- Use division to undo multiplication. - Divide both sides by the coefficient of \(x\). - Check by substitution.

1. Divide both sides by \(4\): \(x = 24 \div 4 = 6\). 2. Divide both sides by \(7\): \(x = 42 \div 7 = 6\). 3. Divide both sides by \(8\): \(x = 64 \div 8 = 8\). 4. Divide both sides by \(9\): \(x = 54 \div 9 = 6\). 5. Divide both sides by \(3\): \(x = 21 \div 3 = 7\).

1) \(x = 6\) 2) \(x = 6\) 3) \(x = 8\) 4) \(x = 6\) 5) \(x = 7\)

- Recall what multiplying by \(0\) or \(1\) does. - Recall what dividing a number by \(1\) does. - Use an inverse operation when the identity property does not give the value immediately.

1. a) Since \(15\) is not \(0\), the product can equal \(0\) only when \(x=0\). 2. b) Multiplying by \(1\) does not change a value, so \(x=789\). 3. c) Dividing by \(1\) does not change a value, so \(x=1\). 4. d) Multiply both sides by \(5\): \(x=0\). 5. e) Divide both sides by \(36\): \(x=1\).

a) \(x=0\) b) \(x=789\) c) \(x=1\) d) \(x=0\) e) \(x=1\)

- Look at the factor attached to the variable. - Use the inverse of multiplication on both sides.

1. Divide both sides by \(5\). 2. The solution is \(a=7\), and \(5\times7=35\).

\(a=7\). Divide by \(5\).

- The fraction bar shows that the unknown was divided. - Use the inverse operation to restore the starting amount.

1. Multiply both sides by \(4\) to undo the division. 2. The solution is \(b=24\), and \(24\div4=6\).

\(b=24\); check: \(24\div4=6\).

- Think of the related multiplication fact. - Check your proposed divisor in the original division equation.

1. The unknown is the divisor. 2. Since \(3\times6=18\), the divisor is \(c=6\).

\(c=6\)

- Put the proposed value into the original equation. - Compare the results on both sides of the equals sign.

1. Substitute \(4\) for \(x\) on the left side. 2. \(7\times4=28\), matching the right side, so it is a solution.

Yes. \(7\times4=28\).

- The variable does not have to be written on the left side of an equation. - How can you undo division by \(3\) while preserving equality?

1. Both sides represent equal amounts. Multiplying equal amounts by \(3\) keeps them equal. 2. Multiply both sides by \(3\) to undo the division, giving \(n=21\). 3. Check: \(21\div3=7\).

\(n=21\). If two amounts are equal, tripling each gives equal amounts again.

- Decide whether the missing value is a factor or a dividend. - Use multiplication and division as inverse operations. - Substitute each result to check the completed equation.

1. For a), divide \(144\) by \(12\): \(\square=12\). 2. For b), divide \(105\) by \(7\): \(\square=15\). 3. For c), divide \(225\) by \(15\): \(\square=15\). 4. For d), multiply \(13\times11=143\), so \(\square=143\).

a) \(12\) b) \(15\) c) \(15\) d) \(143\)

- Identify the role of the missing number in each equation. - Think about the inverse relationship between multiplication and division. - Substitute each result to verify the equation.

1. For a), divide \(48\) by \(8\): \(\square=6\). 2. For b), multiply \(4\times5=20\): \(\square=20\). 3. For c), multiply \(25\times4=100\): \(\square=100\).

a) \(6\) b) \(20\) c) \(100\)

- Identify each blank as a missing factor. - Use the relationship between a product and its factors. - Estimate whether each missing factor should be greater or less than \(1\), then check by multiplication.

1. For a), \(\square=0.21\div0.3=0.7\). 2. For b), \(\square=0.04\div0.2=0.2\). 3. For c), \(\square=0.36\div0.6=0.6\). 4. For d), \(\square=0.008\div0.1=0.08\).

a) \(0.7\) b) \(0.2\) c) \(0.6\) d) \(0.08\)

- Identify whether the blank is a dividend or a factor. - Use the inverse relationship between multiplication and division. - Substitute each result to verify the equation.

1. For a), \(\square=0.12\times4=0.48\). 2. For b), \(\square=4\times0.2=0.8\). 3. For c), \(\square=3.5\div0.5=7\).

a) \(0.48\) b) \(0.8\) c) \(7\)

- Undo division by multiplying by the divisor. - Break \(24\) into \(20+4\) for mental multiplication. - Check the result by dividing it by \(24\).

