Free Math Worksheets

- Substitute the proposed value for every occurrence of \(x\). - Evaluate the left and right sides separately. - The value is a solution only when both sides are equal.

1. Substitute \(x = 3\): \(7 \cdot 3 - 5 = 16\). The equation is true, so \(3\) is a solution. 2. Substitute \(x = 4\): \(18 - 3 \cdot 4 = 6\). The equation is true, so \(4\) is a solution. 3. Substitute \(x = 1\): \(2(1 + 4) = 10\). The equation is true, so \(1\) is a solution. 4. Substitute \(x = 5\): the left side is \(4 \cdot 5 + 1 = 21\), while the right side is \(2 \cdot 5 + 9 = 19\). The equation is false, so \(5\) is not a solution.

1) Yes 2) Yes 3) Yes 4) No

- Use inverse operations to isolate \(x\). - Check the sign of each solution carefully. - Every positive integer is also an integer, and every integer is also rational.

1. For a), \(2x + 5 = -9\), so \(2x = -14\) and \(x = -7\). This value is in \(\mathbb{Z}\) and \(\mathbb{Q}\). 2. For b), \(4x - 3 = -15\), so \(4x = -12\) and \(x = -3\). This value is in \(\mathbb{Z}\) and \(\mathbb{Q}\). 3. For c), \(3x - 4 = 29\), so \(3x = 33\) and \(x = 11\). This value is in \(\mathbb{N}\), \(\mathbb{Z}\), and \(\mathbb{Q}\).

a) \(x = -7\); \(\mathbb{Z}\) and \(\mathbb{Q}\) b) \(x = -3\); \(\mathbb{Z}\) and \(\mathbb{Q}\) c) \(x = 11\); \(\mathbb{N}\), \(\mathbb{Z}\), and \(\mathbb{Q}\)

- Treat the expression inside the grouping symbols as one quantity at first. - Undo the outer operation before the inner operation. - Multiplication and division are inverse operations. - Check each answer in the original equation.

1. For a), divide by \(3\): \(x + 1.2 = 5\). Subtract \(1.2\): \(x = 3.8\). 2. For b), multiply by \(4\): \(18.6 - x = 14\). Subtract \(18.6\): \(-x = -4.6\). Multiply by \(-1\): \(x = 4.6\).

a) \(x = 3.8\) b) \(x = 4.6\)

- Identify the operation performed last on the variable. - Undo addition or subtraction before undoing multiplication or division. - Perform the same inverse operation on both sides. - Follow the sign rules when dividing by a negative number.

1. For a), subtract \(14\): \(5x = 25\). Divide by \(5\): \(x = 5\). 2. For b), subtract \(18\): \(-2x = -8\). Divide by \(-2\): \(x = 4\). 3. For c), subtract \(5\): \(\frac{x}{3} = 7\). Multiply by \(3\): \(x = 21\).

a) \(x = 5\) b) \(x = 4\) c) \(x = 21\)

- In part a, you can divide before working inside the parentheses. - Perform each operation on both sides of the equation. - Pay close attention to a negative coefficient on the variable. - Substitute each result to check it.

1. For a), divide by \(4\): \(x + 2 = 7\). Subtract \(2\): \(x = 5\). 2. For b), subtract \(21\): \(-9 = 3x\). Divide by \(3\): \(x = -3\). 3. For c), subtract \(15\): \(-5x = 20\). Divide by \(-5\): \(x = -4\).

a) \(x = 5\) b) \(x = -3\) c) \(x = -4\)

- Isolate the variable term before dividing. - Perform the same operation on both sides of the equation. - Follow the sign rules when a variable has a negative coefficient.

1. For a), subtract \(12\): \(5x = 25\). Divide by \(5\): \(x = 5\). 2. For b), subtract \(18\): \(-3y = -12\). Divide by \(-3\): \(y = 4\). 3. For c), add \(4\): \(4z = 16\). Divide by \(4\): \(z = 4\). 4. For d), subtract \(7\): \(2x = -18\). Divide by \(2\): \(x = -9\).

a) \(x = 5\) b) \(y = 4\) c) \(z = 4\) d) \(x = -9\)

- Isolate the variable term before dividing. - Follow the sign rules when dividing by a negative number. - Multiplying by a reciprocal can undo multiplication by a fraction. - Apply each operation to both sides.

1. For a), subtract \(14\): \(6x = 36\). Divide by \(6\): \(x = 6\). 2. For b), subtract \(18\): \(-4y = -16\). Divide by \(-4\): \(y = 4\). 3. For c), add \(3\): \(0.5z = 7.5\). Divide by \(0.5\): \(z = 15\). 4. For d), subtract \(4\): \(\frac{2}{3}b = 6\). Multiply by \(\frac{3}{2}\): \(b = 9\).

a) \(x = 6\) b) \(y = 4\) c) \(z = 15\) d) \(b = 9\)

- Work backward from the result. - Undo the last operation first. - Use inverse operations for addition and division. - Check by applying the original steps.

1. Undo the addition: \(30 - 15 = 15\). 2. Undo the division by multiplying: \(15 \cdot 4 = 60\). 3. Check: \(60 \div 4 + 15 = 30\).

The number is \(60\).

- Work backward from \(500\). - Undo the addition before undoing the multiplication. - Translate the riddle into an equation with a variable. - Check your solution in the original statement.

1. Undo the addition: \(500 - 140 = 360\). 2. Divide by \(3\): \(360 \div 3 = 120\). 3. Check: \(3 \cdot 120 + 140 = 500\).

The number is \(120\).

- Work backward from \(40\). - Undo the addition first, then the multiplication. - Use inverse operations. - Check your number by following the original steps.

1. Undo the addition: \(40 - 4 = 36\). 2. Undo the multiplication: \(36 \div 6 = 6\). 3. Check: \(6 \cdot 6 + 4 = 40\).

The number is \(6\).

- Work backward from \(200\). - Undo the subtraction first. - Undo halving by doubling.

1. Undo the subtraction: \(200 + 50 = 250\). 2. Undo division by \(2\): \(250 \cdot 2 = 500\). 3. Check: \(500 \div 2 - 50 = 200\).

The number is \(500\).

- Use inverse operations to isolate \(x\). - Perform the same operation on both sides of the equation. - Substitute each answer into the original equation to check it.

1. In a), add \(5000\) to both sides: \(x=-3500+5000=1500\). 2. In b), add \(2000\) to both sides: \(x=-10{,}000+2000=-8000\).

a) \(x=1500\) b) \(x=-8000\)

- Reverse the magician's steps. - Undo subtraction before undoing multiplication. - Check by applying the original operations.

1. Undo the subtraction: \(100 + 24 = 124\). 2. Undo the multiplication: \(124 \div 4 = 31\). 3. Check: \(4 \cdot 31 - 24 = 100\).

The starting number was \(31\).

- Work backward from \(300\). - Undo the addition first. - Undo multiplication by \(3\) using division. - Check your answer.

1. Undo the addition: \(300 - 120 = 180\). 2. Divide by \(3\): \(180 \div 3 = 60\). 3. Check: \(3 \cdot 60 + 120 = 300\).

The number is \(60\).

- Work backward from \(40\). - Undo division by \(5\) first. - Then undo subtraction of \(50\).

1. Undo the division: \(40 \cdot 5 = 200\). 2. Undo the subtraction: \(200 + 50 = 250\). 3. Check: \((250 - 50) \div 5 = 40\).

The starting number was \(250\).

- Solve each equation by working backward. - Undo addition or subtraction first. - Then undo multiplication by dividing. - Check each solution by substitution.

1. In a), subtract \(6\): \(4x = 24\). Then divide by \(4\): \(x = 6\). 2. In b), add \(5\): \(9x = 45\). Then divide by \(9\): \(x = 5\). 3. In c), subtract \(3\): \(7x = 49\). Then divide by \(7\): \(x = 7\).

a) \(x = 6\) b) \(x = 5\) c) \(x = 7\)

- First find the number that is \(20\) less than \(200\). - Translate “three times a number” into an equation. - Use division to undo multiplication by \(3\).

