Free Math Worksheets

- What happens to the inequality sign when multiplying or dividing by a negative number? - Move variable terms to one side and constants to the other. - Solve an inequality like an equation, while applying the sign-reversal rule.

1. For a), subtract \(2x\) and \(12\): \(3x<-15\). Divide by \(3\): \(x<-5\). 2. For b), subtract \(x\) and \(4\): \(-4x\ge12\). Divide by \(-4\) and reverse the inequality: \(x\le-3\).

a) \(\{x\in\mathbb{R}\mid x<-5\}\) b) \(\{x\in\mathbb{R}\mid x\le-3\}\)

- Convert the fraction to a decimal if that makes the arithmetic easier. - Move all variable terms to one side. - Isolate \(x\) using inverse operations.

1. Since \(\frac{3}{5}=0.6\), rewrite the inequality as \(0.4x+0.6>1.8-0.2x\). 2. Add \(0.2x\): \(0.6x+0.6>1.8\). 3. Subtract \(0.6\): \(0.6x>1.2\). 4. Divide by \(0.6\): \(x>2\).

\(\{x\in\mathbb{R}\mid x>2\}\)

- Solve inequalities with the same balancing steps used for equations. - Reverse the inequality sign when multiplying or dividing by a negative number. - Distribute before collecting variable terms. - Clear a positive denominator without changing the inequality direction.

1. For a), add \(11\): \(4x < 28\). Divide by \(4\): \(x < 7\). 2. For b), subtract \(15\): \(-2x \ge 6\). Divide by \(-2\) and reverse the inequality sign: \(x \le -3\). 3. For c), distribute: \(3x + 12 \le 5x + 2\). Subtract \(3x\) and subtract \(2\): \(10 \le 2x\). Divide by \(2\): \(x \ge 5\). 4. For d), multiply by \(3\): \(x - 2 > 6\). Add \(2\): \(x > 8\).

a) \(x < 7\) b) \(x \le -3\) c) \(x \ge 5\) d) \(x > 8\)

- Use parentheses for an excluded endpoint. - Use a bracket for an included endpoint. - Write the smaller endpoint first. - Translate each compound inequality before writing the interval.

1. For a), the endpoints are excluded: \(-3 < x < 6\), so the interval is \((-3, 6)\). 2. For b), \(4\) is excluded and \(11\) is included: \(4 < x \le 11\), so the interval is \((4, 11]\). 3. For c), both endpoints are excluded: \(-7.5 < x < -2\), so the interval is \((-7.5, -2)\). 4. For d), both endpoints are excluded: \(8.4 < x < 15\), so the interval is \((8.4, 15)\).

a) \((-3, 6)\) b) \((4, 11]\) c) \((-7.5, -2)\) d) \((8.4, 15)\)

- Write a perimeter expression for each rectangle. - Use an inequality to compare the two perimeters. - Solve the inequality even though the variable appears on both sides. - Include the restriction that a side length must be positive.

1. The perimeter of Rectangle A is \(2(x + x + 6) = 4x + 12\). 2. The perimeter of Rectangle B is \(2(x + 3x) = 8x\). 3. Compare the perimeters: \(4x + 12 > 8x\). 4. Subtract \(4x\): \(12 > 4x\). 5. Divide by \(4\): \(3 > x\), or \(x < 3\). 6. Combine this result with \(x > 0\). The solution set is \(0 < x < 3\).

The perimeter of Rectangle A is greater when \(0 < x < 3\).

- Compare the two function values at the same x-value. - At an intersection point, what must be true about the two function values? - To decide when one graph is above the other, compare the two expressions with an inequality.

1. At \(x = -2\), \(f(-2) = -(-2) + 2 = 4\) and \(g(-2) = 0.5 \cdot (-2) - 1 = -2\). Since \(4 > -2\), the graph of \(f\) is higher there. 2. For the intersection, solve \(-x + 2 = 0.5x - 1\). Then \(3 = 1.5x\), so \(x = 2\). Substitution gives \(y = 0\), so \(S = (2, 0)\). 3. To determine when \(g\) is above \(f\), solve \(0.5x - 1 > -x + 2\). This gives \(1.5x > 3\), so \(x > 2\). Because the stated domain is \(\mathbb{Q}\), the solution is all rational \(x > 2\).

a) \(f(-2) = 4\) and \(g(-2) = -2\); the graph of \(f\) is higher. b) \(S = (2, 0)\) c) All rational \(x > 2\)

- Which inequality symbol represents “above”? - Substitute the known coordinate into the inequality. - What happens to an inequality sign when dividing by a negative number?

