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. - 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) \(b = 9\)

- Use a variable for the starting number before doing any arithmetic. - Translate the operations in the order they occur. - Isolate the variable by undoing the outside addition and then the division. - Check in the original equation.

1. Let \(x\) be the number and write \(\frac{x}{4}+15=30\). 2. Subtract \(15\) from both sides: \(\frac{x}{4}=15\). 3. Multiply both sides by \(4\): \(x=60\). 4. Check: \(60\div4+15=30\).

Equation: \(\frac{x}{4}+15=30\) Solution: \(x=60\)

- Decide which quantity is the starting value and which quantity is being subtracted. - Use a variable for the unknown before solving. - Preserve the order of subtraction in each equation. - Check each value in the equation you wrote.

1. For a), write \(120-x=50\). The amount removed is \(120-50\), so \(x=70\). 2. For b), write \(y-45=100\). Add \(45\) to both sides to get \(y=145\). 3. Check each result in the original subtraction statement.

a) Equation: \(120-x=50\); \(x=70\) b) Equation: \(y-45=100\); \(y=145\)

- Represent the unknown with a variable before solving. - Translate “multiply by \(6\), then add \(4\)” in that order. - Undo the added constant before undoing the multiplication. - Substitute the solution back into the equation.

1. Let \(x\) be the number and write \(6x+4=40\). 2. Subtract \(4\) from both sides: \(6x=36\). 3. Divide both sides by \(6\): \(x=6\). 4. Check the value in the original equation.

Equation: \(6x+4=40\) Solution: \(x=6\)

- Use a variable for the starting number. - Preserve the order of multiplication and subtraction when writing the equation. - Undo the subtraction before the multiplication. - Check by substituting into the equation.

1. Let \(x\) be the starting number and write \(4x-24=100\). 2. Add \(24\) to both sides: \(4x=124\). 3. Divide both sides by \(4\): \(x=31\). 4. Check: \(4\cdot31-24=100\).

Equation: \(4x-24=100\) Starting number: \(31\)

- The subtraction happens before the division, so the numerator must represent the result after subtracting. - Use inverse operations on the equation in reverse order. - Keep the grouping intact until the division has been undone. - Check the complete operation sequence.

1. Let \(x\) be the starting number and write \(\frac{x-50}{5}=40\). 2. Multiply both sides by \(5\): \(x-50=200\). 3. Add \(50\) to both sides: \(x=250\). 4. Check: \((250-50)\div5=40\).

Equation: \(\frac{x-50}{5}=40\) Starting number: \(250\)

- 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) \(x = 7\)

- Evaluate the fully known comparison quantity first. - Translate “three times a number” using a variable. - Use the inverse of multiplication to isolate the variable. - Check both parts of the verbal statement.

1. The value \(20\) less than \(200\) is \(180\). 2. Write the one-step equation \(3x=180\). 3. Divide both sides by \(3\): \(x=60\). 4. Check that three times \(60\) is \(20\) less than \(200\).

Equation: \(3x=180\) The number is \(60\).

- Combine the two known changes before solving. - After simplification, identify the single addition operation on the unknown. - Use the inverse operation on both sides. - Check the answer using the original two changes.

1. The net change is \(120-50=70\), so write \(x+70=600\). 2. Subtract \(70\) from both sides: \(x=530\). 3. Check: \(530+120-50=600\).

Equivalent one-step equation: \(x+70=600\) Starting number: \(530\)

- Combine the subtraction and addition into one net change. - Write the simplified relation with a variable before solving. - Use the inverse of the remaining subtraction. - Verify the value with the original sequence.

1. The subtraction of \(245\) is partly offset by adding \(112\). The overall decrease is \(245-112=133\), so write \(x-133=500\). 2. Add \(133\) to both sides: \(x=633\). 3. Check: \(633-245+112=500\).

Equivalent one-step equation: \(x-133=500\) Starting number: \(633\)

- Simplify the side that contains no variable first. - Represent the unknown with a variable and write the equality. - Use the inverse of adding \(325\). - Check both sides after solving.

