Free math worksheets

- Adding the same number to both sides preserves the inequality direction. - Track signs carefully when adding negative numbers. - Adding a negative number is equivalent to subtraction.

1. Add \(7\): \(5 + 7 < 9 + 7\), so \(12 < 16\). 2. Add \(3\): \(-3 + 3 > -8 + 3\), so \(0 > -5\). 3. Add \(-15\): \(12 + (-15) > 4 + (-15)\), so \(-3 > -11\). 4. Add \(-4\): \(-6 + (-4) < -1 + (-4)\), so \(-10 < -5\).

1) \(12 < 16\) 2) \(0 > -5\) 3) \(-3 > -11\) 4) \(-10 < -5\)

- Subtract \(x\) from both sides of each relationship. - Consider positive numbers, negative numbers, and zero separately. - Test one value from each category to check your reasoning.

1. For a), subtract \(x\): \(0 < 2x\). Divide by \(2\): \(x > 0\). 2. For b), subtract \(x\): \(0 = 2x\). Divide by \(2\): \(x = 0\). 3. For c), subtract \(x\): \(0 > 2x\). Divide by \(2\): \(x < 0\).

a) \(x > 0\) b) \(x = 0\) c) \(x < 0\)

- Translate “positive” as greater than \(0\). - Solve each equation or inequality separately. - For part 4), find the overlap of the two solution sets before applying the integer restriction.

1. Solve \(8x-24=0\): \(8x=24\), so \(x=3\). 2. Solve \(8x-24>0\): \(8x>24\), so \(x>3\). 3. Solve \(8x-24<16\): \(8x<40\), so \(x<5\). 4. Both conditions require \(3<x<5\). The only integer in that interval is \(x=4\).

1) \(x=3\) 2) \(x>3\) 3) \(x<5\) 4) \(x=4\)

- Translate “positive” as greater than \(0\). - Isolate the variable term. - Reverse the inequality sign when dividing by a negative number.

1. For a), solve \(4x - 12 > 0\): \(4x > 12\), so \(x > 3\). 2. For b), solve \(18 - 6x > 0\): \(-6x > -18\). Divide by \(-6\) and reverse the inequality sign: \(x < 3\). 3. For c), solve \(-3x - 21 > 0\): \(-3x > 21\). Divide by \(-3\) and reverse the inequality sign: \(x < -7\).

a) \(x > 3\) b) \(x < 3\) c) \(x < -7\)

- Subtract an original side from the corresponding new side. - Check that the same change occurred on both sides. - Adding a negative number moves both sides downward by that amount. - The difference between the two sides remains unchanged.

1. Compute the change: \(4 - (-2) = 6\). The right side also changes by \(11 - 5 = 6\), so \(6\) was added. 2. Compute \(2 - 10 = -8\). The right side changes by \(-5 - 3 = -8\), so \(-8\) was added. 3. Compute \(-15 - (-7) = -8\). The right side changes by \(-9 - (-1) = -8\), so \(-8\) was added. 4. Compute \(0.5 - 0 = 0.5\). The right side changes by \(-3.5 - (-4) = 0.5\), so \(0.5\) was added.

1) \(6\) 2) \(-8\) 3) \(-8\) 4) \(0.5\)

- Use the result of each step as the starting point for the next step. - Dividing by \(0.5\) is equivalent to multiplying by \(2\). - Multiplying or dividing by a positive number preserves the inequality direction.

1. Divide by \(4\): \(12 \div 4 < 24 \div 4\), so \(3 < 6\). 2. Multiply by \(7\): \(3 \cdot 7 < 6 \cdot 7\), so \(21 < 42\). 3. Divide by \(0.5\): \(21 \div 0.5 < 42 \div 0.5\), so \(42 < 84\).

1) \(3 < 6\) 2) \(21 < 42\) 3) \(42 < 84\)

- Isolate \(x\) using equivalent operations. - Pay attention to the stated domain of positive integers. - Test the listed values in the original inequality.

1. Add \(2\): \(3x < 12\). 2. Divide by \(3\): \(x < 4\). 3. The positive integers less than \(4\) are \(1\), \(2\), and \(3\).

\(S = \{1, 2, 3\}\)

- Check both the arithmetic and the inequality direction. - Multiplying by a negative number reverses the inequality sign. - Use the number line to compare the transformed values.