1. Find the value that is \(20\) less than \(200\): \(200 - 20 = 180\). 2. Solve \(3x = 180\): \(x = 180 \div 3 = 60\).

The number is \(60\).

- Work backward from \(100\). - Undo addition before undoing multiplication. - Check the solution in the original statement.

1. Subtract \(40\): \(2x = 60\). 2. Divide by \(2\): \(x = 30\).

The number is \(30\).

- Work backward from \(600\). - Undo the last operation first. - Check your number by applying the original operations.

1. Write the equation \(x + 120 - 50 = 600\). 2. Undo the subtraction: \(600 + 50 = 650\). 3. Undo the addition: \(650 - 120 = 530\). 4. Check: \(530 + 120 - 50 = 600\).

The number is \(530\).

- Work backward from \(450\). - Undo subtraction by adding. - Undo doubling by dividing by \(2\). - Check by following the original steps.

1. Undo the subtraction: \(450 + 150 = 600\). 2. Undo the doubling: \(600 \div 2 = 300\). 3. Check: \(2 \cdot 300 - 150 = 450\).

The number is \(300\).

- Work backward from \(500\). - Undo adding \(112\) by subtracting \(112\). - Undo subtracting \(245\) by adding \(245\). - Write the operations as a chain.

1. Undo the addition: \(500 - 112 = 388\). 2. Undo the subtraction: \(388 + 245 = 633\). 3. Check: \(633 - 245 + 112 = 500\).

The starting number is \(633\).

- Evaluate \(900 - 150\) first. - Write an equation with a variable. - Use subtraction to find the missing addend. - Check both sides of the equation.

1. Evaluate the right side: \(900 - 150 = 750\). 2. Solve \(x + 325 = 750\): \(x = 750 - 325 = 425\).

The number is \(425\).

- Simplify the numerical expression first. - Replace that side with its value. - Then use an inverse operation to isolate \(x\). - Check each solution.

1. \(400 + 100 = 500\), so \(x - 150 = 500\) and \(x = 650\). 2. \(250 + 50 = 300\), so \(800 - x = 300\) and \(x = 500\). 3. \(900 - 300 = 600\), so \(x + 220 = 600\) and \(x = 380\). 4. \(120 + 180 = 300\), so \(1000 - x = 300\) and \(x = 700\).

1) \(x = 650\) 2) \(x = 500\) 3) \(x = 380\) 4) \(x = 700\)

- Evaluate \(135 + 165\) first. - Write an equation with a variable. - Use addition to undo subtraction.

1. Evaluate the right side: \(135 + 165 = 300\). 2. Solve \(x - 257 = 300\): \(x = 300 + 257 = 557\).

The number is \(557\).

- Work backward from \(920\), starting with the last operation. - Undo adding \(120\), then undo multiplication by \(4\). - Check your solution by following the original steps.

1. Subtract \(120\): \(4x = 800\). 2. Divide by \(4\): \(x = 200\). 3. Check: \(4 \cdot 200 + 120 = 920\).

The number is \(200\).

- Work backward from \(100\). - Undo division by \(3\) first. - Then undo the increase of \(120\). - Check your result in the original steps.

1. Undo the division by multiplying: \(100 \cdot 3 = 300\). 2. Undo the addition by subtracting: \(300 - 120 = 180\). 3. Check: \((180 + 120) \div 3 = 300 \div 3 = 100\).

The mystery number is \(180\).

- First determine the exact value of three times the number. - Translate “\(40\) less than \(250\)” into a calculation. - Use division as the inverse of multiplication to find the number. - Check the result against both conditions.

1. Find the value of three times the number: \(250 - 40 = 210\). 2. Divide by \(3\): \(210 \div 3 = 70\). 3. Check: \(3 \cdot 70 = 210\), and \(210\) is \(40\) less than \(250\).

The number is \(70\).

- First evaluate \(120 + 130\). - Write the situation step by step as an equation. - Treat the unknown as the missing number being subtracted in \(1000 - x\). - Ask what must be removed from \(1000\) to reach the calculated result.

1. Simplify the known expression: \(120 + 130 = 250\). 2. Write the equation \(1000 - x = 250\). 3. Solve for the subtracted number: \(x = 1000 - 250 = 750\). 4. Check: \(1000 - 750 = 250\).

The number is \(750\).

- Simplify the expression on the right first. - Write the resulting simpler equation. - Isolate \(x\) using subtraction. - Compare your result with \(400\).

1. Simplify the right side: \(900 - 140 = 760\). 2. Solve \(260 + x = 760\) by subtracting \(260\): \(x = 760 - 260 = 500\). 3. Since \(500 > 400\), the missing number is greater than \(400\).

The missing number is \(500\). It is greater than \(400\).

- Evaluate \(150 + 90\) first. - Then use addition to undo subtracting \(360\). - Check your answer in the original statement.

1. Simplify the known expression: \(150 + 90 = 240\). 2. Write the equation \(x - 360 = 240\). 3. Add \(360\) to both sides: \(x = 240 + 360 = 600\). 4. Check: \(600 - 360 = 240\).

The number is \(600\).

- First find twice \(110\). - Then write a subtraction equation with the unknown in the subtracted position. - Check your answer.

1. Find the difference: \(2 \cdot 110 = 220\). 2. Write \(720 - x = 220\). 3. Solve for the number subtracted: \(x = 720 - 220 = 500\). 4. Check: \(720 - 500 = 220\).

The number subtracted is \(500\).

- Evaluate \(200 + 120\) first. - Translate “eight times a number” into an equation. - Use division to undo multiplication by \(8\).

1. Simplify the known expression: \(200 + 120 = 320\). 2. Write \(8x = 320\). 3. Divide by \(8\): \(x = 320 \div 8 = 40\). 4. Check: \(8 \cdot 40 = 320\).

The number is \(40\).

- Work backward from the final result. - Undo the subtraction first. - Then undo division by \(4\). - Check the original sequence.

1. Work backward from \(20\). 2. Undo subtracting \(80\): \(20 + 80 = 100\). 3. Undo dividing by \(4\): \(100 \cdot 4 = 400\). 4. Check: \(400 \div 4 - 80 = 100 - 80 = 20\).

The number is \(400\).

- Isolate the variable term before dividing. - Substitute the proposed value into the original equation. - A value is a solution only when both sides are equal.

1. Add \(3\): \(4y = 20\). 2. Divide by \(4\): \(y = 5\). 3. For \(y = 4\), the left side is \(4 \cdot 4 - 3 = 13\). 4. Since \(13 \ne 17\), \(y = 4\) is not a solution.

a) \(y = 5\) b) \(y = 4\) is not a solution because the left side equals \(13\), not \(17\).

- Move the constant term away from the variable term first. - Divide by the coefficient of the variable. - Follow the sign rules when dividing by a negative number. - Apply each operation to both sides.

1. For a), add \(12\): \(5x = 30\). Divide by \(5\): \(x = 6\). 2. For b), subtract \(7\): \(-2y = 6\). Divide by \(-2\): \(y = -3\). 3. For c), subtract \(4\): \(\frac{1}{2}z = -3\). Multiply by \(2\): \(z = -6\). 4. For d), add \(0.4\): \(1.2w = 2.4\). Divide by \(1.2\): \(w = 2\).

a) \(x = 6\) b) \(y = -3\) c) \(z = -6\) d) \(w = 2\)

- Solve each equation independently. - Isolate the variable term before dividing. - Compare the three final values.

1. For equation 1), add \(2.8\): \(4x = 8\). Divide by \(4\): \(x = 2\). 2. For equation 2), subtract \(15\): \(-3x = -9\). Divide by \(-3\): \(x = 3\). 3. For equation 3), subtract \(1.5\): \(0.5x = 1\). Divide by \(0.5\): \(x = 2\). 4. Equations 1) and 3) both have solution \(x = 2\).