1. A point is above the line when its y-coordinate is greater than the line's output: \(10>-2x+6\). 2. Subtract \(6\): \(4>-2x\). 3. Divide by \(-2\) and reverse the inequality: \(-2<x\), or \(x>-2\).

The point lies above the line when \(x>-2\).

- A vertical displacement keeps the x-coordinate unchanged. - First find the point on the line at \(x=6\). - Write an inequality that represents “below the line.”

1. At \(x=6\), the line has value \(f(6)=\frac{2}{3} \cdot 6-1=3\). Four units above gives \(y_A=3+4=7\). 2. For \(B\) to be below the line, \(3<\frac{2}{3}x_B-1\). 3. Add \(1\): \(4<\frac{2}{3}x_B\). Multiply by \(\frac{3}{2}\): \(6<x_B\), so \(x_B>6\).

a) \(y_A=7\) b) \(x_B>6\)

- Multiplying by a common denominator can eliminate fractions. - Distribute the factor to every term inside the parentheses. - How is a strict “greater than” condition written in interval notation?

1. Distribute on the right: \(\frac{3}{4}x-2>\frac{1}{2}x+2\). 2. Subtract \(\frac{1}{2}x\) and add \(2\): \(\frac{1}{4}x>4\). 3. Multiply by \(4\): \(x>16\).

\((16, \infty)\)

- Work from the innermost grouping symbols outward. - How do signs change when a minus sign is directly before brackets? - Combine like terms on each side before moving terms across the inequality.

1. Simplify inside the brackets: \(2-[3x-x-5]\le8-4x+1\). 2. Combine terms: \(2-[2x-5]\le9-4x\). 3. Distribute the negative sign: \(2-2x+5\le9-4x\), so \(7-2x\le9-4x\). 4. Add \(4x\) and subtract \(7\): \(2x\le2\). 5. Divide by \(2\): \(x\le1\).

\(\{x\in\mathbb{R}\mid x\le1\}\)

- Write the perimeter of a rectangle in terms of its height. - Express all lengths in the same unit. - Which inequality represents “cannot exceed”?

1. Let \(h\) be the height in inches. Then the width is \(2h\). 2. The perimeter is \(2(h+2h)=6h\). 3. Convert the rod length: \(16\,\text{ft}=192\,\text{in}\). The material constraint is \(6h\le192\). 4. Divide by \(6\): \(h\le32\). The greatest whole-number height is \(32\) inches.

The greatest possible height is \(32\) inches.

- Count the edges in both triangular bases and the lateral edges. - Write the total wire length using only one variable. - Remember that a side length must be positive.

1. The two triangular bases have \(6\) edges of length \(a\), and the prism has \(3\) lateral edges of height \(h\). 2. Since \(h=a+10\), the total wire length is \(6a+3(a+10)=9a+30\). 3. The constraint is \(9a+30\le120\). Subtract \(30\): \(9a\le90\). Divide by \(9\): \(a\le10\). 4. Since \(a\) is a positive integer, \(a\in\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}\).

The possible side lengths are all whole numbers from \(1\) inch through \(10\) inches.

- There are separate maximum-size and minimum-length conditions. - The girth is the perimeter of the square end. - Solve both inequalities and find their overlap.

1. The length is \(l=3x\), and the girth is \(4x\). 2. The size limit gives \(3x+4x\le165\), so \(7x\le165\) and \(x\le\frac{165}{7}\approx23.57\). 3. The minimum length gives \(3x\ge48\), so \(x\ge16\). 4. Combining the conditions gives \(16\le x\le23.57\). The integer values are \(16, 17, 18, 19, 20, 21, 22, 23\).

The possible integer side lengths are \(16\) inches through \(23\) inches.

- Write a cost expression for each plan. - How can “Power is less expensive” be written as an inequality? - Find when the two plans have equal costs, then compare beyond that point. - Include both fixed and variable costs.

1. The cost functions are \(C_S(x)=4.95+0.12x\) and \(C_P(x)=9.95+0.08x\). 2. At \(80\) minutes, \(C_S(80)=\$4.95+\$0.12 \cdot 80=\$14.55\), and \(C_P(80)=\$9.95+\$0.08 \cdot 80=\$16.35\). 3. Solve \(9.95+0.08x<4.95+0.12x\). Subtract \(4.95\) and \(0.08x\): \(5.00<0.04x\). Divide by \(0.04\): \(x>125\).

a) Smart costs \(\$14.55\), and Power costs \(\$16.35\). b) Power is less expensive for more than \(125\) minutes.