1. For a), multiplying gives \(-20\) and \(-30\), but the inequality sign must reverse. The correct result is \(-20 > -30\), so the transformation is incorrect. 2. For b), multiplying by \(-0.5\) gives \(2\) and \(4\), and the sign correctly reverses from \(>\) to \(<\). The transformation is correct. 3. For c), multiplying by \(-1\) gives \(-5\) and \(-9\), and the sign correctly reverses from \(\le\) to \(\ge\). The transformation is correct.

a) Incorrect; \(-20 > -30\) b) Correct c) Correct

- Positive multipliers and divisors preserve the inequality direction. - Negative multipliers and divisors reverse the inequality direction. - Compare the resulting numbers on a number line.

1. Divide by the positive number \(6\): \(18 \div 6 < 30 \div 6\), so \(3 < 5\). 2. Divide by the positive number \(4\): \(-12 \div 4 > -20 \div 4\), so \(-3 > -5\). 3. Multiply by the negative number \(-3\) and reverse the sign: \(5 \cdot (-3) < (-2) \cdot (-3)\), so \(-15 < 6\). 4. Divide by the negative number \(-2\) and reverse the sign: \(-8 \div (-2) > -4 \div (-2)\), so \(4 > 2\).

1) \(3 < 5\) 2) \(-3 > -5\) 3) \(-15 < 6\) 4) \(4 > 2\)

- Locate both boundary values on a number line. - List only integers strictly between the boundaries. - Include \(0\), which is an integer.

1. The integers between \(-2.7\) and \(3.5\) are \(-2, -1, 0, 1, 2, 3\). 2. The least integer is \(-2\), and the greatest integer is \(3\). 3. Their sum is \(-2 + 3 = 1\).

The integers are \(-2, -1, 0, 1, 2, 3\), and the requested sum is \(1\).

- Decide whether each endpoint is included. - Remember that integers include negative numbers and \(0\). - Rewrite the fractional bounds as decimals if useful. - Count each set only after listing its elements.

1. For a), the integer solutions are \(-3, -2, -1, 0, 1\), for a total of \(5\). 2. For b), the integer solutions are \(-2, -1, 0, 1, 2, 3\), for a total of \(6\). 3. For c), the bounds are \(-0.5\) and \(2.5\), so the integer solutions are \(0, 1, 2\), for a total of \(3\). 4. Condition b) has the greatest number of integer solutions.

a) \(\{-3, -2, -1, 0, 1\}\) b) \(\{-2, -1, 0, 1, 2, 3\}\) c) \(\{0, 1, 2\}\) Condition b) has the most solutions.

- Subtract the left sides and the right sides in corresponding order. - View subtraction as adding the opposite inequality. - Distribute subtraction signs through parentheses. - Combine like terms before solving further.

1. For a), subtract corresponding sides: \(18 - 3 > 10 - 12\), so \(15 > -2\). 2. For b), subtract corresponding sides: \(-5 - 7 < 4 - 2\), so \(-12 < 2\). 3. For c), subtract corresponding sides: \(8x - 3x > 24 - 15\), so \(5x > 9\). Divide by \(5\): \(x > \frac{9}{5} = 1.8\). 4. For d), subtract corresponding sides: \((a + 5) - (b - 2) < 9 - 4\). Simplify: \(a - b + 7 < 5\). Subtract \(7\): \(a - b < -2\).

a) \(15 > -2\) b) \(-12 < 2\) c) \(x > \frac{9}{5}\) d) \(a - b < -2\)

- Compute each product before comparing. - Track the sign and magnitude separately. - A product of numbers with the same sign is positive. - Use the signed multiplication rule explicitly in part c).

1. a) \(9\cdot3=27\) and \(4\cdot2=8\), so \(27>8\). 2. b) \(6\cdot4=24\) and \((-3)\cdot1=-3\), so \(24>-3\). 3. c) \((-5)(-4)=20\) and \(2\cdot3=6\), so \(20>6\). 4. In part c), the product of two negative numbers is positive.

a) \(27>8\) b) \(24>-3\) c) \(20>6\) The key rule in part c) is that a negative times a negative is positive.

- Read the x-value at the open endpoint. - Decide which direction the blue ray extends. - An open endpoint means the boundary value is not included.

1. The open endpoint is at \(-2\), and the solution extends to the right, so the inequality is \(x>-2\). 2. The boundary value \(-2\) is not included because the endpoint is an open circle. 3. For example, \(0\) is a solution and \(-3\) is not a solution.

a) \(x>-2\) b) No. The open circle means \(-2\) is excluded. c) For example, \(0\) is a solution and \(-3\) is not.