1) \(x = 2\) 2) \(x = 3\) 3) \(x = 2\) Equations 1) and 3) have the same solution.

- Use a variable for the points per correct answer. - Write an expression that combines the question points and the bonus. - Remove the bonus before dividing by the number of correct answers.

1. Let \(x\) be the points earned for each correct answer. 2. Write the equation \(6x + 25 = 145\). 3. Subtract the bonus: \(6x = 120\). 4. Divide by \(6\): \(x = 20\).

Each correct answer is worth \(20\) points.

- Move the constant term away from the variable term first. - Divide by the coefficient of \(x\). - Apply each operation to both sides. - Remember that \(-x\) means \(-1 \cdot x\).

1. Subtract \(9\): \(3x = -9\). Divide by \(3\): \(x = -3\). 2. Add \(4\): \(-2x = 10\). Divide by \(-2\): \(x = -5\). 3. Add \(5\): \(20 = 4x\). Divide by \(4\): \(x = 5\). 4. Subtract \(10\): \(-x = 3\). Multiply by \(-1\): \(x = -3\).

1) \(x = -3\) 2) \(x = -5\) 3) \(x = 5\) 4) \(x = -3\)

- Combine like terms before solving. - Use an inverse operation to remove the constant term. - Divide by the variable coefficient at the end.

1. For a), combine like terms: \(9z - 14 = 4\). 2. Add \(14\): \(9z = 18\). Divide by \(9\): \(z = 2\). 3. For b), combine like terms: \(11w - 5 = 50\). 4. Add \(5\): \(11w = 55\). Divide by \(11\): \(w = 5\).

a) \(z = 2\) b) \(w = 5\)

- Combine terms that contain the same variable. - Combine constant terms separately. - Isolate the variable term before dividing. - Apply each operation to both sides.

1. Combine like terms: \(9x = 81\). Divide by \(9\): \(x = 9\). 2. Combine the constants: \(3y + 9 = 24\). Subtract \(9\): \(3y = 15\). Divide by \(3\): \(y = 5\). 3. Combine like terms: \(7z - 4 = 31\). Add \(4\): \(7z = 35\). Divide by \(7\): \(z = 5\).

1) \(x = 9\) 2) \(y = 5\) 3) \(z = 5\)

- Combine variable terms and constant terms separately. - Isolate the variable term before dividing. - Track negative signs carefully. - Check each solution in the original equation.

1. Combine like terms: \(10x - 5 = 25\). 2. Add \(5\): \(10x = 30\). Divide by \(10\): \(x = 3\). 3. Combine like terms: \(-8y + 12 = 36\). 4. Subtract \(12\): \(-8y = 24\). Divide by \(-8\): \(y = -3\).

1) \(x = 3\) 2) \(y = -3\)

- Combine the terms containing \(x\) first. - Combine the constant terms. - Isolate the variable term before dividing. - Use division to undo multiplication by \(11\).

1. Combine like terms: \(11x - 6 = 27\). 2. Add \(6\) to both sides: \(11x = 33\). 3. Divide both sides by \(11\): \(x = 3\).

\(x = 3\)

- Divide by the factor outside the parentheses first. - Use the inverse operation to remove the constant inside the parentheses. - Keep the equation balanced at every step. - Check each answer by substitution.

1. Divide by \(6\): \(x + 4 = 8.5\). Subtract \(4\): \(x = 4.5\). 2. Divide by \(9\): \(x - 12 = 2\). Add \(12\): \(x = 14\). 3. Divide by \(4\): \(3x + 2 = 14\). Subtract \(2\): \(3x = 12\). Divide by \(3\): \(x = 4\). 4. Divide by \(0.4\): \(x - 5 = 5\). Add \(5\): \(x = 10\).

1) \(x = 4.5\) 2) \(x = 14\) 3) \(x = 4\) 4) \(x = 10\)

- Use Anna’s amount as the variable so the other two amounts are easy to express. - Add the three expressions and set the sum equal to the total. - Check the result by adding all three amounts.

1. Let \(x\) be the number of pounds Anna collects. 2. Beth collects \(x - 10\), and Clara collects \(x + 10\). 3. Write the total equation \(x + (x - 10) + (x + 10) = 120\). 4. Combine like terms: \(3x = 120\). 5. Divide by \(3\): \(x = 40\). 6. Beth collects \(40 - 10 = 30\,\text{lb}\), and Clara collects \(40 + 10 = 50\,\text{lb}\). 7. Check: \(40 + 30 + 50 = 120\).

Anna collected \(40\,\text{lb}\), Beth collected \(30\,\text{lb}\), and Clara collected \(50\,\text{lb}\).

- Use Mia’s card count as the base value. - Express the other two card counts in terms of Mia’s count. - Add the three expressions and set the sum equal to \(200\). - Check that the three answers total \(200\).

1. Let \(x\) be the number of cards Mia has. 2. Lucas has \(x + 20\), and Noah has \(2x\). 3. Write the total equation \(x + (x + 20) + 2x = 200\). 4. Combine like terms: \(4x + 20 = 200\). 5. Subtract \(20\): \(4x = 180\). Divide by \(4\): \(x = 45\). 6. Lucas has \(45 + 20 = 65\) cards, and Noah has \(2 \cdot 45 = 90\) cards.

Mia has \(45\) cards, Lucas has \(65\) cards, and Noah has \(90\) cards.

- Find a common denominator for each equation. - Multiply every term on both sides by that denominator. - Combine the terms containing \(x\) after clearing the fractions.

1. For a), multiply every term by \(20\): \(8x + 5x = 260\). 2. Combine like terms: \(13x = 260\). Divide by \(13\): \(x = 20\). 3. For b), multiply every term by \(6\): \(7x - 3x = 48\). 4. Combine like terms: \(4x = 48\). Divide by \(4\): \(x = 12\).

a) \(x = 20\) b) \(x = 12\)

- Use the lemon-juice amount as the base value. - Translate “three times as much” and “\(4\,\text{cups}\) more” into expressions. - Add the three amounts to equal \(24\,\text{cups}\).

1. Let \(x\) be the number of cups of lemon juice. 2. The amount of orange juice is \(3x\), and the amount of apple juice is \(x + 4\). 3. Write the total-volume equation \(x + 3x + (x + 4) = 24\). 4. Combine like terms: \(5x + 4 = 24\). 5. Subtract \(4\): \(5x = 20\). Divide by \(5\): \(x = 4\). 6. The punch contains \(3 \cdot 4 = 12\,\text{cups}\) of orange juice and \(4 + 4 = 8\,\text{cups}\) of apple juice.

The punch contains \(4\,\text{cups}\) of lemon juice, \(12\,\text{cups}\) of orange juice, and \(8\,\text{cups}\) of apple juice.

- Identify which symbol has no assigned value. - Substitute the known values into the equation. - Undo the operations in reverse order. - Dividing by \(0.25\) is equivalent to multiplying by \(4\).

1. The remaining unknown variable is \(x\). 2. Substitute the known values: \(3 = 0.25(x - 5) + 8\). 3. Subtract \(8\): \(-5 = 0.25(x - 5)\). 4. Divide by \(0.25\): \(-20 = x - 5\). 5. Add \(5\): \(x = -15\).

The unknown variable is \(x\), and \(x = -15\).

- Undo the subtraction before undoing the multiplication. - Use the array dimensions to determine the value of \(x\). - Explain which dots are removed and which remain.

1. The full array has \(49\) dots arranged in \(7\) rows of \(7\). 2. The gray row contains the \(7\) dots that are subtracted, leaving \(42\) blue dots. 3. Thus \(7 \cdot 7 - 7 = 42\), so \(x = 7\).

\(x = 7\). The gray row represents the \(7\) dots being subtracted.

- Treat the unknown as a variable and use inverse operations. - In part a), how can you isolate the number being subtracted? - In part b), simplify the constant terms before isolating \(x\). - Use a common denominator when an equation contains unlike fractions.