- Does Plan B's cost change with mileage? - Evaluate Plan A at \(100\) miles and compare. - In part c, use an unknown for the new daily fee. - “Exactly the same cost” means set the expressions equal.

1. The cost functions are \(C_A(x)=24+0.25x\) and \(C_B(x)=44\), with \(x\ge0\). 2. Solve \(24+0.25x<44\): \(0.25x<20\), so \(x<80\). Thus, Plan A is less expensive for \(0\le x<80\). 3. At \(100\) miles, Plan A costs \(\$24+\$0.25 \cdot 100=\$49\), while Plan B costs \(\$44\). Plan B is less expensive. 4. Let \(F\) be the new daily fee. Solve \(F+0.25 \cdot 100=44\): \(F+25=44\), so \(F=\$19\).

a) Plan A is less expensive for \(0\le x<80\) miles. At \(80\) miles, the costs are equal. b) Plan B; Plan A costs \(\$49\), while Plan B costs \(\$44\). c) The daily fee should be \(\$19\).

- What happens to the inequality sign when dividing by a negative number? - How do open and closed endpoints differ in interval notation? - Move all variable terms to one side.

1. For a), \(4x-9\le2x+5\) gives \(2x\le14\), so \(x\le7\). 2. For b), \(18-3x>33\) gives \(-3x>15\). Divide by \(-3\) and reverse the inequality: \(x<-5\). 3. For c), \(\frac{2}{3}x+4<2x-2\). Rearranging gives \(6<\frac{4}{3}x\), so \(x>\frac{9}{2}=4.5\).

a) \((-\infty, 7]\) b) \((-\infty, -5)\) c) \((4.5, \infty)\)

- Solve the inequalities separately. - Pay attention when dividing the second inequality by a negative number. - Compare the resulting intervals.

1. Solve the first inequality: \(6x+12\le36\), so \(6x\le24\) and \(x\le4\). Its solution is \((-\infty, 4]\). 2. Solve the second inequality: \(-3x\ge-12\). Dividing by \(-3\) reverses the inequality, giving \(x\le4\). Its solution is also \((-\infty, 4]\). 3. The intervals are identical, so the inequalities have the same solution set.

Yes. Both solution sets are \((-\infty, 4]\).

- Write an inequality comparing the monthly totals. - Movie counts must be whole numbers. - How does lowering Service A's per-movie price affect the break-even point?

1. For part a, solve \(\$18.50<\$7.50+\$2.00x\). This gives \(11<2x\), so \(x>5.5\). Since the number of movies is a whole number, Service B is less expensive starting at \(6\) movies. 2. For part b, solve \(\$18.50<\$7.50+\$1.50x\). This gives \(11<1.50x\), so \(x>7.333\ldots\). Therefore, Service B is less expensive starting at \(8\) movies.

a) \(6\) movies b) \(8\) movies

- Write an inequality that compares the two expressions. - Move variable terms to one side and constants to the other. - You may use fractions or decimals consistently.

1. Write the inequality \(\frac{1}{2}x+4>2x-5\). 2. Subtract \(\frac{1}{2}x\): \(4>\frac{3}{2}x-5\). 3. Add \(5\): \(9>\frac{3}{2}x\). 4. Multiply by \(\frac{2}{3}\): \(6>x\), so \(x<6\).

\(\{x\in\mathbb{R}\mid x<6\}\)

- Simplify both sides before trying to isolate \(x\). - What does it mean if all variable terms cancel? - Decide whether the remaining statement is always true or always false.

1. For a), distribute: \(6x-8<6x-5\). Subtract \(6x\): \(-8<-5\), which is always true. Therefore, every real number is a solution. 2. For b), simplify: \(4-x-5\ge10-x\), so \(-1-x\ge10-x\). Add \(x\): \(-1\ge10\), which is false. Therefore, there are no solutions.

a) \(\mathbb{R}\) b) \(\varnothing\)

- Simplify both sides before isolating \(x\). - Distribute carefully. - Which bracket is used when an endpoint is included?

1. Distribute: \(3x-2\le2x+10+1\). 2. Combine terms: \(3x-2\le2x+11\). 3. Subtract \(2x\): \(x-2\le11\). 4. Add \(2\): \(x\le13\).