- Read the x-value at the filled endpoint. - Decide which direction the blue ray extends. - A filled endpoint means the boundary value is included. - Compare \(4\) and \(-1\) with the boundary value \(3\).

1. The filled endpoint is at \(3\), and the solution extends to the left, so the inequality is \(x\le3\). 2. The value \(3\) is included because the endpoint is a filled dot. 3. Since \(4>3\), \(4\) is not a solution. Since \(-1\le3\), \(-1\) is a solution.

a) \(x\le3\) b) Yes. The filled dot means \(3\) is included. c) \(4\) is not a solution; \(-1\) is a solution.

- Apply the same operation to all three parts of each compound inequality. - Undo the division before undoing the addition. - List only the natural numbers strictly between the final bounds.

1. For a), multiply every part by \(5\): \(10<n+5<20\). Subtract \(5\) from every part: \(5<n<15\). Thus \(n\in\{6,7,8,9,10,11,12,13,14\}\). 2. For b), multiply every part by \(10\): \(5<n+2<8\). Subtract \(2\) from every part: \(3<n<6\). Thus \(n\in\{4,5\}\).

a) \(n\in\{6,7,8,9,10,11,12,13,14\}\) b) \(n\in\{4,5\}\)

- Distinguish carefully between \(<\) and \(\le\). - Apply the same operation to all three parts of each inequality. - Use inverse operations in reverse order, then list the natural numbers in the resulting interval.

1. For a), multiply every part by \(12\): \(8<x+3\le15\). Subtract \(3\) from every part: \(5<x\le12\). Thus \(x\in\{6,7,8,9,10,11,12\}\). 2. For b), multiply every part by \(8\): \(6<x-2<12\). Add \(2\) to every part: \(8<x<14\). Thus \(x\in\{9,10,11,12,13\}\).

a) \(x\in\{6,7,8,9,10,11,12\}\) b) \(x\in\{9,10,11,12,13\}\)

- Apply the same operation to every part of a compound inequality. - Use two inverse-operation steps to isolate \(k\). - After finding the interval, list the natural numbers it contains.

1. For a), multiply every part by \(7\): \(7<2k<14\). Divide every part by \(2\): \(3.5<k<7\). Thus \(k\in\{4,5,6\}\). 2. For b), multiply every part by \(9\): \(3\le k+2<6\). Subtract \(2\) from every part: \(1\le k<4\). Thus \(k\in\{1,2,3\}\).

a) \(k\in\{4,5,6\}\) b) \(k\in\{1,2,3\}\)

- Use inverse operations to rewrite each condition as an inequality for \(x\). - Apply the same operation to both sides of each inequality. - Choose a fraction strictly between the resulting bounds.

1. For the first inequality, subtract \(\frac{1}{3}\): \(x\cdot\frac{2}{3}>\frac{2}{3}\). Divide by \(\frac{2}{3}\): \(x>1\). 2. For the second inequality, add \(\frac{1}{2}\): \(x\cdot\frac{2}{3}<\frac{3}{2}\). Divide by \(\frac{2}{3}\): \(x<\frac{3}{2}\cdot\frac{3}{2}=\frac{9}{4}\). 3. Choose a fraction between \(1\) and \(\frac{9}{4}\), such as \(x=\frac{3}{2}\). 4. Check: \(\frac{3}{2}\cdot\frac{2}{3}+\frac{1}{3}=\frac{4}{3}>1\), and \(\frac{3}{2}\cdot\frac{2}{3}-\frac{1}{2}=\frac{1}{2}<1\).

One possible value is \(x=\frac{3}{2}\).

- First find the total cost of the water. - The total cost may equal the budget but cannot exceed it. - Subtract the fixed cost before dividing by the price per juice bottle.

1. The water costs \(12 \cdot \$0.75 = \$9.00\). 2. The total cost cannot exceed the budget, so \(9 + 1.40x \leq 30\). 3. Subtract \(9\): \(1.40x \leq 21\). 4. Divide by \(1.40\): \(x \leq 15\). 5. Since \(x\) counts bottles, the greatest possible value is \(15\).

The inequality is \(12 \cdot 0.75 + 1.40x \leq 30\), or \(9 + 1.40x \leq 30\). The group can buy at most \(15\) bottles of apple juice.

- Translate each verbal quantity into an algebraic expression. - Test the two specified values by substitution. - One counterexample is enough to disprove an “always” claim.