1. Evaluate the known side: \(900-150=750\). 2. Write \(x+325=750\). 3. Subtract \(325\) from both sides: \(x=425\). 4. Check that both sides equal \(750\).

Equation: \(x+325=750\) 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) \(x=700\)

- Simplify the fully numerical side first. - Write an equation with a variable for the unknown number. - Use the inverse of subtraction to isolate the variable. - Substitute the result to check.

1. Evaluate the known side: \(135+165=300\). 2. Write \(x-257=300\). 3. Add \(257\) to both sides: \(x=557\). 4. Check the equality in the original statement.

Equation: \(x-257=300\) The number is \(557\).

- The increase happens before the division, so group the increased quantity together. - Undo the outer division before undoing the addition. - Keep the equation balanced at each step. - Check the original operation order.

1. Let \(x\) be the mystery number and write \(\frac{x+120}{3}=100\). 2. Multiply both sides by \(3\): \(x+120=300\). 3. Subtract \(120\): \(x=180\). 4. Check: \((180+120)\div3=100\).

Equation: \(\frac{x+120}{3}=100\) Mystery number: \(180\)

- Simplify the known sum before writing the equation. - Keep the variable in the subtracted position shown by the sentence. - Ask what amount must be removed from \(1000\) to reach the known difference. - Check the subtraction after solving.

1. Evaluate \(120+130=250\). 2. Write \(1000-x=250\). 3. The amount removed is \(1000-250\), so \(x=750\). 4. Check: \(1000-750=250\).

Equation: \(1000-x=250\) The number subtracted 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 \(400\).

- Simplify the side without the variable first. - Translate the remaining statement into an equation. - Use the inverse of subtracting \(360\). - Check the value in the equation.

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

Equation: \(x-360=240\) The number is \(600\).

- Evaluate the fully known expression for the difference first. - Represent the unknown number being subtracted with a variable. - Preserve the order of subtraction when you write the equation. - Check by substituting the result.

1. The known difference is \(2\times110=220\). 2. Write \(720-x=220\). 3. Solve for the subtracted number: \(x=720-220=500\). 4. Check: \(720-500=220\).

Equation: \(720-x=220\) The number subtracted is \(500\).

- Simplify the side that contains only known numbers. - Translate “eight times a number” into an equation. - Use division to undo multiplication. - Check the equation after solving.

1. Evaluate \(200+120=320\). 2. Write \(8x=320\). 3. Divide both sides by \(8\): \(x=40\). 4. Check: \(8\times40=320\).

Equation: \(8x=320\) The number is \(40\).

- Translate the division and subtraction in the order stated. - Undo the subtraction before undoing the division. - Apply the same inverse operation to both sides. - Check the value in the original equation.

1. Let \(x\) be the number and write \(\frac{x}{4}-80=20\). 2. Add \(80\) to both sides: \(\frac{x}{4}=100\). 3. Multiply both sides by \(4\): \(x=400\). 4. Check: \(400\div4-80=20\).

Equation: \(\frac{x}{4}-80=20\) Solution: \(x=400\)

- 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. - After simplifying, use two inverse operations to isolate the variable.

1. Combine like terms: \(9x+9=81\). Subtract \(9\): \(9x=72\). Divide by \(9\): \(x=8\). 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) \(x=8\) 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 central variable because the other two amounts are symmetric around it. - Express Beth and Clara before adding the three amounts. - Notice what happens to the constant terms when the equation is simplified. - Check the total and both difference relationships.

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: \(40 + 30 + 50 = 120\).

Equation: \(x+(x-10)+(x+10)=120\), which simplifies to \(3x=120\). Anna: \(40\,\text{lb}\); Beth: \(30\,\text{lb}\); Clara: \(50\,\text{lb}\)

- Use Mia’s count as the base variable for both relationships. - Translate “\(20\) more” and “twice as many” into separate expressions. - Add all three quantities before setting the total equal to \(200\). - Check both relationships after solving.

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.