1. For a), add \(x\) to both sides and subtract \(2 \frac{3}{4}\) from both sides: \(x=5 \frac{1}{2}-2 \frac{3}{4}=2 \frac{3}{4}\). 2. For b), combine the constant terms: \(1 \frac{1}{3}-2 \frac{1}{6}=\frac{8}{6}-\frac{13}{6}=-\frac{5}{6}\). The equation becomes \(x-\frac{5}{6}=3\). 3. Add \(\frac{5}{6}\) to both sides: \(x=3 \frac{5}{6}\).

a) \(x=2 \frac{3}{4}\) b) \(x=3 \frac{5}{6}\)

- Use inverse operations to isolate the unknown. - For part b), think about how to find a missing subtrahend. - In part c), undo the outside operation first.

1. For a), \(\square=30.10-12.45=17.65\). 2. For b), \(\square=45.60-12.78=32.82\). 3. For c), first undo the subtraction: \(\square+2.5=5.8+1.2=7.0\). Then \(\square=7.0-2.5=4.5\).

a) \(17.65\) b) \(32.82\) c) \(4.5\)

- Rewrite all fractions with denominator \(12\). - What expression appears in the numerator of the second equivalent fraction? - Solve the resulting equation for \(x\).

1. Rewrite the fractions with denominator \(12\): \(\frac{1}{6}=\frac{2}{12}\) and \(\frac{x}{4}=\frac{3x}{12}\). 2. The numerator equation is \(2+3x=11\). 3. Subtract \(2\): \(3x=9\). Divide by \(3\): \(x=3\). 4. Check: \(\frac{1}{6}+\frac{3}{4}=\frac{2}{12}+\frac{9}{12}=\frac{11}{12}\).

\(x=3\)

- Undo addition or subtraction before undoing multiplication or division. - Apply the same operation to both sides. - Check each result by substitution.

1. For a), add \(1.5\): \(2\Box=-5\). Divide by \(2\): \(\Box=-2.5\). 2. For b), subtract \(0.4\): \((-0.2)\cdot\Box=1\). Divide by \(-0.2\): \(\Box=-5\). 3. For c), subtract \(0.5\): \(\Box\div(-10)=0.25\). Multiply by \(-10\): \(\Box=-2.5\). 4. For d), subtract \(0.8\): \(-2\Box=0.8\). Divide by \(-2\): \(\Box=-0.4\).

a) \(-2.5\) b) \(-5\) c) \(-2.5\) d) \(-0.4\)

- Undo addition or subtraction before undoing multiplication. - Apply the same operation to both sides. - Substitute each result to check it.

1. For a), subtract \(1.5\): \(2\Box=-4\). Divide by \(2\): \(\Box=-2\). 2. For b), add \(2.4\): \(6\Box=-2.4\). Divide by \(6\): \(\Box=-0.4\). 3. For c), subtract \(1\): \(\Box\cdot\left(-\frac{2}{3}\right)=4\). Divide by \(-\frac{2}{3}\): \(\Box=4\cdot\left(-\frac{3}{2}\right)=-6\).

a) \(-2\) b) \(-0.4\) c) \(-6\)

- Isolate the variable term before dividing. - Use the sign rules for multiplication and division. - A terminating decimal is a rational number.

1. For a), subtract \(2\): \(-1.5x = 0.75\). Divide by \(-1.5\): \(x = -0.5 = -\frac{1}{2}\). This value is in \(\mathbb{Q}\) only. 2. For b), add \(1\): \(0.5x = 3\). Divide by \(0.5\): \(x = 6\). This value is in \(\mathbb{N}\), \(\mathbb{Z}\), and \(\mathbb{Q}\). 3. For c), add \(1.2\): \(-x = 3.6\). Multiply by \(-1\): \(x = -3.6 = -\frac{18}{5}\). This value is in \(\mathbb{Q}\) only.

a) \(x = -0.5 = -\frac{1}{2}\); \(\mathbb{Q}\) b) \(x = 6\); \(\mathbb{N}\), \(\mathbb{Z}\), and \(\mathbb{Q}\) c) \(x = -3.6 = -\frac{18}{5}\); \(\mathbb{Q}\)

- Simplify each side before isolating the variable. - A subtraction sign before parentheses changes the signs inside. - Check each solution by substitution.

1. For a), simplify: \(-15+z-10=5\), so \(z-25=5\) and \(z=30\). 2. For b), distribute the subtraction: \(30-z-5=15\), so \(25-z=15\) and \(z=10\). 3. For c), simplify: \(z+4-12=-20\), so \(z-8=-20\) and \(z=-12\).

a) \(z=30\) b) \(z=10\) c) \(z=-12\)

- Simplify double negative signs. - Combine fractions using a common denominator. - If the coefficient of \(x\) is \(-1\), multiply both sides by \(-1\).

1. For a), the left side is \(x+\frac{1}{6}\), and the right side is \(-\frac{7}{6}\). Thus \(x=-\frac{7}{6}-\frac{1}{6}=-\frac{4}{3}\). 2. For b), the right side is \(-\frac{4}{8}=-0.5\). Then \(-0.75-x=-0.5\), so \(-x=0.25\) and \(x=-0.25\).

a) \(x=-\frac{4}{3}\), or \(x=-1\frac{1}{3}\) b) \(x=-0.25\), or \(x=-\frac{1}{4}\)

- Simplify expressions inside parentheses first. - Pay attention to a negative sign before parentheses. - Convert between fractions and decimals when useful.

1. For a), \(\frac{1}{5}-0.4=0.2-0.4=-0.2\). The right side is \(-0.5-(-0.2)=-0.3\). Thus \(x+0.3=-0.3\), so \(x=-0.6\). 2. For b), the left side is \(\frac{2}{3}\). On the right, \(-\frac{1}{4}-\frac{5}{12}=-\frac{3}{12}-\frac{5}{12}=-\frac{2}{3}\). Thus \(\frac{2}{3}=x-\frac{2}{3}\), so \(x=\frac{4}{3}\).

a) \(x=-0.6\), or \(x=-\frac{3}{5}\) b) \(x=\frac{4}{3}\), or \(x=1\frac{1}{3}\)

- Undo the division first. - Determine the required value of the expression in parentheses. - Then use addition to isolate the missing number.

1. Multiply both sides by \(0.5\): \(-12.4+\Box=-5\). 2. Add \(12.4\): \(\Box=7.4\).

\(7.4\)

- Work backward from the final result. - Undo each operation in reverse order. - Subtracting \(-8\) is the same as adding \(8\).

1. Start at \(-10\) and undo subtraction of \(14\): \(-10+14=4\). 2. Undo addition of \(-8\): \(4-(-8)=12\). 3. Therefore, \(x=12\).

\(x=12\)

- Write a separate equation for each person. - Work backward from the same ending balance. - Keep careful track of deposits and payments. - Interpret a negative starting value in context.

1. Let \(x\) be Luke's starting balance. Write \(x + 15.50 - 40.00 = 5.00\). 2. Simplify: \(x - 24.50 = 5.00\), so \(x = 29.50\). 3. Let \(y\) be Sarah's starting balance. Write \(y + 25.00 - 10.50 = 5.00\). 4. Simplify: \(y + 14.50 = 5.00\), so \(y = -9.50\). 5. A negative account balance represents money owed, so Sarah had the negative starting balance.

a) Luke's starting balance was \(\$29.50\), and Sarah's starting balance was \(-\$9.50\). b) Sarah had a negative balance because her starting balance was less than \(\$0\).

- Solve each equation without considering the domain first. - Then check whether the result is in \(\{1, 2, 3, \ldots\}\). - If the result is outside the stated domain, the solution set is empty. - Substitute any accepted solution to verify it.