1. Three times the number is \(3x\), and the sum of the number and \(10\) is \(x + 10\). 2. At \(x = 2\), \(3 \cdot 2 = 6\) and \(2 + 10 = 12\). Since \(6 < 12\), the claim is false for \(x = 2\). 3. At \(x = 10\), \(3 \cdot 10 = 30\) and \(10 + 10 = 20\). Since \(30 > 20\), the claim is true for \(x = 10\). 4. The claim is not true for all positive integers because \(x = 2\) is a counterexample. 5. In fact, \(3x > x + 10\) simplifies to \(2x > 10\), so the inequality is true exactly when \(x > 5\).

a) \(3x\) and \(x + 10\) b) The claim is false at \(x = 2\) and true at \(x = 10\). c) No. The counterexample \(x = 2\) shows that the claim is not true for all positive integers.

- Write an expression for the total amount saved after \(n\) months. - “Enough money” means the amount must be at least the price. - Round up to the next whole month after solving.

1. Let \(n\) be the number of months. 2. After \(n\) months, Lena will have \(155 + 25n\) dollars. 3. She needs at least \(\$540\), so \(155 + 25n \geq 540\). 4. Subtract \(155\): \(25n \geq 385\). 5. Divide by \(25\): \(n \geq 15.4\). 6. Since only full months count, the least possible value is \(16\).

Lena will have enough money after \(16\) months.

- Read the two places where the blue line meets the dashed horizontal boundaries. - Follow the line from left to right and describe its position relative to the strip in each region. - An open strip excludes points exactly on either boundary. - Use the overall left-to-right direction of the line to describe whether the function is increasing or decreasing.

1. The line crosses the lower dashed boundary \(y=-3\) at \(x=-4\) and the upper dashed boundary \(y=2\) at \(x=6\). 2. Because the line is increasing, it lies below the strip for \(x<-4\), inside the strip for \(-4<x<6\), and above the strip for \(x>6\). 3. The strip is open, so the crossing points themselves are excluded. 4. The graph rises from left to right, so it must meet the lower boundary before the upper boundary.

a) Below for \(x<-4\); inside for \(-4<x<6\); above for \(x>6\) b) Crossings at \(x=-4\) and \(x=6\); inside interval: \(-4<x<6\) c) \(g\) is increasing; it crosses the lower boundary first and the upper boundary second.

- Identify the longest side. - Compare the sum of the two shorter sides with the longest side. - After finding the least integer value of \(k\), add all three side lengths.

1. For positive \(k\), the longest side is \(k + 4\). The sum of the two shorter sides must be greater than the longest side. 2. Write \(k + (k + 2) > k + 4\). 3. Simplify: \(2k + 2 > k + 4\), so \(k > 2\). 4. The other two triangle inequalities are satisfied whenever \(k\) is positive. 5. The least integer greater than \(2\) is \(3\). 6. The perimeter is \(k + (k + 2) + (k + 4) = 3k + 6\). For \(k = 3\), the perimeter is \(3 \cdot 3 + 6 = 15\,\text{in.}\).

The triangle exists when \(k > 2\). If \(k\) is an integer, the least possible perimeter is \(15\,\text{in.}\).

- Include both the fixed bus cost and the per-person cost in the inequality. - “Does not exceed” means \(\le\). - Solve the inequality before applying the whole-number restriction. - The algebraic solution set contains decimals, but a person count does not.

1. The complete cost is \(135+5.20x\), and it may not exceed \(280\): \( 135+5.20x\le280. \) 2. Subtract \(135\): \( 5.20x\le145. \) 3. Divide by \(5.20\): \( x\le27.8846\ldots \) 4. Since \(x\) counts people, the greatest feasible whole number is \(27\).

a) \(135+5.20x\le280\) b) \(x\le27.8846\ldots\) c) At most \(27\) people can attend.

- Translate “positive” as being greater than \(0\). - Isolate the variable term first. - Recall the rule for dividing an inequality by a negative number.

1. A positive value is represented by \(-4x - 12 > 0\). 2. Add \(12\): \(-4x > 12\). 3. Divide by \(-4\) and reverse the inequality sign: \(x < -3\). 4. The sign reverses because multiplying or dividing both sides of an inequality by a negative number reverses the order.

The expression is positive when \(x < -3\). The inequality sign reverses because the inequality is divided by \(-4\).

- Subtract the common term \(a\) from both sides. - Subtracting a negative number is equivalent to adding a positive number. - In part b), collect all terms containing \(b\) on one side.