\((-\infty, 13]\)

- A test value is a solution only if it makes the inequality true. - Solve step by step as you would solve an equation. - Check whether part a agrees with the final solution set.

1. Substitute \(x=5\): \(5(5-2)\ge3 \cdot 5+4\), which becomes \(15\ge19\). This is false, so \(5\) is not a solution. 2. Distribute: \(5x-10\ge3x+4\). 3. Subtract \(3x\) and add \(10\): \(2x\ge14\). 4. Divide by \(2\): \(x\ge7\).

a) No, because \(15\ge19\) is false. b) \([7, \infty)\)

- Multiply by a common denominator to eliminate the fractions. - Pay attention to the minus sign before the second fraction. - Multiplying by a positive number does not reverse the inequality.

1. Multiply every term by the least common denominator \(6\): \(2(2x+1)-3(x-2)\le12\). 2. Distribute: \(4x+2-3x+6\le12\). 3. Combine terms: \(x+8\le12\). 4. Subtract \(8\): \(x\le4\).

\(\{x\in\mathbb{R}\mid x\le4\}\)

- When is division undefined? - What sign must the denominator have for a negative numerator to produce a positive quotient? - A quotient is negative when its numerator and denominator have opposite signs.

1. The expression is undefined when the denominator is zero: \(x + 4 = 0\), so \(x = -4\). 2. Because the numerator is negative, the quotient is positive when the denominator is negative. Solve \(x + 4 < 0\) to get \(x < -4\), or \((-\infty, -4)\). 3. The quotient is negative when the negative numerator is divided by a positive denominator. Solve \(x + 4 > 0\) to get \(x > -4\), or \((-4, \infty)\).

1. \(\{-4\}\). 2. \((-\infty, -4)\). 3. \((-4, \infty)\).

- Find where the denominator equals zero. - With a positive numerator, when is the quotient negative? - Changing only the sign of the numerator multiplies every defined output by \(-1\).

1. The denominator is zero when \(2x - 10 = 0\), so \(x = 5\). Therefore, the domain is \(D = \mathbb{R} \setminus \{5\}\). 2. Because the numerator \(8\) is positive, the quotient is negative when the denominator is negative. Solve \(2x - 10 < 0\) to get \(x < 5\), so the solution is \((-\infty, 5)\). 3. With numerator \(-8\), every nonzero output changes sign. The new expression is positive on \((-\infty, 5)\) and negative on \((5, \infty)\). The domain restriction at \(x = 5\) does not change.

a) \(D = \mathbb{R} \setminus \{5\}\). b) \((-\infty, 5)\). c) The signs reverse: the new expression is positive on \((-\infty, 5)\) and negative on \((5, \infty)\).

- What can you say about the value of a real square? - When the denominator is always positive, which part determines the sign of the quotient? - Can \(x^2 + 5\) ever equal zero for a real number?

1. For every real number \(x\), \(x^2 \ge 0\), so \(x^2 + 1 \ge 1\). The denominator is never zero, and the domain is \(\mathbb{R}\). 2. Because the denominator is always positive, the sign of \(Q(x)\) is determined by \(x - 5\). 3. The expression is negative when \(x - 5 < 0\), so \(x < 5\), or \((-\infty, 5)\). It is positive when \(x - 5 > 0\), so \(x > 5\), or \((5, \infty)\). It equals zero at \(x = 5\). 4. If the numerator were \(x^2 + 5\), then it would be at least \(5\) for every real \(x\). Since a rational expression can equal zero only when its numerator is zero and its denominator is nonzero, the modified expression would never equal zero.

a) The domain is \(\mathbb{R}\) because \(x^2 + 1\) is always positive. b) Negative on \((-\infty, 5)\); positive on \((5, \infty)\); zero at \(x = 5\). c) The numerator \(x^2 + 5\) is always positive, so it can never make the expression equal to zero.

- Multiply by a common denominator to clear the fractions. - Distribute the subtraction before the second numerator carefully. - Reverse the inequality sign when multiplying by \(-1\). - Use the final solution set to identify the greatest integer.