Equation: \(x+(x+20)+2x=200\) Mia: \(45\) cards; Lucas: \(65\) cards; Noah: \(90\) 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) \(x = 12\)

- Identify which symbol has no assigned value. - Substitute the known values before solving. - Undo multiplication before undoing the subtraction inside the parentheses.

1. The remaining unknown variable is \(x\). 2. Substitute the known values: \(3=0.25(x-5)\). 3. Divide by \(0.25\): \(12=x-5\). 4. Add \(5\): \(x=17\).

The unknown variable is \(x\), and \(x=17\).

- Identify the operation applied to the variable term last. - Undo that operation on both sides first. - Substitute your solution into the original equation to check it.

1. Add \(7\) to both sides: \(7x=49\). 2. Divide both sides by \(7\): \(x=7\). 3. The expression \(7x-7\) means \(7\) is subtracted after the product \(7x\) is formed. Reversing the operations in reverse order isolates the variable term first.

\(x=7\). Add \(7\) first to isolate \(7x\), then divide by \(7\).

- Use inverse addition or subtraction to isolate the unknown. - For part b), combine the known fractional terms before solving. - Use a common denominator for the mixed-number and fraction calculations. - Check each value in the original equation.

1. For a), the amount subtracted is the difference between the starting value and the result: \( x=5\frac12-2\frac34=2\frac34. \) 2. For b), combine the known positive rational terms: \( 1\frac13-\frac16=\frac86-\frac16=\frac76. \) The equation becomes \(x+\frac76=3\). 3. Subtract \(\frac76\) from both sides: \(x=\frac{18}{6}-\frac76=\frac{11}{6}=1\frac56\).

a) \(x=2\frac34\) b) \(x=1\frac56\)

- Undo addition or subtraction before undoing multiplication in forms such as \(px+q=r\). - In a form such as \(p(x+q)=r\), undo the outside multiplication first. - Keep track of the negative coefficient in part b. - Check each result by substitution.

1. For a), subtract \(12.45\): \(2.5x=17.75\). Divide by \(2.5\): \(x=7.1\). 2. For b), subtract \(45.6\): \(-1.2x=-33.6\). Divide by \(-1.2\): \(x=28\). 3. For c), divide by \(1.2\): \(x+2.5=7\). Subtract \(2.5\): \(x=4.5\).

a) \(x=7.1\) b) \(x=28\) c) \(x=4.5\)

- 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 the fully known side before isolating the variable. - Use a common denominator in part a. - In part b, interpret \(0.75-x=0.25\) as a subtraction equation with an unknown amount removed. - Check each solution in its original equation.

1. For a), simplify the right side: \(\frac23+\frac12=\frac76\). Then add \(\frac16\) to both sides: \(x=\frac86=\frac43\). 2. For b), simplify the right side: \(\frac18+\frac18=\frac14=0.25\). The equation is \(0.75-x=0.25\), so the amount subtracted is \(x=0.75-0.25=0.50\).

a) \(x=\frac43\), or \(1\frac13\) b) \(x=0.5\), or \(\frac12\)

- Simplify the known quantities before applying an inverse operation to the variable. - Convert decimals and fractions only when it makes a calculation clearer. - Use a common denominator in part b. - Substitute the result to check each equation.

1. For a), \(\frac15+0.1=0.2+0.1=0.3\), so the right side is \(0.8\). Thus \(x+0.3=0.8\), giving \(x=0.5\). 2. For b), combine the known constants: \(\frac14-\frac1{12}=\frac3{12}-\frac1{12}=\frac16\). The equation is \(\frac23=x+\frac16\). 3. Subtract \(\frac16\) from \(\frac23\): \(x=\frac46-\frac16=\frac12\).

a) \(x=0.5\), or \(\frac12\) b) \(x=\frac12\)

- 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 \(12.4\): \(\Box=7.4\).

\(7.4\)

- Combine the two known changes before solving for the variable. - After simplifying, identify the single operation being performed on \(x\). - Use the inverse operation on both sides, then check in the original sequence.