1. For a), subtract \(a\) from both sides of \(a - b > a\): \(-b > 0\). This is equivalent to \(b < 0\). 2. Subtracting a negative number increases the original value. For example, \(7 - (-3) = 10\), and \(10 > 7\). 3. For b), subtract \(a\) from both sides: \(-b = b\). 4. Add \(b\): \(0 = 2b\). Divide by \(2\): \(b = 0\). The value of \(a\) can be any rational number.

a) \(b < 0\). For example, \(7 - (-3) = 10 > 7\). b) \(b = 0\)

- Reverse the inequality sign when dividing by a negative number. - Solve each condition separately. - Use the sign of the coefficient of \(x\) to describe the trend.

1. Solve \(15 - 3x < 0\): \(-3x < -15\). Divide by \(-3\) and reverse the inequality sign: \(x > 5\). 2. Solve \(15 - 3x > 21\): \(-3x > 6\). Divide by \(-3\) and reverse the sign: \(x < -2\). 3. Solve \(15 - 3x = -3\): \(-3x = -18\), so \(x = 6\). 4. The coefficient of \(x\) is negative, so the expression decreases as \(x\) increases.

1) \(x > 5\) 2) \(x < -2\) 3) \(x = 6\) 4) The expression decreases as \(x\) increases.

- Translate “less than \(0\)” and “not positive” into inequality symbols. - Clear the fraction in part a) with multiplication. - Decide whether the boundary value belongs to the solution set. - Reverse the inequality sign when dividing by a negative number.

1. For a), solve \(\frac{1}{4}z + 2 < 0\). Subtract \(2\): \(\frac{1}{4}z < -2\). Multiply by \(4\): \(z < -8\). 2. For b), solve \(15 - 3z \le 0\). Subtract \(15\): \(-3z \le -15\). Divide by \(-3\) and reverse the inequality sign: \(z \ge 5\).

a) \(z < -8\) b) \(z \ge 5\)

- Solve each inequality separately. - List or describe the integer solutions to each one. - Find the intersection of the two solution sets.

1. Solve (I): subtract \(5\) to get \(x < -3\). Over the integers, the solution set is \(\{\ldots, -6, -5, -4\}\). 2. Solve (II): add \(1\) to get \(x > -5\). Over the integers, the solution set is \(\{-4, -3, -2, \ldots\}\). 3. The only integer that is both less than \(-3\) and greater than \(-5\) is \(-4\).

Yes. The common integer solution is \(x = -4\).

- Use an equation when the expression values are equal. - Use an inequality when one expression value is smaller. - Move variable terms to one side and constants to the other. - Dividing by a positive number preserves the inequality direction.

1. For a), set the expressions equal: \(5x - 3 = 2x + 9\). 2. Subtract \(2x\) and add \(3\): \(3x = 12\). Divide by \(3\): \(x = 4\). 3. For b), write \(5x - 3 < 2x + 9\). 4. Subtract \(2x\) and add \(3\): \(3x < 12\). Divide by \(3\): \(x < 4\).

a) \(x = 4\) b) \(x < 4\)

- Translate positive, negative, and zero into comparison symbols. - Multiplying by the positive denominator \(3\) preserves the inequality direction. - Reverse the sign when dividing by the negative coefficient \(-2\). - The zero value is the boundary between the positive and negative cases.

1. For a), solve \(\frac{5 - 2x}{3} > 0\). Multiply by \(3\): \(5 - 2x > 0\). Subtract \(5\): \(-2x > -5\). Divide by \(-2\) and reverse the sign: \(x < 2.5\). 2. For b), solve \(\frac{5 - 2x}{3} < 0\). Multiplying by \(3\) and solving gives \(-2x < -5\), so \(x > 2.5\). 3. For c), solve \(\frac{5 - 2x}{3} = 0\). Then \(5 - 2x = 0\), so \(x = 2.5\).

a) \(x < 2.5\) b) \(x > 2.5\) c) \(x = 2.5\)

- Adding the same value preserves order. - Multiplying by a positive value preserves order. - Multiplying or dividing by a negative value reverses order. - Picture the values on a number line if needed.

1. Adding \(10\) to both sides preserves the order: \(x + 10 < y + 10\). 2. Multiplying by the positive number \(5\) preserves the order: \(5x < 5y\). 3. Multiplying by \(-1\) reverses the order: \(-x > -y\). 4. Dividing by \(-2\) reverses the order: \(\frac{x}{-2} > \frac{y}{-2}\).

a) \(<\) b) \(<\) c) \(>\) d) \(>\)

- Translate “five times the number minus \(7\)” into an expression. - Use inequality symbols that do not include the endpoints. - Perform the same operation on all three parts of the compound inequality. - Keep only whole-number solutions at the end.

