: 6(3x−2)/2 < 6(x+4)/3, giving 3(3x−2)<2(x+4). Expand to obtain 9x−6<2x+8. Subtract 2x and add 6: 7x<14. Dividing by positive 7 gives x<2. Option A matches. Options B and D reverse the direction without justification, while C has the wrong boundary and direction. Since the original inequality is strict, x=2 is excluded.

Because both inequalities must hold simultaneously, their solution sets must be intersected. First solve 2x − 3 ≤ 5: adding 3 gives 2x ≤ 8, and division by positive 2 gives x ≤ 4. Next solve x + 1 > 0: subtracting 1 gives x > −1. The numbers satisfying both conditions are therefore greater than −1 and at most 4, written as −1 < x ≤ 4. The lower endpoint is open because x = −1 makes the second inequality false, while the upper endpoint is closed because x = 4 satisfies the first inequality with equality. Option B incorrectly includes −1 and excludes 4; C and D retain only one incorrect boundary region. Hence A is correct.

A positive expression satisfies \(3-2x>0\). Thus, \(-2x>-3\). Dividing both sides by the negative number \(-2\) reverses the inequality sign, giving \(x<\frac{3}{2}\). Option B results from forgetting this reversal. Exam tip: Always reverse an inequality sign when multiplying or dividing by a negative number.

‘Non-negative’ means \(5x+2\geq 0\). Therefore, \(5x\geq -2\), and dividing by 5 gives \(x\geq -\frac{2}{5}\). Option B is incorrect because it excludes the endpoint \(x=-\frac{2}{5}\). Exam tip: whenever an expression is described as ‘not negative,’ include the equality case.

Multiplying \(\frac{x+5}{2}>3\) by the positive number 2 gives \(x+5>6\), or \(x>1\). Since \(x\) is an integer, the least integer greater than 1 is 2. Therefore, the correct answer is 2. Exam tip: because the inequality is strict (\(>\)), \(x=1\) is not included.

For option B, \(2x-8\geq 0\) gives \(2x\geq 8\), so \(x\geq 4\). Hence, the solution set is \([4,\infty)\). Option A gives \(x>4\), whose solution set is \((4,\infty)\), so 4 is excluded. Exam tip: a square bracket indicates that the endpoint is included, which corresponds to an inequality containing equality.

‘Not positive’ means the expression can be zero or negative, so \(2x+5\leq 0\). Subtracting 5 from both sides gives \(2x\leq -5\), and dividing by 2 gives \(x\leq -\frac{5}{2}\). Therefore, option D is correct. Option C incorrectly uses \(<\) and excludes \(x=-\frac{5}{2}\), even though the expression equals zero at that value and is therefore not positive. Exam tip: ‘not positive’ includes zero, whereas ‘negative’ excludes zero.

Adding the same expression \(3x-2\) to both sides preserves the inequality, so A is always equivalent. In B and C, the factor or divisor may change sign; moreover, C is undefined at \(x=-2\). Exam tip: do not multiply or divide by an expression whose sign is unknown.

Subtracting 1 from both sides gives \\(10\leq 2x\\). Dividing both sides by the positive number 2 yields \\(5\leq x\\), which is equivalent to \\(x\geq 5\\). The inequality sign does not reverse when dividing by a positive number. Exam tip: Reverse the sign only when multiplying or dividing an inequality by a negative number.

From \(3x+2\geq 17\), we get \(3x\geq 15\), and hence \(x\geq 5\). Since \(x\) is an integer, the least possible value is 5. Option 4 is incorrect because \(3(4)+2=14<17\). Exam tip: dividing an inequality by a positive number does not reverse its sign.

“Not greater than” means \(\leq\), so the inequality is \(4x-7\leq 13\). Adding 7 to both sides gives \(4x\leq 20\), and dividing by 4 gives \(x\leq 5\). Therefore, option C is correct. Option D is incorrect because it excludes \(x=5\), even though \(4(5)-7=13\). Exam tip: interpret “not greater than” as \(\leq\) and “not less than” as \(\geq\).

Expanding the bracket gives \(6-3x>x+10\), so \(-4x>4\). Dividing both sides by \(-4\) reverses the inequality sign, yielding \(x<-1\). Option B incorrectly keeps the original direction after division by a negative number. Exam tip: Always reverse the inequality sign when multiplying or dividing by a negative number.

The first inequality gives x ≥ −4. The second gives 2x < 10, so x < 5. Together, these mean −4 ≤ x < 5, which is the interval [−4, 5).

Multiplying both sides by \(-2\) reverses the inequality sign because \(-2\) is negative: \(x-4\leq -6\). Adding 4 to both sides gives \(x\leq -2\), so option B is correct. Exam tip: always reverse an inequality sign when multiplying or dividing by a negative number.

In \(2<x<7\), both inequalities are strict, so neither 2 nor 7 is included; the solution is the open interval \((2,7)\). B and C are half-open intervals. Exam tip: use an open endpoint for \(<\) or \(>\).

In option C, both boundary signs are ‘\(\leq\)’, so both endpoints are included. Solving it gives \(-5\leq x\leq 2\), the closed interval \([-5,2]\). A and B give half-infinite intervals. Exam tip: two non-strict bounds indicate a closed interval.

Subtracting \\(2x\\) from both sides gives \\(-3<3x+12\\). Subtracting 12 and then dividing by the positive number 3 gives \\(-15<3x\\), hence \\(x>-5\\). The inequality sign does not reverse because 3 is positive; therefore, option B is incorrect. Exam tip: reverse the inequality sign only when multiplying or dividing by a negative number.

Rearranging the inequality gives \(9-3x\geq 2x-16\Rightarrow 25\geq 5x\). Dividing both sides by the positive number 5 yields \(x\leq 5\), so option A is correct. The inequality sign would reverse only if we divided or multiplied by a negative number; that does not happen here. Exam tip: after collecting the variable terms, always check the sign of the number used to divide.