1. Multiply both sides by \(24\): \(x=125\times24\). 2. Calculate \(125\times24=125\times20+125\times4=2500+500=3000\).

\(x=3000\)

- Use the inverse operation in each equation. - To find a missing factor, divide the product by the known factor. - To find a missing dividend, multiply the divisor by the quotient.

1. For a), divide the product by the known factor: \(2.4\div0.4=6\). 2. For b), multiply the quotient by the divisor: \(6\times2.5=15\). 3. For c), multiply the quotient by the divisor: \(2.5\times6=15\).

a) \(\square=6\) b) \(\square=15\) c) \(\square=15\)

- Use multiplication and division as inverse operations. - For equations of the form \(x\div p=q\), multiply \(q\) by \(p\). - Check each result by substitution.

1. For a), divide by \(14\): \(a=126\div14=9\). 2. For b), multiply by \(12\): \(b=9\times12=108\). 3. For c), multiply by \(15\): \(c=11\times15=165\). 4. Substitution confirms each solution.

a) \(a=9\) b) \(b=108\) c) \(c=165\)

- Use multiplication and division as inverse operations. - For \(px=q\), divide by \(p\). - For \(x\div p=q\), multiply by \(p\).

1. In a), \(x=81\div9=9\). 2. In b), \(x=5\times9=45\). 3. In c), \(x=64\div8=8\). 4. In d), \(x=7\times6=42\). 5. In e), \(x=40\div4=10\).

a) \(x=9\) b) \(x=45\) c) \(x=8\) d) \(x=42\) e) \(x=10\)

- Find Luke's number using multiplication. - Find Marie's number the same way. - Compare the results.

1. Luke's number is \(20 \times 5 = 100\). 2. Marie's number is \(10 \times 10 = 100\). 3. Both equations give the same number.

Yes. Luke's number and Marie's number are both \(100\).

- For a missing dividend, multiply the quotient by the divisor. - For a missing factor, divide the product by the known factor. - Check each completed equation.

1. For a), \(\square=130\times5=650\). 2. For b), \(\square=60\times8=480\). 3. For c), \(\square=720\div8=90\). 4. For d), \(\square=150\times4=600\).

a) \(650\) b) \(480\) c) \(90\) d) \(600\)

- A missing dividend is the quotient multiplied by the divisor. - A missing factor is the product divided by the known factor. - Check each equation after filling the blank.

1. For a), \(\square=6\times11=66\). 2. For b), \(\square=4\times18=72\). 3. For c), \(\square=85\div17=5\). 4. For d), \(\square=4\times15=60\). 5. For e), \(\square=51\div17=3\).

a) \(66\) b) \(72\) c) \(5\) d) \(60\) e) \(3\)

- 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 \(60\).

- Use the inverse operation to isolate \(x\). - For part a), divide the product by the known factor. - For part b), multiply the quotient by the divisor.

1. For a), divide by \(8\): \(x=960\div8=120\). 2. For b), multiply by \(70\): \(x=6\times70=420\).

a) \(x=120\) b) \(x=420\)

- What happens when a number is multiplied by \(0\)? - When does division give a quotient of \(1\)? - Which factor leaves a number unchanged when multiplying? - What dividend gives a quotient of \(0\) with a nonzero divisor?

1. a) A product is \(0\) when at least one factor is \(0\). Since \(12 \times 9 \ne 0\), \(x = 0\). 2. b) A quotient is \(1\) when the dividend equals the nonzero divisor, so \(x = 48\). 3. c) Multiplying by \(1\) leaves a number unchanged, so \(x = 1\). 4. d) A quotient with a nonzero divisor is \(0\) when the dividend is \(0\), so \(x = 0\).

a) \(x = 0\) b) \(x = 48\) c) \(x = 1\) d) \(x = 0\)

- Use inverse operations in both equations. - In part a), recover the dividend from the divisor and quotient. - In part b), recover the missing factor from the product.

1. For a), multiply: \(x=42\times56=2352\). 2. For b), divide: \(x=8820\div105=84\). 3. Check: \(2352\div56=42\) and \(105\times84=8820\).

a) \(x=2352\) b) \(x=84\)

- 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 \(40\).

- For a missing product, multiply the two given factors. - For a missing factor, use division as the inverse operation. - Use place value to work with multiples of \(10\).