1. Translate the statement into the compound inequality \(15 < 5n - 7 < 35\). 2. Add \(7\) to all three parts: \(22 < 5n < 42\). 3. Divide all three parts by \(5\): \(\frac{22}{5} < n < \frac{42}{5}\), or \(4.4 < n < 8.4\). 4. The whole numbers in this interval are \(5\), \(6\), \(7\), and \(8\).

The values are \(n \in \{5, 6, 7, 8\}\).

- Mark each open interval on a number line. - Place the endpoints so the desired integers are inside and the neighboring integers are outside. - For part 3), slide an interval of fixed length along the number line. - Check both the interval length and the integers it contains.

1. The integers satisfying \(-3.2 < x < 2.1\) are \(-3, -2, -1, 0, 1, 2\), so there are \(6\). 2. The lower endpoint can be between \(-2\) and \(-1\), and the upper endpoint can be between \(1\) and \(2\). One example is \(-1.5 < x < 1.5\), whose integer solutions are exactly \(-1, 0, 1\). 3. The interval \(0.1 < x < 1.6\) has length \(1.5\) and contains only the integer \(1\). 4. The interval \(0.9 < x < 2.4\) also has length \(1.5\), but it contains the integers \(1\) and \(2\).

1) \(6\) integers 2) One possible answer is \(a = -1.5\) and \(b = 1.5\). 3) For example, \((0.1, 1.6)\) contains one integer, while \((0.9, 2.4)\) contains two integers.

- Add the left sides together and the right sides together. - The inequality signs must point in the same direction before adding. - Combine like terms after the addition. - Continue solving if the result contains one variable.

1. Add corresponding sides: \(22 + 5 > 14 + (-2)\), so \(27 > 12\). 2. Add corresponding sides: \(-10 + 3 < -4 + 8\), so \(-7 < 4\). 3. Add corresponding sides: \((4x + 7) + (x - 5) > (2x - 1) + (6 - x)\). Simplify: \(5x + 2 > x + 5\). Subtract \(x\) and \(2\): \(4x > 3\), so \(x > \frac{3}{4} = 0.75\). 4. Add corresponding sides: \((3a - 2b) + (b - a) < (a + 5) + (4 - 2a)\). Simplify: \(2a - b < 9 - a\). Add \(a\): \(3a - b < 9\).

1) \(27 > 12\) 2) \(-7 < 4\) 3) \(x > \frac{3}{4}\) 4) \(3a - b < 9\)

- Subtracting an upper bound from a lower bound can produce a valid lower bound for a difference. - For the counterexample, choose \(x\) and \(y\) close together. - A proposed rule is false if one valid pair violates its conclusion.

1. Because \(x > 15\) and \(y < 6\), subtracting corresponding sides gives \(x - y > 15 - 6\), so \(x - y > 9\). 2. A possible counterexample is \(x = 16\) and \(y = 14\). Both original conditions hold because \(16 > 15\) and \(14 > 10\). 3. However, \(x - y = 16 - 14 = 2\), and \(2\) is not greater than \(5\). Therefore, subtracting inequalities with the same direction does not justify \(x - y > 5\).

1) \(x - y > 9\) 2) One counterexample is \(x = 16\), \(y = 14\), for which \(x - y = 2\).

- Calculate both products before choosing a comparison sign. - Compare the final sign with the signs in the original pair. - Pay close attention to products involving negative numbers.

1. For a), \(2 \cdot 4 = 8\) and \(5 \cdot 6 = 30\), so \(8 < 30\). The direction remains \(<\). 2. For b), \(5 \cdot (-4) = -20\) and \((-2)(-8) = 16\), so \(-20 < 16\). The direction changes from \(>\) to \(<\). 3. For c), \((-3)(-5) = 15\) and \(4 \cdot (-1) = -4\), so \(15 > -4\). The direction changes from \(<\) to \(>\). 4. The direction changes in b) and c). Therefore, term-by-term multiplication does not generally preserve the direction of inequalities.

a) \(8 < 30\); direction unchanged b) \(-20 < 16\); direction changed c) \(15 > -4\); direction changed The direction changes in b) and c).

- Solve the inequality step by step first. - Decide whether the endpoint at \(5\) should be open or filled. - Decide whether values greater than \(5\) lie to the left or to the right on the number line. - Remember that multiplying or dividing an inequality by a negative number reverses the inequality sign.