1. For a), \(4x - 7 = 13\), so \(4x = 20\) and \(x = 5\). Since \(5\) is a positive integer, \(S = \{5\}\). 2. For b), \(3x + 12 = 5\), so \(3x = -7\) and \(x = -\frac{7}{3}\). This is not a positive integer, so \(S = \varnothing\). 3. For c), \(2x + 1 = 8\), so \(2x = 7\) and \(x = 3.5\). This is not a positive integer, so \(S = \varnothing\). 4. For d), \(15 - 5x = 10\), so \(-5x = -5\) and \(x = 1\). Since \(1\) is a positive integer, \(S = \{1\}\).

a) \(S = \{5\}\) b) \(S = \varnothing\) c) \(S = \varnothing\) d) \(S = \{1\}\)

- Solve each two-step equation first. - Integers include negative whole numbers, zero, and positive whole numbers. - A noninteger result is not in the stated domain. - Use the empty-set symbol when no value in the domain solves the equation.

1. For a), \(3x + 12 = 15\), so \(3x = 3\) and \(x = 1\). Since \(1 \in \mathbb{Z}\), \(S = \{1\}\). 2. For b), \(5x - 8 = 7\), so \(5x = 15\) and \(x = 3\). Since \(3 \in \mathbb{Z}\), \(S = \{3\}\). 3. For c), \(2x + 9 = 2\), so \(2x = -7\) and \(x = -3.5\). Since \(-3.5 \notin \mathbb{Z}\), \(S = \varnothing\). 4. For d), \(4x - 2 = 8\), so \(4x = 10\) and \(x = 2.5\). Since \(2.5 \notin \mathbb{Z}\), \(S = \varnothing\).

a) \(S = \{1\}\) b) \(S = \{3\}\) c) \(S = \varnothing\) d) \(S = \varnothing\)

- Solve the equation before considering either domain. - Decide whether the result is an integer or a rational number. - An equation may have a numerical solution that is excluded by a stated domain. - Use the empty set when no value in the domain works.

1. Solve the equation: \(2x + 6 = 11\), so \(2x = 5\) and \(x = 2.5\). 2. For the integer domain, \(2.5 \notin \mathbb{Z}\). Therefore, \(S = \varnothing\), and the student’s claim is correct. 3. For the rational-number domain, \(2.5 \in \mathbb{Q}\). Therefore, \(S = \{2.5\}\), and the student’s claim is not correct.

1. Over \(\mathbb{Z}\), \(S = \varnothing\), so the claim is correct. 2. Over \(\mathbb{Q}\), \(S = \{2.5\}\), so the claim is not correct.

- Isolate the variable term first. - Use decimal points and track negative signs carefully. - For a grouped expression, undo the outer operation first. - Check each solution in the original equation.

1. For a), add \(4\): \(0.5x = 5\). Divide by \(0.5\): \(x = 10\). 2. For b), subtract \(7\): \(-1.5x = 6\). Divide by \(-1.5\): \(x = -4\). 3. For c), divide by \(2\): \(3x + 4 = -2\). Subtract \(4\): \(3x = -6\). Divide by \(3\): \(x = -2\).

a) \(x = 10\) b) \(x = -4\) c) \(x = -2\)

- Simplify each side before isolating the variable. - A minus sign before parentheses changes the signs inside. - In some equations, dividing first removes the grouping symbols efficiently.

1. For a), divide by \(2\): \(x - 5 = 7\). Add \(5\): \(x = 12\). 2. For b), combine like terms: \(4x + 8 = 20\). Subtract \(8\), then divide by \(4\): \(x = 3\). 3. For c), divide by \(3\): \(5 = y + 2\). Subtract \(2\): \(y = 3\). 4. For d), remove the parentheses: \(5 - x + 2 = 10\). Then \(7 - x = 10\), so \(-x = 3\) and \(x = -3\).

a) \(x = 12\) b) \(x = 3\) c) \(y = 3\) d) \(x = -3\)

- Try transforming the first equation into the second using the same operation on both sides. - When distributing, multiply every term inside the parentheses. - You may solve both equations and compare their solution sets.

1. For a), divide every term in \(7x - 14 = 21\) by \(7\). This gives \(x - 2 = 3\), exactly the second equation. Both equations have solution \(x = 5\), so they are equivalent. 2. For b), distributing in the first equation gives \(3a + 12 = 15\), not \(3a + 4 = 15\). The first equation has solution \(a = 1\), while the second has solution \(a = \frac{11}{3}\). Therefore, they are not equivalent.

a) Yes. The equations are equivalent. b) No. The equations are not equivalent.

- Apply an operation to every term on both sides of an equation. - Multiplying by \(2\) must affect both terms on the left in part b. - Solving both equations can confirm whether their solution sets match.

1. For a), subtract \(12\) from both sides of the first equation to get \(-4y = -4\). Multiplying both sides by \(-1\) gives \(4y = 4\). The transformation is valid, and both equations have solution \(y = 1\). 2. For b), multiplying the first equation by \(2\) must multiply every term on both sides. The correct result is \(z + 10 = 22\), not \(z + 5 = 22\). The transformation is not valid.

a) Yes, the equations are equivalent. b) No, the equations are not equivalent.

- An equivalent transformation must be applied to both sides and to every affected term. - Solve the original equation first. - You can test whether each new equation has the same solution.

1. The original equation gives \(5x = 25\), so \(x = 5\). 2. For a), adding \(10\) to both sides gives \(5x = 25\). This is equivalent. 3. For b), only the term \(5x\) was divided by \(5\). Dividing the entire left side would give \(x - 2\), not \(x - 10\). This equation has solution \(x = 13\), so it is not equivalent. 4. For c), dividing every term of the original equation by \(5\) gives \(x - 2 = 3\). This is equivalent. 5. For d), subtracting \(15\) from both sides gives \(5x - 25 = 0\). This is equivalent.

a) Yes; add \(10\) to both sides. b) No; the entire left side was not divided by \(5\). c) Yes; divide every term by \(5\). d) Yes; subtract \(15\) from both sides.

- How can you represent twice a number? - Which operation is indicated by “increased by”? - Evaluate the fixed difference before solving the equation.

1. Translate the statement into an equation: \(2x + (15.4 - 8.4) = 25\). 2. Evaluate the difference: \(15.4 - 8.4 = 7\). 3. Simplify: \(2x + 7 = 25\). 4. Subtract \(7\): \(2x = 18\). 5. Divide by \(2\): \(x = 9\).

The equation is \(2x + (15.4 - 8.4) = 25\), and \(x = 9\).

- Solve Equations A and B separately. - A known solution makes an equation true when substituted. - Substitute \(x = 4\) into the equation with the missing number.

1. For Equation A, add \(8\): \(5x = 20\). Divide by \(5\): \(x = 4\). 2. For Equation B, divide by \(2\): \(x + 1 = 5\). Subtract \(1\): \(x = 4\). 3. Substitute \(x = 4\) into \(4x + \square = 20\): \(16 + \square = 20\). Subtract \(16\): \(\square = 4\).

a) Both equations have solution \(x = 4\). b) The missing number is \(4\).

- Combine known terms before isolating the box. - Undo addition or subtraction first, then undo multiplication or division. - Write fractions with a common denominator when needed.

1. For part a), combine the known terms: \(12.5+2.5=15\). Then \(15-\Box=10\), so \(\Box=5\). 2. For part b), combine the known fractions: \(\frac{3}{8}-\frac{1}{8}=\frac{2}{8}\). Then \(\Box=\frac{7}{8}-\frac{2}{8}=\frac{5}{8}\). 3. For part c), subtract \(0.8\): \(0.4\cdot\Box=2\). Divide by \(0.4\): \(\Box=5\). 4. For part d), add \(0.5\): \(\frac{\Box}{4}=2\). Multiply by \(4\): \(\Box=8\).

a) \(5\) b) \(\frac{5}{8}\) c) \(5\) d) \(8\)

- Rewrite the equation so that the box is isolated on one side. - Follow the order of operations when working with parentheses or a fraction bar. - Converting fractions to decimals may make the calculations easier.