1. Adding \(8\) and then subtracting \(14\) produces an overall decrease of \(14-8=6\), so the puzzle is equivalent to \(x-6=10\). 2. Add \(6\) to both sides: \(x=16\). 3. Check in the original sequence: \(16+8-14=10\).

Equivalent one-step equation: \(x-6=10\) Solution: \(x=16\)

- 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 constants before isolating the variable term. - Undo addition or subtraction before undoing multiplication or division. - Use a common denominator when combining fractions.

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

a) \(x=10\) b) \(x=\frac{5}{2}\), or \(x=2.5\) c) \(x=5\) d) \(x=8\)

- Rewrite fractions as decimals when that makes the constants easier to combine. - Undo the outer operation before the inner operation when parentheses are present. - Keep track of the negative coefficient in part c).

1. For a), \(\frac{3}{4}=0.75\), and \(0.75-0.25=0.5\). The equation becomes \(0.5+0.5x=1\). Subtract \(0.5\): \(0.5x=0.5\). Divide by \(0.5\): \(x=1\). 2. For b), divide both sides by \(2\): \(x+1.5=3.5\). Subtract \(1.5\): \(x=2\). 3. For c), subtract \(10\): \(-\frac{x}{2}=-1.5\). Multiply by \(-2\): \(x=3\).

a) \(x=1\) b) \(x=2\) c) \(x=3\)

- Represent each unknown with a variable before calculating. - Translate “sum” as addition and “difference between a number and \(20\)” as subtraction from the unknown. - Use the inverse addition or subtraction operation to isolate the variable. - Check each value in the equation you wrote.

1. For a), write \(x+30=75\). Subtract \(30\) from both sides to get \(x=45\). 2. For b), write \(y-20=10\). Add \(20\) to both sides to get \(y=30\). 3. Substitute each value into its equation to verify it.

a) Equation: \(x+30=75\); \(x=45\) b) Equation: \(y-20=10\); \(y=30\)

- 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: \(1200 - (350 + 150) = 1200 - 500 = 700\).

\(x = 350\)

- Undo addition or subtraction first. - Then undo multiplication to isolate \(x\). - Check each solution in the original equation.

1. For a), add \(6\): \(7x=35\). Divide by \(7\): \(x=5\). 2. For b), subtract \(4\): \(9x=54\). Divide by \(9\): \(x=6\). 3. For c), add \(14\): \(8x=64\). Divide by \(8\): \(x=8\). 4. For d), subtract \(27\): \(5x=25\). Divide by \(5\): \(x=5\).

a) \(x=5\) b) \(x=6\) c) \(x=8\) d) \(x=5\)

- Translate each statement into its own equation before solving. - In each equation, undo the added or subtracted constant before the multiplication. - Keep the variables separate because the two people may have different numbers. - Compare only after both equations are solved.

1. Luke’s statement gives \(6x+4=40\). Subtract \(4\) and divide by \(6\): \(x=6\). 2. Sara’s statement gives \(5y-3=32\). Add \(3\) and divide by \(5\): \(y=7\). 3. Since \(7>6\), Sara chose the greater number.

Luke: \(6x+4=40\), so \(x=6\). Sara: \(5y-3=32\), so \(y=7\). Sara chose the greater number.

- Write one equation for each operation sequence before calculating. - Luke’s variable is multiplied before a subtraction; Sarah’s is divided before an addition. - Use inverse operations in the reverse order from the equation. - Compare the solutions after both are found.

1. Luke’s equation is \(2x-20=80\). Add \(20\) and divide by \(2\): \(x=50\). 2. Sarah’s equation is \(\frac{y}{2}+20=80\). Subtract \(20\) and multiply by \(2\): \(y=120\). 3. Since \(120>50\), Sarah chose the greater number.

Luke: \(2x-20=80\), so \(x=50\). Sarah: \(\frac{y}{2}+20=80\), so \(y=120\). 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 when a factor multiplies parentheses. - In part c), isolate the variable term before dividing by its coefficient. - Check each solution in the original equation.

