- 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) \(-10 < -5\)

- What is being added to the variable term? - Which operation will remove that number from the left side?

1. Subtract \(3\) from both sides to undo the addition. 2. The inequality becomes \(2x>8\). Subtracting the same number preserves its direction.

Subtract \(3\) from both sides: \(2x>8\).

- What operation will leave \(x\) alone? - Does dividing by a positive number reverse an inequality? - What does the line under the inequality symbol mean?

1. Divide both sides of \(3x\le12\) by positive \(3\): \(x\le4\). 2. The value \(4\) is allowed because \(3\cdot4-2=10\), equal to the permitted upper limit.

\(x\le4\). Yes, \(x=4\) is allowed.

- Check a sample negative and positive value of \(x\) to see which side works. - Think about what multiplying an inequality by a negative does to its order.

1. Divide both sides by \(-2\), which is negative. 2. Reverse the inequality direction: \(x>-3\).

\(x>-3\). The inequality reverses because both sides were divided by a negative number.

- Is the boundary value itself part of the solution? - Where are greater numbers located on a number line?

1. The sign \(\ge\) includes equality, so the point at \(-1\) is filled. 2. Values greater than \(-1\) lie to the right, so the ray goes right.

Filled point at \(-1\), because equality is included; shade right for numbers greater than \(-1\).

- Undo the subtraction to isolate \(x\). - Does a strict inequality include the boundary? - Which side of the boundary has smaller numbers?

1. Add \(2\) to both sides: \(x<5\). 2. The endpoint is open at \(5\), since equality is not allowed. 3. Shade to the left because the solutions are less than \(5\).

\(x<5\); open point at \(5\), shaded ray to the left.

- 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 \(0\). - 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 restriction to positive integers. - Check that the boundary itself is not included.

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

\(x=1,2,3\).

- 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) \(4 > 2\)

- Compare how an inequality changes under the same operation on both sides. - Pay special attention to whether the multiplier or divisor is positive or negative. - Check your choices with a simple example satisfying \(x<y\).

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) \(>\)

- 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.

- 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.

- Get the term containing \(x\) alone first. - What happens to the inequality direction when you divide by a negative? - Decide if the boundary itself is included.

1. Subtract \(5\) from both sides: \(-x\ge3\). 2. Divide by \(-1\) and reverse the direction: \(x\le-3\). 3. The point at \(-3\) is filled and the ray extends left.

\(x\le-3\); filled point at \(-3\), ray to the left.

- Check the exact boundary in the original inequality. - Compare the original inequality with the student's answer. - Check one number on either side of the proposed boundary.

1. The original inequality at \(x=4\) gives \(2\cdot4-1=7\), which is not less than \(7\). 2. Add \(1\): \(2x<8\). Divide by \(2\): \(x<4\). 3. The endpoint must not be included.

No: at \(x=4\), \(2\cdot4-1=7\), which is not less than \(7\). The correct solution is \(x<4\).

- Solve without the integer restriction first. - Which integers lie in both the allowed range and your solution? - Can you name a solution outside the stated range?

1. Add \(2\): \(3x\ge6\). 2. Divide by \(3\): \(x\ge2\). 3. Among the allowed integers, only \(2\), \(3\), and \(4\) satisfy the inequality. 4. The full solution has every real number at least \(2\), but the list is restricted to the stated integers.

Full solution: \(x\ge2\). Allowed integers: \(2,3,4\). The list excludes other solutions because only integers from \(-4\) through \(4\) are allowed.

- Solve the two inequalities separately. - Look for the numbers that appear in both solution sets. - What do the strict and inclusive signs say about the endpoints?

1. The first inequality gives \(x>1\). 2. The second gives \(x\le4\). 3. Values satisfying both are \(1<x\le4\). 4. The endpoint \(1\) is open and \(4\) is filled; shade only between them.

\(1<x\le4\); open at \(1\), filled at \(4\), shade between them.