1. For part a), \(\frac{3}{4}=0.75\), and \(0.75-0.25=0.5\). The equation becomes \(0.5+\Box=1\), so \(\Box=0.5\). 2. For part b), divide both sides by \(2\): \(\Box+1.5=3.5\). Subtract \(1.5\): \(\Box=2\). 3. For part c), isolate the fraction: \(\frac{\Box}{2}=10-8.5=1.5\). Multiply by \(2\): \(\Box=3\).

a) \(0.5\), or \(\frac{1}{2}\) b) \(2\) c) \(3\)

- Substitute the stated solution for \(x\). - After substitution, solve the resulting equation for \(k\). - Check that the original equation is true with both values inserted.

1. For a), substitute \(x = 4\): \(7 \cdot 4 - k = 15\). Then \(28 - k = 15\), so \(k = 13\). 2. For b), substitute \(x = 6\): \(k(6 + 2) = 24\). Then \(8k = 24\), so \(k = 3\). 3. For c), substitute \(x = 10\): \(10 + k = 20 - 5\). Then \(10 + k = 15\), so \(k = 5\).

a) \(k = 13\) b) \(k = 3\) c) \(k = 5\)

- Use inverse operations to isolate the missing value. - Rewrite subtracting a negative as addition. - Check each answer by substitution.

1. In a), \(x-10=-3\), so \(x=7\). 2. In b), \(-5-x=2\), so \(-x=7\) and \(x=-7\). 3. In c), \(x+8=5\), so \(x=-3\). 4. In d), \(4-x=-6\), so \(-x=-10\) and \(x=10\).

a) \(7\) b) \(-7\) c) \(-3\) d) \(10\)

- Translate each sentence into an equation. - Use inverse operations to isolate the unknown. - Substitute your answer back into the equation to check it.

1. In a), solve \(x+(-30)=-75\). Adding \(30\) to both sides gives \(x=-45\). 2. In b), solve \(-20-y=10\). Adding \(20\) gives \(-y=30\), so \(y=-30\).

a) \(-45\) b) \(-30\)

- Identify the starting number and the number being subtracted. - Represent each missing number with a variable. - Use inverse operations to solve the equations. - Check each result in the original statement.

1. In a), \(-250-750=-1000\). 2. In b), solve \(-120-x=-50\). Then \(-x=70\), so \(x=-70\). 3. In c), solve \(x-45=-100\). Add \(45\) to both sides to get \(x=-55\).

a) \(-1000\) b) \(-70\) c) \(-55\)

- Work backward from \(500\) one operation at a time. - Use the inverse operation at each step. - You may also combine \(-45 + 120\) before solving.

1. Undo adding \(120\): \(500 - 120 = 380\). 2. Undo subtracting \(45\): \(380 + 45 = 425\). 3. Check: \(425 - 45 + 120 = 500\).

\(\square = 425\)

- Treat the entire expression in parentheses as one unknown quantity first. - What must be subtracted from \(1200\) to get \(700\)? - Once you know the value of the parentheses, solve for \(x\).

1. Treat the expression in parentheses as one quantity. Since \(1200 - 700 = 500\), the equation becomes \(x + 150 = 500\). 2. Subtract \(150\): \(x = 500 - 150 = 350\). 3. Check: \(1200 - (350 + 150) = 1200 - 500 = 700\).

\(x = 350\)

- Evaluate any numerical product first. - Use inverse operations to isolate \(x\). - Check each solution in the original equation.

1. a) Evaluate \(7 \cdot 6 = 42\). Then \(42 - x = 30\), so \(x = 12\). 2. b) Evaluate \(9 \cdot 4 = 36\). Then \(36 + x = 50\), so \(x = 14\). 3. c) Add \(14\) to both sides: \(8x = 64\). Divide by \(8\): \(x = 8\). 4. d) Subtract \(27\) from both sides: \(5x = 25\). Divide by \(5\): \(x = 5\).

a) \(x = 12\) b) \(x = 14\) c) \(x = 8\) d) \(x = 5\)

- Write one equation for each person. - Remember that “five times a number” means multiply the number by \(5\). - Undo addition or subtraction before undoing multiplication. - Compare the two solutions.

1. For Luke, solve \(6x + 4 = 40\): \(6x = 36\), so \(x = 6\). 2. For Sara, solve \(5y - 3 = 32\): \(5y = 35\), so \(y = 7\). 3. Since \(7 > 6\), Sara chose the greater number.

Luke chose \(6\). Sara chose \(7\). Sara chose the greater number.

- Represent each blank with a variable. - Use inverse operations to isolate the variable. - Substitute each answer into the original equation to verify it.

1. In a), \(x=-4000-(-1500)=-2500\). 2. In b), solve \(-2800-x=-1200\), which gives \(x=-1600\). 3. In c), \(x=2100-5400=-3300\). 4. In d), \(x+750=-250\), so \(x=-1000\).

a) \(-2500\) b) \(-1600\) c) \(-3300\) d) \(-1000\)

- Isolate the missing number using inverse operations. - Pay attention to subtracting negative values. - Substitute each answer to verify it.

1. In part a), \(20-64=-44\). 2. In part b), \(-40-(-15)=-25\). 3. In part c), if \(50-x=85\), then \(-x=35\), so \(x=-35\).

a) \(-44\) b) \(-25\) c) \(-35\)

- Use inverse operations to isolate the variable. - In part a), account for the negative coefficient on \(x\). - Check each solution in the original equation.

1. In part a), add \(30\) to both sides to get \(-x=45\), so \(x=-45\). 2. In part b), subtract \(42\): \(x=-18-42=-60\). 3. In part c), add \(12\): \(x=-50+12=-38\).

a) \(x=-45\) b) \(x=-60\) c) \(x=-38\)

- Find Luke’s starting number first. - Then find Sarah’s starting number. - Work backward from \(80\) for both equations. - Compare the two solutions.

1. For Luke, solve \(2x - 20 = 80\): \(2x = 100\), so \(x = 50\). 2. For Sarah, solve \(y \div 2 + 20 = 80\): \(y \div 2 = 60\), so \(y = 120\). 3. Since \(120 > 50\), Sarah chose the greater number.

Luke chose \(50\), and Sarah chose \(120\). Sarah chose the greater number.

- Undo the operation outside the parentheses first. - Use inverse operations to isolate \(x\). - Check each solution in the original equation.

1. a) Divide by \(4\): \(x + 8 = 10\). Then subtract \(8\): \(x = 2\). 2. b) Divide by \(6\): \(x - 2 = 6\). Then add \(2\): \(x = 8\). 3. c) Subtract \(15\): \(3x = 30\). Then divide by \(3\): \(x = 10\). 4. d) Subtract \(7\): \(4x = 24\). Then divide by \(4\): \(x = 6\).

a) \(x = 2\) b) \(x = 8\) c) \(x = 10\) d) \(x = 6\)

- Undo the outside operation first. - Continue using inverse operations until \(x\) is isolated. - Check each solution in the original equation.

1. a) Divide both sides by \(6\): \(x + 5 = 12\). Subtract \(5\): \(x = 7\). 2. b) Divide both sides by \(4\): \(x - 12 = 12\). Add \(12\): \(x = 24\). 3. c) Subtract \(15\): \(60 \div x = 6\). Therefore, \(x = 10\), because \(60 \div 10 = 6\).

a) \(x = 7\) b) \(x = 24\) c) \(x = 10\)

- Undo the multiplication outside the parentheses first. - Then use addition or subtraction to isolate \(x\). - Check each solution by substitution.

1. a) Divide by \(2\): \(x + 18 = 25\). Subtract \(18\): \(x = 7\). 2. b) Divide by \(2\): \(x - 3 = 17\). Add \(3\): \(x = 20\). 3. c) Divide by \(5\): \(x - 4 = 5\). Add \(4\): \(x = 9\).

a) \(x = 7\) b) \(x = 20\) c) \(x = 9\)

- When the difference is missing, subtract directly. - When one of the numbers is missing, write an equation. - Use inverse operations and check each answer in the original equation.