1. For part a), \(14 \times 20 = 14 \times 2 \times 10 = 280\). 2. For part b), divide to find the missing factor: \(900 \div 30 = 30\). 3. For part c), \(500 \div 25 = 20\). 4. For part d), \(12 \times 40 = 12 \times 4 \times 10 = 480\). 5. For part e), \(360 \div 18 = 20\).

a) \(280\) b) \(30\) c) \(20\) d) \(480\) e) \(20\)

- Rewrite each division equation as a related multiplication equation. - Identify the dividend, divisor, and quotient. - Which factor multiplied by the quotient gives the dividend?

1. a) \(\square = 96 \div 8 = 12\). 2. b) \(\square = 144 \div 12 = 12\). 3. c) \(\square = 225 \div 15 = 15\). 4. In general, \(\text{divisor} = \text{dividend} \div \text{quotient}\).

a) \(\square = 12\) b) \(\square = 12\) c) \(\square = 15\) Rule: \(\text{divisor} = \text{dividend} \div \text{quotient}\).

- The same number of tiles is repeated in each equal box. - Use the operation that undoes multiplication by the number of boxes.

1. Divide both sides by \(6\): \(t=42\div6\). 2. Calculate: \(t=7\). 3. Each box holds \(7\) tiles.

\(t=7\) tiles

- Think about which operation reverses dividing the unknown by the rate. - Check that the resulting volume is consistent with a little less than five minutes of dispensing.

1. Multiply both sides by \(2.5\): \(v=4.8\times2.5\). 2. Calculate: \(v=12\). 3. The machine releases \(12\) liters.

\(v=12\,\text{L}\)

- Represent one jug's amount with the variable. - Ask how many equal copies of that amount make the total. - Use the inverse operation to isolate the variable.

1. Five equal jug amounts make the total, so the equation is \(5j=17.5\). 2. Divide the total by \(5\): \(j=17.5\div5=3.5\). 3. Check: \(5\times3.5=17.5\).

\(5j=17.5\), so \(j=3.5\,\text{L}\).

- Count how many equal side lengths make the perimeter. - Write the total perimeter as a multiple of one side length. - Use division to undo that multiplication.

1. A regular hexagon has six equal sides, so its perimeter is \(6s\). 2. The equation is \(6s=45.6\). 3. Divide by \(6\): \(s=7.6\). 4. Check: \(6\times7.6=45.6\).

\(6s=45.6\), so \(s=7.6\,\text{cm}\).

- Identify which quantity is the total and which is the equal share. - Think about the inverse of dividing a total by \(7\). - Interpret the solved variable using the units in the context.

1. The equation says the total amount divided equally across \(7\) days is \(6.5\) gallons per day. 2. Multiply both sides by \(7\): \(w=6.5\times7=45.5\). 3. The solution means the camp uses \(45.5\) gallons over the \(7\) days.

\(w=45.5\,\text{gal}\). The camp uses \(45.5\) gallons in total.

- Look at the side between the top corner and bottom-right corner. - Which multiplication fact includes the number in that side field?

1. The top corner and bottom-right corner multiply to give \(42\). 2. Since the top corner is \(6\), the other corner is \(42\div6=7\).

\(7\)

- What does one box contribute to the total? - Which operation undoes multiplying by the number of bars per box?

1. The number of boxes times bars per box equals the total, so \(8b=48\). 2. Divide by \(8\): \(b=6\).

\(8b=48\); \(b=6\) boxes.

- Which operation is being performed on \(m\) in the original equation? - Does multiplying by the same factor again undo that operation?

1. Multiplying by \(4\) gives \(16m=144\), so the variable still has a coefficient. 2. Dividing both sides of the original equation by \(4\) gives \(m=9\). 3. Check: \(4\times9=36\).

Multiplying by \(4\) leaves \(16m\), not \(m\). Divide by \(4\); \(m=9\).

- Use the inverse operation to isolate \(x\). - Dividing by the fraction multiplying \(x\) will leave \(x\) by itself. - Check each result by substituting it into the original equation.

1. For a), divide both sides by \(\frac{2}{9}\): \(x=\frac{2}{3}\div\frac{2}{9}=3\). 2. For b), \(x=\frac{9}{10}\div\frac{3}{20}=6\). 3. For c), \(x=\frac{5}{2}\div\frac{5}{12}=6\).

a) \(x=3\) b) \(x=6\) c) \(x=6\)

- Use inverse operations to isolate the unknown. - Convert the mixed number to an improper fraction in part a). - For part b), a dividend equals the quotient multiplied by the divisor.