1. For a), divide both sides by \(6\): \(x+5=12\). Subtract \(5\): \(x=7\). 2. For b), divide both sides by \(4\): \(x-12=12\). Add \(12\): \(x=24\). 3. For c), subtract \(60\): \(-3x=-39\). Divide by \(-3\): \(x=13\).

a) \(x=7\) b) \(x=24\) c) \(x=13\)

- 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) \(x = 9\)

- Use long division one place at a time. - After each multiplication, subtract to find the partial remainder before bringing down the next digit. - A final remainder must be smaller than the divisor. - Verify each result with divisor \(\times\) quotient \(+\) remainder.

1. For a), \(23\) goes into \(124\) five times: \(5\times23=115\), leaving \(9\). Bring down \(7\) to get \(97\). Then \(23\) goes into \(97\) four times: \(4\times23=92\), leaving \(5\). Thus \(1247\div23=54\text{ R }5\). 2. Check: \(23\times54+5=1242+5=1247\). 3. For b), \(37\) goes into \(238\) six times: \(6\times37=222\), leaving \(16\). Bring down \(5\) to get \(165\). Then \(37\) goes into \(165\) four times: \(4\times37=148\), leaving \(17\). Thus \(2385\div37=64\text{ R }17\). 4. Check: \(37\times64+17=2368+17=2385\).

a) \(54\text{ R }5\) b) \(64\text{ R }17\) c) \(1247=23\times54+5\) and \(2385=37\times64+17\)

- Combine the two known additions before solving. - After simplification, identify the single operation on \(x\). - Use the inverse operation on both sides. - Check using the original two additions.

1. Combine the known changes: \(15+22=37\). 2. Write the equivalent one-step equation \(x+37=44\). 3. Subtract \(37\) from both sides: \(x=7\). 4. Check in the original sequence: \(7+15+22=44\).

Equivalent one-step equation: \(x+37=44\) \(x=7\)

- Follow the order of operations described in each statement. - A sum that is multiplied needs parentheses. - Simplify fixed numerical expressions, then use two inverse operations to isolate the variable.

1. Write \(2n+17=45\). Subtract \(17\): \(2n=28\). Divide by \(2\): \(n=14\). 2. Write \(100-2z=4\cdot18\). Simplify to \(100-2z=72\). Subtract \(100\): \(-2z=-28\). Divide by \(-2\): \(z=14\). 3. Write \(5(k+2.2)=25\). Divide by \(5\): \(k+2.2=5\). Subtract \(2.2\): \(k=2.8\).

1) \(n=14\) 2) \(z=14\) 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 \(18\).

- Choose the second piece as the base quantity because the other two are described from it. - Express all three lengths with the same variable before adding them. - Set their sum equal to the original rope length. - Use the solved variable to recover all three piece lengths.

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 \(42 - 5 = 37\,\text{in.}\).

Equation: \(2x+x+(2x-5)=100\) Piece lengths: \(42\,\text{in.}\), \(21\,\text{in.}\), and \(37\,\text{in.}\)

- Use the first-day distance as the base variable. - Build the second- and third-day expressions from that same variable. - Add the three daily expressions and equate them to the total distance. - Check that the solved distances satisfy both stated relationships.

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 \(18 - 2 = 16\,\text{miles}\).

Equation: \(x+1.5x+(1.5x-2)=46\) Day 1: \(12\,\text{miles}\); Day 2: \(18\,\text{miles}\); Day 3: \(16\,\text{miles}\)

- 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.

- In the arithmetic approach, ask what one of the five equal groups must equal. - In the algebraic approach, identify the outside operation on \(x-2\). - Compare the order in which the two methods reverse the original operations.

1. Arithmetic reasoning: if \(5\) equal groups total \(35\), each group is \(35\div5=7\). Since each group is \(x-2\), \(x\) must be \(2\) more than \(7\), so \(x=9\). 2. Algebraically, divide both sides by \(5\): \(x-2=7\). Add \(2\): \(x=9\). 3. Both methods first undo multiplication by \(5\) and then undo subtraction of \(2\).

a) \(x=9\) b) \(x=9\) c) Both methods undo multiplication by \(5\) first and subtraction of \(2\) second.