1. Multiply every term by the positive least common denominator, \(12\): \(3(x + 1) - 4(x - 2) \ge 6\). 2. Distribute: \(3x + 3 - 4x + 8 \ge 6\). 3. Combine like terms: \(-x + 11 \ge 6\). 4. Subtract \(11\): \(-x \ge -5\). 5. Multiply by \(-1\) and reverse the inequality sign: \(x \le 5\). 6. The greatest integer satisfying \(x \le 5\) is \(5\).

a) \(x \le 5\) b) \(5\)

- When does a rational expression equal zero? - What sign does \(x^2 + 1\) always have? - What happens to \(4x - 12\) when \(x > 3\)? - Use parentheses when substituting a negative number.

1. The denominator \(x^2 + 1\) is positive for every real \(x\), so \(T(x) = 0\) exactly when the numerator is zero. Solve \(4x - 12 = 0\) to get \(x = 3\). 2. If \(x > 3\), then \(4x > 12\), so \(4x - 12 > 0\). The denominator is also positive, so the quotient is positive. 3. Substitute \(x = -1\): \(T(-1) = \frac{4(-1) - 12}{(-1)^2 + 1} = \frac{-16}{2} = -8\).

a) The zero is \(x = 3\). b) For \(x > 3\), both \(4x - 12\) and \(x^2 + 1\) are positive, so \(T(x) > 0\). c) \(T(-1) = -8\).

- Multiply by a common denominator to eliminate all fractions. - Combine like terms carefully. - Substitute the test value into the original inequality. - Why does multiplying by a positive common denominator preserve the inequality sign?

1. Multiply by \(10\): \(2(2x-3)\le5(x+2)-10\). 2. Distribute and simplify: \(4x-6\le5x\). 3. Subtract \(4x\): \(-6\le x\), so \(x\ge-6\). 4. For \(x=0\), the left side is \(-\frac{3}{5}=-0.6\), and the right side is \(0\). Since \(-0.6\le0\), \(0\) is a solution.

a) \(\{x\in\mathbb{R}\mid x\ge-6\}\) b) Yes, because \(-0.6\le0\).

- Expand both expressions before comparing them. - What happens to the quadratic terms? - Pay attention to the strict inequality. - Which integer comes immediately after the boundary value?

1. Write \((x+4)^2<x(x+10)-2\). 2. Expand: \(x^2+8x+16<x^2+10x-2\). 3. Subtract \(x^2\): \(8x+16<10x-2\). 4. Subtract \(8x\) and add \(2\): \(18<2x\). Divide by \(2\): \(x>9\). 5. The smallest integer greater than \(9\) is \(10\).

a) \(x>9\) b) \(10\)

- Eliminate fractions using a common denominator. - Distribute a negative factor carefully. - Reverse the inequality when dividing by a negative number. - Move variable terms to one side and constants to the other.

1. Multiply the first inequality by \(8\): \(2(x-3)-4(x+1)\le x\). Distribute: \(2x-6-4x-4\le x\), so \(-2x-10\le x\). Add \(2x\): \(-10\le3x\). Divide by \(3\): \(x\ge-\frac{10}{3}\). 2. Distribute in the second inequality: \(7-6x+2>5x+20\), so \(9-6x>5x+20\). Subtract \(5x\) and \(9\): \(-11x>11\). Divide by \(-11\) and reverse the inequality: \(x<-1\).

1) \(x\ge-\frac{10}{3}\) 2) \(x<-1\)

- Solve each inequality separately. - The word “and” means take the intersection of the solution sets. - A number line can help compare the two sets. - Identify where both conditions are true.

1. For the first inequality, multiply by \(6\): \(6-2(2x-5)\le3(3x+1)\). 2. Simplify: \(6-4x+10\le9x+3\), so \(16-4x\le9x+3\). Then \(13\le13x\), giving \(x\ge1\). 3. For the second inequality, \(6x-2>5x-4\), so \(x>-2\). 4. The intersection of \(x\ge1\) and \(x>-2\) is \(x\ge1\).

\(x\ge1\)

- Solve each inequality separately. - Because both conditions must hold, take the intersection of the two solution sets. - Include only integers in the final answer.

1. Solve the first inequality: \(5x-3>2x+6\), so \(3x>9\) and \(x>3\). 2. Solve the second inequality: \(\frac{2x+8}{4}\leq5\), so \(2x+8\leq20\), \(2x\leq12\), and \(x\leq6\). 3. The intersection is \(3<x\leq6\). 4. The integers in this interval are \(4\), \(5\), and \(6\).

\(\{4, 5, 6\}\)

- Multiply the first inequality by the least common denominator to eliminate the fractions. - Distribute carefully when a minus sign is in front of parentheses. - Find the overlap of the two solution sets, then keep only integer values.