1. For a), write \(2\frac{1}{2}=\frac{5}{2}\). Then \(x=\frac{5}{2}\div\frac{5}{6}=3\). 2. For b), multiply the quotient by the divisor: \(x=21\times\frac{4}{21}=4\).

a) \(x=3\) b) \(x=4\)

- Use the inverse operation to undo multiplication. - Rewrite the decimal as a fraction before dividing. - Dividing by a fraction is equivalent to multiplying by its reciprocal.

1. For a), divide both sides by \(\frac{5}{6}\): \(x=\frac{1}{3}\div\frac{5}{6}=\frac{2}{5}\). 2. For b), rewrite \(0.4\) as \(\frac{2}{5}\). Then \(y=\frac{2}{3}\div\frac{2}{5}=\frac{5}{3}\). 3. For c), \(z=1\div\frac{3}{4}=\frac{4}{3}\).

a) \(x=\frac{2}{5}\) b) \(y=\frac{5}{3}\) c) \(z=\frac{4}{3}\)

- Treat each missing amount as an unknown factor. - What operation undoes multiplication? - Rewrite the mixed number as an improper fraction before dividing. - How can you divide by a fraction?

1. For a), let the missing factor be \(x\): \(x\times\frac{8}{9}=\frac{2}{3}\). Then \(x=\frac{2}{3}\div\frac{8}{9}=\frac{3}{4}\). 2. For b), let \(x\) be the number of pounds: \(\frac{3}{5}\times x=1 \frac{1}{5}=\frac{6}{5}\). Then \(x=\frac{6}{5}\div\frac{3}{5}=2\), so the amount is \(2\,\text{lb}\). 3. For c), let \(x\) be the number of hours: \(\frac{3}{4}\times x=\frac{1}{2}\). Then \(x=\frac{1}{2}\div\frac{3}{4}=\frac{2}{3}\), so the time is \(\frac{2}{3}\,\text{h}\).

a) \(\frac{3}{4}\) b) \(2\,\text{lb}\) c) \(\frac{2}{3}\,\text{h}\)

- Identify whether the missing value is a factor or a dividend. - Use the inverse operation that matches each equation. - When multiplying fractions, simplify common factors when possible.

1. For a), \(\square=\frac{1}{3}\div\frac{5}{6}=\frac{2}{5}\). 2. For b), multiply: \(\square=\frac{9}{8}\times\frac{2}{3}=\frac{3}{4}\). 3. For c), multiply \(\frac{3}{10}\) by \(2\): \(\square=\frac{3}{5}\). 4. For d), divide both sides by \(\frac{2}{3}\): \(\square=1\).

a) \(\frac{2}{5}\) b) \(\frac{3}{4}\) c) \(\frac{3}{5}\) d) \(1\)

- Convert each mixed number to an improper fraction. - Use the inverse operation to isolate \(x\). - Simplify common factors before multiplying.

1. For a), \(x=\frac{3}{8}\div\frac{9}{4}=\frac{3}{8}\times\frac{4}{9}=\frac{1}{6}\). 2. For b), multiply both sides by \(1\frac{2}{3}=\frac{5}{3}\): \(x=\frac{2}{5}\times\frac{5}{3}=\frac{2}{3}\).

a) \(x=\frac{1}{6}\) b) \(x=\frac{2}{3}\)

- Each equation has a missing factor, so divide the product by the known factor. - Compare the decimal place values before calculating. - Express each resulting factor of \(10\) with exponent notation.

1. For a), \(x=40\div0.04=1000=10^3\). 2. For b), \(x=5.8\div0.058=100=10^2\). 3. For c), \(x=0.7\div0.0007=1000=10^3\). 4. For d), \(x=12\div0.0012=10{,}000=10^4\).

a) \(1000=10^3\) b) \(100=10^2\) c) \(1000=10^3\) d) \(10{,}000=10^4\)

- A missing dividend can be found by multiplying the quotient by the divisor. - First evaluate any side whose numbers are all known. - Check that both sides have the same value after substitution.

1. For a), \(\square=20\times0.3=6\). 2. For b), \(\square=9\times0.2=1.8\). 3. For c), \(0.75\div0.25=3\). Then \(\square\div2.5=3\), so \(\square=7.5\). 4. For d), \(0.04\div0.008=5\). Then \(\square\div8=5\), so \(\square=40\).

a) \(6\) b) \(1.8\) c) \(7.5\) d) \(40\)

- Use inverse operations to isolate \(x\). - A missing factor is found by division. - A missing dividend is found by multiplying the quotient by the divisor.