- 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 \(x=\frac{3}{2}\).

- Write the two quantities on opposite sides of an inequality. - To identify all working integers, solve the inequality, not just two examples. - Compare the two given test values with the boundary you find.

1. The claim is true when \(3x>x+10\). 2. Subtract \(x\): \(2x>10\). Divide by \(2\): \(x>5\). 3. Therefore the positive integers for which the comparison is true are \(6,7,8,\ldots\). 4. At \(x=2\), \(6\) is not greater than \(12\); at \(x=10\), \(30\) is greater than \(20\). The “always” claim is false.

a) \(3x>x+10\) b) \(x>5\), so positive integers \(6,7,8,\ldots\). c) False at \(x=2\) and true at \(x=10\); it is not always true.

- 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 \(-4\).

- 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) \(x=4\)

- 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) \(z \ge 5\)

- What happens when you undo the constant in each inequality? - Test the integers nearest each boundary in the original inequalities. - Which values pass both tests?

1. For (I), subtract \(5\): \(x<-3\), so the integer solutions are \(\ldots,-6,-5,-4\). 2. For (II), add \(1\): \(x>-5\), so the integer solutions are \(-4,-3,-2,\ldots\). 3. The only integer in both lists is \(-4\).

(I) Integer solutions: \(\ldots,-6,-5,-4\). (II) Integer solutions: \(-4,-3,-2,\ldots\). Both: \(x=-4\).

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

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

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) \(x = 2.5\)

- 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 \(n \in \{5, 6, 7, 8\}\).

- Place the endpoints on a number line and identify integers between them. - For part 2), choose endpoints just outside the desired integers. - For part 3), move a fixed-length interval to a different position. - Check the distance between the two endpoints.

1. The integers between \(-3.2\) and \(2.1\) are \(-3,-2,-1,0,1,2\), so there are \(6\). 2. For example, \(-1.5<x<1.5\) includes exactly \(-1,0,1\). 3. The interval \(0.1<x<1.6\) has length \(1.5\) and contains only \(1\). 4. The interval \(0.9<x<2.4\) also has length \(1.5\) and contains \(1\) and \(2\).

1) \(6\) integers. 2) For example, \(a=-1.5\), \(b=1.5\). 3) For example, \(0.1<x<1.6\) contains one integer, while \(0.9<x<2.4\) contains two.

- Can you remove the added \(6\) without changing which values work? - Test a value larger than your proposed boundary in the original inequality. - Should the boundary itself count as a solution?

1. Subtract \(6\) from both sides: \(-2x<-10\). 2. Divide both sides by \(-2\) and reverse the sign: \(x>5\). 3. The graph has an open point at \(5\) and shading to the right because \(5\) is excluded and greater values work. 4. Dividing both sides by a negative number reverses their order. For instance, \(-2<-1\), but \(2>1\) after dividing by \(-1\).

\(x>5\); open point at \(5\), shade right. The sign reverses because division by a negative reverses the order of numbers.

- Solve the inequality independently before judging the graph. - Compare the inequality symbol with the meaning of an open versus filled endpoint. - Check the boundary value in the original inequality.

1. Add \(3\) to both sides: \(5x \le 20\). 2. Divide both sides by \(5\): \(x \le 4\). 3. Elif’s solution 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 before finding their overlap. - Check your work by trying a value in each original inequality. - Decide which endpoints belong to both sets and describe the segment yourself.

1. From \(2x+5>-1\), subtract \(5\) to get \(2x>-6\), then divide by \(2\): \(x>-3\). 2. From \(-3x+6\ge0\), subtract \(6\) to get \(-3x\ge-6\); divide by \(-3\) and reverse the sign: \(x\le2\). 3. Both inequalities hold when \(-3<x\le2\). 4. Place an open point at \(-3\), a filled point at \(2\), and shade only between them.

\(-3<x\le2\); open at \(-3\), filled at \(2\), shade between them.

- 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 \(x \ge -1\).