1. Multiply the first inequality by \(6\): \(3(3x-5)-2(x+1)<12\). 2. Distribute and combine like terms: \(9x-15-2x-2<12\), so \(7x<29\) and \(x<\frac{29}{7}\). 3. Solve the second inequality: \(1-3(2-x)\geq-5\) becomes \(-5+3x\geq-5\), so \(x\geq0\). 4. The intersection is \(0\leq x<\frac{29}{7}\). The integer solutions are \(0\), \(1\), \(2\), \(3\), and \(4\).

\(\{0, 1, 2, 3, 4\}\)

- Use a common denominator to eliminate the fractions in inequality I. - Distribute the negative factor carefully in inequality II. - Reverse the inequality sign when dividing by a negative number. - Take the intersection because both inequalities must be true.

1. Multiply inequality I by \(6\): \(2(2x+1)-(x-2)\leq12\). 2. Simplify: \(4x+2-x+2\leq12\), so \(3x\leq8\) and \(x\leq\frac{8}{3}\). 3. For inequality II, distribute and simplify: \(4-3x-3<2x+11\), so \(1-3x<2x+11\). 4. Subtract \(1\) and \(2x\): \(-5x<10\). Divide by \(-5\) and reverse the inequality: \(x>-2\). 5. The intersection is \(-2<x\leq\frac{8}{3}\).

\(\left(-2, \frac{8}{3}\right]\)

- Solve each inequality separately. - Reverse the inequality sign when dividing by a negative number. - Compare the two solution sets on a number line. - What does it mean if the two sets do not overlap?

1. Multiply inequality I by \(2\): \(10-(x+2)<2\). 2. Simplify: \(8-x<2\), so \(-x<-6\). Divide by \(-1\) and reverse the inequality: \(x>6\). 3. Multiply inequality II by \(2\): \(4(x-3)\leq x+4\). 4. Simplify: \(4x-12\leq x+4\), so \(3x\leq16\) and \(x\leq\frac{16}{3}\). 5. No real number can satisfy both \(x>6\) and \(x\leq\frac{16}{3}\), because \(\frac{16}{3}<6\).

There is no solution: \(\varnothing\).

- Translate “at most” and “nonnegative” into inequality symbols. - Include the boundary value when the wording allows equality. - Move variable terms to one side when comparing two expressions. - Reverse the sign when dividing by a negative coefficient.

1. “At most \(10\)” gives \(4(x + 2) \le 10\). Distribute: \(4x + 8 \le 10\). Subtract \(8\): \(4x \le 2\). Divide by \(4\): \(x \le 0.5\). 2. “Nonnegative” gives \(15 - 5x \ge 0\). Subtract \(15\): \(-5x \ge -15\). Divide by \(-5\) and reverse the sign: \(x \le 3\). 3. Write \(5x - 8 > 2x + 1\). Subtract \(2x\) and add \(8\): \(3x > 9\). Divide by \(3\): \(x > 3\).

1) \(x \le 0.5\) 2) \(x \le 3\) 3) \(x > 3\)

- Solve the inequality for \(x\) while leaving \(k\) in the expression. - Match the resulting boundary with the desired interval endpoint. - Solve the resulting equation for \(k\).

1. Solve for \(x\): \(2x+k>10\), so \(x>\frac{10-k}{2}\). 2. For the solution boundary to be \(7\), require \(\frac{10-k}{2}=7\). 3. Then \(10-k=14\), so \(-k=4\) and \(k=-4\).

\(k=-4\)

- Factor the equation \(ab - a = 0\). - Consider the zero-product property. - Dividing an inequality by a positive number preserves its direction. - Dividing an inequality by a negative number reverses its direction.

1. Rewrite \(ab = a\) as \(ab - a = 0\), then factor: \(a(b - 1) = 0\). Therefore, either \(a = 0\), with any rational value of \(b\), or \(b = 1\). 2. If \(a > 0\), divide \(ab < a\) by the positive number \(a\). The inequality sign stays the same, giving \(b < 1\). 3. If \(a < 0\), divide \(ab < a\) by the negative number \(a\). The inequality sign reverses, giving \(b > 1\).

1) \(a = 0\) with any rational \(b\), or \(b = 1\) 2) If \(a > 0\), then \(b < 1\). 3) If \(a < 0\), then \(b > 1\).