1. Subtract \(6\) from both sides: \(-2x < -10\). 2. Divide both sides by \(-2\): \(x > 5\). 3. The correct graph is b), because it shows an open endpoint at \(5\) and the ray extends to the right. 4. The inequality sign changes direction because dividing both sides of an inequality by a negative number reverses the order.

\(x > 5\). Graph b) is correct. The sign reverses because both sides are divided by \(-2\), which is negative.

- Solve the inequality before judging the graph. - The blue solution segment is drawn above the x-axis so it stays visible. - The dashed vertical guide connects the endpoint to its x-coordinate. - Compare the symbol \(\le\) with the meaning of an open versus filled endpoint. - Substitute \(x = 4\) into the original inequality to check whether the boundary value belongs.

1. Add \(3\) to both sides: \(5x \le 20\). 2. Divide both sides by \(5\): \(x \le 4\). 3. The student's blue segment extends in the correct direction, toward values less than \(4\). 4. However, the endpoint at \(x = 4\) is shown as open. 5. Because the inequality is \(x \le 4\), equality is included, so the endpoint at \(4\) must be filled. 6. Substituting \(x = 4\) into the original inequality confirms this: \(5(4) - 3 = 17\), and \(17 \le 17\) is true.

a) \(x \le 4\) b) The endpoint at \(x = 4\) should be filled rather than open. c) Yes. \(x = 4\) satisfies the original inequality and is part of the solution set.

- Solve each inequality separately. - The final solution set is the overlap of the two individual solution sets. - Use the inequality symbols to decide whether each endpoint is open or closed. - Compare the graph with the interval you found.

1. Solve \(2x + 5 > -1\): subtract \(5\) to get \(2x > -6\), then divide by \(2\) to get \(x > -3\). 2. Solve \(-3x + 6 \ge 0\): subtract \(6\) to get \(-3x \ge -6\), then divide by \(-3\) and reverse the sign to get \(x \le 2\). 3. The values satisfying both inequalities are those in the overlap: \(-3 < x \le 2\). 4. The graph shows an open endpoint at \(-3\) because \(-3\) is not included. 5. The graph shows a filled endpoint at \(2\) because \(2\) is included.

\(-3 < x \le 2\). The endpoint at \(-3\) is open because it is excluded, and the endpoint at \(2\) is filled because it is included.

- Read the graph first: decide the solved inequality shown by the endpoint and direction. - Test each possible symbol in the original inequality. - Watch carefully for a sign reversal when dividing by \(-3\). - A filled endpoint means the boundary value is included.

1. The graph shows a filled endpoint at \(-1\) and extends to the right, so the solved inequality is \(x \ge -1\). 2. We need \(-3x + 2 \square 5\) to solve to \(x \ge -1\). 3. If we choose \(\le\), then \(-3x + 2 \le 5\). 4. Subtract \(2\): \(-3x \le 3\). 5. Divide by \(-3\) and reverse the sign: \(x \ge -1\). 6. This matches the graph, so the correct symbol is \(\le\). 7. The symbol changes because dividing an inequality by a negative number reverses the direction.

The correct symbol is \(\le\). Solving \(-3x + 2 \le 5\) gives \(x \ge -1\).

- Start with cost per notebook times the number of notebooks. - Add the fixed shipping charge. - Use “at most” to decide the inequality symbol. - The graph is continuous, but the context is discrete. - Count only whole-number values that satisfy the inequality.

1. Let \(n\) be the number of notebooks. 2. The total cost is \(4n + 8\), and it must be at most \(28\), so the inequality is \(4n + 8 \le 28\). 3. Subtract \(8\): \(4n \le 20\). 4. Divide by \(4\): \(n \le 5\). 5. The continuous graph shows all real values \(n \le 5\), but a notebook count must be a whole number and cannot be negative. 6. Therefore, the feasible notebook counts are \(0, 1, 2, 3, 4,\) and \(5\).

a) \(4n + 8 \le 28\) b) A notebook count must be a nonnegative whole number, so fractional and negative values are not possible. c) \(0, 1, 2, 3, 4, 5\)

- Find how much of each sale remains after paying the per-waffle topping cost. - Represent the number of waffles with one variable and use the net amount as its coefficient. - Translate “at least” with a lower-bound inequality. - Interpret a decimal boundary using the fact that the number of waffles must be whole.