1. For a), \(-40-25=-65\). 2. For b), solve \(15-b=-10\), which gives \(b=25\). 3. For c), solve \(c-60=-100\), which gives \(c=-40\).

a) \(-65\) b) \(25\) c) \(-40\)

- Simplify the known terms first. - Use inverse operations to isolate the unknown. - Work one operation at a time. - Substitute each value to check it.

1. a) Combine the known terms: \(450 + 280 = 730\). Then \(730 - x = 600\), so \(x = 130\). 2. b) Undo adding \(320\): \(550 - 320 = 230\). Then undo subtracting \(140\): \(230 + 140 = 370\). 3. c) Compute in order: \(800 - 330 = 470\), and \(470 - 170 = 300\). 4. d) Combine the known addends: \(260 + 390 = 650\). Then \(x = 1000 - 650 = 350\).

a) \(x = 130\) b) \(x = 370\) c) \(x = 300\) d) \(x = 350\)

- Represent each blank with a variable. - Use inverse operations to isolate the variable. - Rewrite subtraction of a negative number as addition before solving.

1. In a), \(x=-60-(-35)=-25\). 2. In b), solve \(-45-y=-15\), which gives \(y=-30\). 3. In c), \(z+20=5\), so \(z=-15\).

a) \(-25\) b) \(-30\) c) \(-15\)

- Write the steps as an equation. - Combine the known integer changes first. - Undo the remaining addition.

1. Write \(x+(-15)+22=4\). 2. Combine the known terms: \(-15+22=7\), so \(x+7=4\). 3. Subtract \(7\): \(x=4-7=-3\).

\(x=-3\)

- Follow the order of operations described in each statement. - A sum that is multiplied needs parentheses. - Simplify fixed numerical expressions before isolating the variable.

1. Write \(2n + 17 = 45\). Subtract \(17\): \(2n = 28\). Divide by \(2\): \(n = 14\). 2. Write \(100 - z = 4 \cdot 18\). Simplify to \(100 - z = 72\), so \(z = 100 - 72 = 28\). 3. Write \(5\left(k + 2.2\right) = 25\). Divide by \(5\): \(k + 2.2 = 5\). Subtract \(2.2\): \(k = 2.8\).

1) \(n = 14\) 2) \(z = 28\) 3) \(k = 2.8\), or \(2\frac{4}{5}\)

- How can you rewrite the mixed number as an improper fraction? - Which term should you remove first to isolate the variable term? - How do you divide by a fraction? - How can substitution verify your solution?

1. Rewrite the mixed number: \(2\frac{1}{6} = \frac{13}{6}\). 2. Subtract \(\frac{1}{2}\) from both sides: \(\frac{2}{3}x = \frac{13}{6} - \frac{3}{6} = \frac{10}{6} = \frac{5}{3}\). 3. Divide by \(\frac{2}{3}\): \(x = \frac{5}{3} \div \frac{2}{3} = \frac{5}{3} \cdot \frac{3}{2} = \frac{5}{2} = 2.5\). 4. Check by substitution: \(\frac{2}{3} \cdot \frac{5}{2} + \frac{1}{2} = \frac{5}{3} + \frac{1}{2} = \frac{10}{6} + \frac{3}{6} = \frac{13}{6} = 2\frac{1}{6}\). The equation is true.

\(x = \frac{5}{2} = 2.5\)

- Use a variable for the original amount. - Express double the amount and half the amount algebraically. - Combine the variable terms before solving.

1. Let \(x\) be the original amount of money. 2. Write the equation \(2x + 0.5x + 12 = 107\). 3. Combine like terms: \(2.5x + 12 = 107\). 4. Subtract \(12\): \(2.5x = 95\). 5. Divide by \(2.5\): \(x = 38\).

The piggy bank originally contained \(\$38\).

- Convert both sides of the scale to the same unit. - Represent the four identical boxes with a variable term. - A level scale means the expressions on both sides are equal. - Remove the known weight before dividing among the four boxes.

1. Convert the right-side weight to ounces: \(3\,\text{lb} = 48\,\text{oz}\). 2. Let \(x\) be the weight of one candy box in ounces. Write \(4x + 8 = 48\). 3. Subtract \(8\): \(4x = 40\). 4. Divide by \(4\): \(x = 10\).

Each box of candy weighs \(10\,\text{oz}\).

- First isolate the term containing the variable. - Track the sign when dividing by a negative number. - You may rewrite a mixed number as a fraction or decimal. - Use the inverse of the operation applied to the variable.

1. Subtract \(12\): \(-5x = -30\). Divide by \(-5\): \(x = 6\). 2. Add \(3.2\): \(\frac{a}{8} = -1.5\). Multiply by \(8\): \(a = -12\). 3. Rewrite \(2\frac{1}{4}\) as \(2.25\). Subtract \(7\): \(2.25y = -4.5\). Divide by \(2.25\): \(y = -2\).

1) \(x = 6\) 2) \(a = -12\) 3) \(y = -2\)

- Combine the variable terms first. - Track the signs carefully in part b). - Isolate the variable term before dividing.

1. For a), combine like terms: \(5k - 4.2 = 15.8\). 2. Add \(4.2\): \(5k = 20\). Divide by \(5\): \(k = 4\). 3. For b), combine like terms: \(20 - 4m = 8\). 4. Subtract \(20\): \(-4m = -12\). Divide by \(-4\): \(m = 3\).

a) \(k = 4\) b) \(m = 3\)

- Group terms containing the same variable. - Combine constant terms separately. - Treat decimal coefficients the same way as whole-number coefficients. - Substitute your answers to check them.

1. Combine like terms: \(2z - 8 = -12\). 2. Add \(8\): \(2z = -4\). Divide by \(2\): \(z = -2\). 3. Combine like terms: \(2k - 8 = 14\). 4. Add \(8\): \(2k = 22\). Divide by \(2\): \(k = 11\).

1) \(z = -2\) 2) \(k = 11\)

- Recall the sum of two angles that form a straight angle. - Represent the smaller angle with a variable. - Express the larger angle using the given difference. - Add the two expressions and solve.

1. Let \(x\) degrees be the smaller angle. The larger angle is \(x + 44\) degrees. 2. Their sum is \(180^\circ\), so \(x + (x + 44) = 180\). 3. Combine like terms: \(2x + 44 = 180\). 4. Subtract \(44\): \(2x = 136\). 5. Divide by \(2\): \(x = 68\). 6. The larger angle is \(68^\circ + 44^\circ = 112^\circ\).

The angle measures are \(68^\circ\) and \(112^\circ\).

- Express each integer after the first in terms of \(x\). - Add the four expressions and set their sum equal to \(66\). - Combine like terms before solving.

1. Let \(x\) be the first integer. 2. The next three integers are \(x + 1\), \(x + 2\), and \(x + 3\). 3. Write the equation \(x + (x + 1) + (x + 2) + (x + 3) = 66\). 4. Combine like terms: \(4x + 6 = 66\). 5. Subtract \(6\): \(4x = 60\). 6. Divide by \(4\): \(x = 15\). 7. The integers are \(15\), \(16\), \(17\), and \(18\).

The four integers are \(15\), \(16\), \(17\), and \(18\).

- Consecutive odd integers differ by \(2\). - Let the middle integer be \(m\), then express the integers before and after it. - Add the three expressions and compare the middle value with \(105 \div 3\).

1. Let \(m\) be the middle odd integer. 2. The previous and next odd integers are \(m - 2\) and \(m + 2\). 3. Write the equation \((m - 2) + m + (m + 2) = 105\). 4. The constants cancel, giving \(3m = 105\). 5. Divide by \(3\): \(m = 35\). 6. The integers are \(33\), \(35\), and \(37\). 7. Since \(105 \div 3 = 35\), the claim is correct for this case.

The integers are \(33\), \(35\), and \(37\). The claim is correct in this case because \(105 \div 3 = 35\), the middle integer.

- Consecutive odd integers differ by \(2\). - For part b, determine whether a sum of even numbers is even or odd. - Compare that result with the parity of \(153\).