1. Let \(w\) be the number of waffles sold. The net amount from each waffle after toppings is \(1.50-0.35=1.15\) dollars. 2. To avoid a loss, the net waffle revenue must cover the fixed costs: \(1.15w\ge42\). 3. Solve: \(w\ge36.521\ldots\). Because \(w\) is a whole number, at least \(37\) waffles are needed. 4. A profit of at least \(\$120.00\) means net waffle revenue minus fixed costs is at least \(120\): \(1.15w-42\ge120\). 5. Add \(42\): \(1.15w\ge162\), so \(w\ge140.869\ldots\). 6. Therefore, at least \(141\) waffles are needed.

a) Net amount per waffle: \(\$1.15\). Inequality: \(1.15w\ge42\); at least \(37\) waffles. b) Inequality: \(1.15w-42\ge120\); at least \(141\) waffles.

- Use the triangle inequality: the sum of any two sides must be greater than the third side. - Write all three inequalities. - Solve each inequality for \(x\). - Find the overlap of the solution sets before choosing the least integer.

1. Apply the triangle inequality to each pair of sides. 2. From \(x + 12 > 3x - 2\), obtain \(14 > 2x\), so \(x < 7\). 3. From \(x + (3x - 2) > 12\), obtain \(4x - 2 > 12\), so \(x > \frac{7}{2}\). 4. From \(12 + (3x - 2) > x\), obtain \(10 + 3x > x\), so \(x > -5\). 5. Side lengths must also be positive. Those restrictions are weaker than \(x > \frac{7}{2}\). 6. Combining all conditions gives \(\frac{7}{2} < x < 7\). The least integer in this interval is \(4\).

The range is \(\frac{7}{2} < x < 7\), and the least possible integer value is \(4\).

- Write one triangle inequality for each pair of sides. - Solve all three inequalities and find their common solution. - Check that every side length is positive. - Count the integers strictly between the two bounds.

1. Apply the triangle inequality to each pair of sides. 2. From \(2x + (x + 10) > 30\), obtain \(3x + 10 > 30\), so \(x > \frac{20}{3}\). 3. From \(2x + 30 > x + 10\), obtain \(x > -20\). 4. From \((x + 10) + 30 > 2x\), obtain \(x < 40\). 5. The positivity requirements for the side lengths are weaker than \(x > \frac{20}{3}\). Therefore, \(\frac{20}{3} < x < 40\). 6. The integer values are \(7, 8, \ldots, 39\). Their number is \(39 - 7 + 1 = 33\).

The range is \(\frac{20}{3} < x < 40\). There are \(33\) possible integer values of \(x\).

- Express the ones digit using the tens digit first. - Write the two-digit number as tens value plus ones value. - Translate “greater than \(20\) and less than \(60\)” into one compound inequality. - After solving, enforce that the tens digit must be an integer digit.

1. The ones digit is \(t+4\), so the number is \(10t+(t+4)=11t+4\). 2. The range condition is \(20<11t+4<60\). 3. Subtract \(4\): \(16<11t<56\). 4. Divide by \(11\): \(\frac{16}{11}<t<\frac{56}{11}\). 5. Since \(t\) is a digit, \(t\in\{2,3,4,5\}\). The corresponding ones digits are \(6,7,8,9\). 6. The numbers are \(26,37,48,59\).

Compound inequality: \(20<11t+4<60\), so \(\frac{16}{11}<t<\frac{56}{11}\). Possible numbers: \(26,37,48,59\).

- Add corresponding sides first. - Also solve each original inequality to find the smallest integer values of \(x\) and \(y\). - The added inequality may be weaker than the original conditions. - For part b), search for examples that produce \(<\), \(>\), and \(=\).

1. For a), add corresponding sides: \((x + 3) + (y - 5) > 10 + 2\). 2. Simplify: \(x + y - 2 > 12\), so \(x + y > 14\). 3. The original inequalities give the stronger individual conditions \(x > 7\) and \(y > 7\). Since \(x\) and \(y\) are integers, their least possible values are \(8\) and \(8\). Therefore, the least possible sum is \(16\). 4. For b), opposite directions do not determine one relationship. For example, \(5 < 12\) and \(10 > 3\) give equal sums: \(15 = 15\). 5. In contrast, \(1 < 2\) and \(10 > 1\) give \(11 > 3\), while \(1 < 10\) and \(2 > 1\) give \(3 < 11\). 6. Therefore, adding inequalities with opposite directions can produce \(<\), \(>\), or \(=\), depending on the values.

a) The added inequality is \(x + y > 14\), and the least possible integer value of \(x + y\) is \(16\). b) No single relationship is guaranteed when the inequality signs point in opposite directions.