1. Let \(x\) be the smallest odd integer. The next two are \(x + 2\) and \(x + 4\). 2. Write \(x + (x + 2) + (x + 4) = 153\). 3. Combine like terms: \(3x + 6 = 153\). 4. Subtract \(6\): \(3x = 147\). Divide by \(3\): \(x = 49\). 5. The three integers are \(49\), \(51\), and \(53\). 6. Three even integers always have an even sum. Since \(153\) is odd, it cannot be the sum of three even integers.

a) The integers are \(49\), \(51\), and \(53\). b) The sum of three even integers is even, but \(153\) is odd, so this is impossible.

- Let the second piece be \(x\) so the other two pieces are easy to express. - The three piece lengths must add to the original rope length. - Combine the three expressions into one equation.

1. Let \(x\) be the length of the second piece in inches. 2. The first piece is \(2x\), and the third piece is \(2x - 5\). 3. Write the total-length equation \(2x + x + (2x - 5) = 100\). 4. Combine like terms: \(5x - 5 = 100\). 5. Add \(5\): \(5x = 105\). Divide by \(5\): \(x = 21\). 6. The first piece is \(2 \cdot 21 = 42\,\text{in.}\), the second is \(21\,\text{in.}\), and the third is \(42 - 5 = 37\,\text{in.}\).

The three pieces are \(42\,\text{in.}\), \(21\,\text{in.}\), and \(37\,\text{in.}\) long.

- Use the nonfiction count as a convenient base value \(x\). - Translate “twice as many” and “\(40\) more” into expressions. - Add the three category expressions and set the sum equal to \(280\).

1. Let \(x\) be the number of nonfiction books. 2. The number of novels is \(2x\), and the number of graphic novels is \(x + 40\). 3. Write the total equation \(x + 2x + (x + 40) = 280\). 4. Combine like terms: \(4x + 40 = 280\). 5. Subtract \(40\): \(4x = 240\). Divide by \(4\): \(x = 60\). 6. There are \(60\) nonfiction books, \(2 \cdot 60 = 120\) novels, and \(60 + 40 = 100\) graphic novels.

There are \(60\) nonfiction books, \(120\) novels, and \(100\) graphic novels.

- Use a variable for the first-day distance. - Express \(1.5\) times that distance for the second day. - Use the second-day expression to write the third-day distance. - Add all three daily distances to equal \(46\,\text{miles}\).

1. Let \(x\) be the distance hiked on the first day. 2. The second-day distance is \(1.5x\), and the third-day distance is \(1.5x - 2\). 3. Write the total equation \(x + 1.5x + (1.5x - 2) = 46\). 4. Combine like terms: \(4x - 2 = 46\). 5. Add \(2\): \(4x = 48\). Divide by \(4\): \(x = 12\). 6. The second-day distance is \(1.5 \cdot 12 = 18\,\text{miles}\), and the third-day distance is \(18 - 2 = 16\,\text{miles}\).

The group hiked \(12\,\text{miles}\) on the first day, \(18\,\text{miles}\) on the second day, and \(16\,\text{miles}\) on the third day.

- Represent side \(a\) with a variable. - Express sides \(b\) and \(c\) using that variable. - Add the three side expressions to form the perimeter equation. - Solve the resulting two-step equation.

1. Let \(x\) inches be the length of side \(a\). 2. Then side \(b\) is \(2x\), and side \(c\) is \(2x - 5\). 3. Use the perimeter: \(x + 2x + (2x - 5) = 45\). 4. Combine like terms: \(5x - 5 = 45\). 5. Add \(5\) and divide by \(5\): \(5x = 50\), so \(x = 10\). 6. The side lengths are \(a = 10\,\text{in.}\), \(b = 20\,\text{in.}\), and \(c = 15\,\text{in.}\).

The side lengths are \(a = 10\,\text{in.}\), \(b = 20\,\text{in.}\), and \(c = 15\,\text{in.}\).

- Use Bag B’s weight as the variable. - Bag C is compared directly with Bag A, not Bag B. - Add all three weight expressions to equal \(10.5\,\text{lb}\). - Include the correct units in the final answer.

1. Let \(x\) be the weight of Bag B in pounds. 2. Bag A weighs \(2x\), and Bag C weighs \(2x - 1.5\). 3. Write the total-weight equation \(2x + x + (2x - 1.5) = 10.5\). 4. Combine like terms: \(5x - 1.5 = 10.5\). 5. Add \(1.5\): \(5x = 12\). Divide by \(5\): \(x = 2.4\). 6. Bag A weighs \(2 \cdot 2.4 = 4.8\,\text{lb}\), and Bag C weighs \(4.8 - 1.5 = 3.3\,\text{lb}\).

Bag A weighs \(4.8\,\text{lb}\), Bag B weighs \(2.4\,\text{lb}\), and Bag C weighs \(3.3\,\text{lb}\).

- The third-day distance in parts a and b is compared with the second-day distance. - Add the three daily expressions to equal the total distance. - In part c, change only the expression for the third day.

1. For part a, the three daily distances are \(x\), \(2x\), and \(2x - 15\). Write \(x + 2x + (2x - 15) = 180\). 2. Combine like terms: \(5x - 15 = 180\). 3. Add \(15\): \(5x = 195\). Divide by \(5\): \(x = 39\). 4. The second-day distance is \(2 \cdot 39 = 78\,\text{miles}\), and the third-day distance is \(78 - 15 = 63\,\text{miles}\). 5. For part c, replace the third-day expression with \(x + 10\). The new equation is \(x + 2x + (x + 10) = 180\), or \(4x + 10 = 180\).

a) \(x + 2x + (2x - 15) = 180\) b) The first-day distance is \(39\,\text{miles}\), the second-day distance is \(78\,\text{miles}\), and the third-day distance is \(63\,\text{miles}\). c) The equation becomes \(x + 2x + (x + 10) = 180\), or \(4x + 10 = 180\).

- Express the top brick in terms of the three bottom bricks. - The two outside bottom bricks each contribute once to the top, while the middle bottom brick contributes twice. - Subtract the outside bricks from the top. The remainder must equal twice the middle brick. - Check whether the remainder is divisible by \(2\) and gives a nonnegative result. - Also check whether the sum of the two outside bricks already exceeds the top brick.

1. Let the bottom row be \(a\), \(b\), and \(c\), and let the top be \(T\). Then \(T = a + 2b + c\), so \(2b = T - a - c\). 2. Wall a): \(2b = 500 - 150 - 250 = 100\), so \(b = 50\). The middle row is \(200\) and \(300\). 3. Wall b): \(2b = 500 - 150 - 249 = 101\). This would give \(b = 50.5\), which is not a nonnegative integer, so the wall cannot be completed under the stated condition. 4. Wall c): \(2b = 800 - 300 - 450 = 50\), so \(b = 25\). The middle row is \(325\) and \(475\). 5. Wall d): \(2b = 800 - 300 - 520 = -20\), so \(b = -10\). This is not nonnegative, so the wall cannot be completed under the stated condition.

1. Walls a) and c) can be completed. a) Bottom middle: \(50\); middle row: \(200\), \(300\) c) Bottom middle: \(25\); middle row: \(325\), \(475\) 2. Walls b) and d) cannot be completed using nonnegative integers. b) The equation gives \(2b = 101\), so \(b = 50.5\), which is not an integer. d) The equation gives \(2b = -20\), so \(b = -10\), which is not nonnegative.

- Translate each division statement into the dividend formula. - Isolate \(x\) using inverse operations. - Check that every remainder is less than its divisor.

1. Write \(x = 5 \cdot 24 + 2\). Then \(x = 122\). 2. Write \(130 = 18x + 4\). Subtract \(4\): \(126 = 18x\). Divide by \(18\): \(x = 7\). 3. Write \(212 = 7 \cdot 30 + x\). Since \(7 \cdot 30 = 210\), \(x = 2\).

a) \(x = 122\) b) \(x = 7\) c) \(x = 2\)