1. For a), \(x=1.44\div0.2=7.2\). 2. For b), \(x=0.8\times0.7=0.56\). 3. For c), \(x=0.04\times1.5=0.06\).

a) \(x=7.2\) b) \(x=0.56\) c) \(x=0.06\)

- Evaluate one side first when possible. - Use inverse operations to solve for a missing factor or divisor. - Recall the values of powers such as \(10^2\), \(10^3\), and \(10^4\).

1. For a), \(0.0007\times10^4=7\). Solve \(100x=7\): \(x=0.07\). 2. For b), \(15.2\div1000=0.0152\). Solve \(0.152\div x=0.0152\): \(x=10\). 3. For c), the first dividend is scaled from \(10^2\) to \(10^4\), a factor of \(100\). Scale the divisor by the same factor: \(8.5\times100=850\).

a) \(0.07\) b) \(10\) c) \(850\)

- Convert the mixed number to an improper fraction. - Recall the reciprocal of a nonzero number. - Use multiplication to reverse division.

1. Write \(1\frac{1}{2}=\frac{3}{2}\). For a), its reciprocal is \(\frac{2}{3}\), and \(\frac{3}{2}\times\frac{2}{3}=1\). 2. For b), a nonzero number divided by itself is \(1\), so the divisor is \(1\frac{1}{2}\). 3. For c), let the dividend be \(x\). Then \(x=2\times\frac{3}{2}=3\).

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

- Identify whether the missing value is a factor, dividend, or product. - Use multiplication and division as inverse operations. - Simplify fraction calculations before multiplying when possible.

1. For a), \(\square=15\div\frac{3}{4}=20\). 2. For b), \(\square=10\div\frac{2}{5}=25\). 3. For c), \(\square=4\times3=12\). 4. For d), \(\frac{7}{9}\times18=14\).

a) \(20\) b) \(25\) c) \(12\) d) \(14\)

- Identify the operation involving the unknown in each equation. - Use the inverse operation to isolate the unknown. - Divide by a fraction by multiplying by its reciprocal.

1. For part a), divide by \(\frac{4}{5}\): \(12\div\frac{4}{5}=12\times\frac{5}{4}=15\). 2. For part b), multiply both sides by \(\frac{2}{3}\): \(\frac{9}{10}\times\frac{2}{3}=\frac{18}{30}=\frac{3}{5}\). 3. For part c), divide by \(\frac{7}{3}\): \(\frac{7}{6}\div\frac{7}{3}=\frac{7}{6}\times\frac{3}{7}=\frac{1}{2}\).

a) \(15\) b) \(\frac{3}{5}\) c) \(\frac{1}{2}\)

- Use the completed products to reason about the unknown row and column headers. - A fraction-to-decimal form may make one comparison easier. - Check the completed table by multiplying each row header by each column header.

1. For the missing column header, solve \(\frac{2}{5}x=1\). Since \(\frac{2}{5}=0.4\), \(x=1\div0.4=2.5\). 2. For the missing row header, solve \(y\times1.5=0.3\). Thus \(y=0.3\div1.5=0.2\). 3. Fill the remaining cells: \(0.4\times1.5=0.6\) and \(0.2\times2.5=0.5\).

<table> <tr><td>\(\times\)</td><td>\(1.5\)</td><td>\(2.5\)</td></tr> <tr><td>\(\frac{2}{5}\)</td><td>\(0.6\)</td><td>\(1\)</td></tr> <tr><td>\(0.2\)</td><td>\(0.3\)</td><td>\(0.5\)</td></tr> </table>

- Compare what happens when the unknown factor is replaced by \(10\). - Use the comparison with the target product to decide on which side of \(10\) the answer lies. - Check each solved value in the original equation.

1. Since \(2.5\times10=25<26.25\), the missing factor must be greater than \(10\). Then \(26.25\div2.5=10.5\). 2. Since \(0.5\times10=5<6\), the missing factor must be greater than \(10\). Then \(6\div0.5=12\). 3. Since \(\frac{1}{3}\times10=\frac{10}{3}>3\), the missing factor must be less than \(10\). Then \(3\div\frac{1}{3}=9\).

a) Greater than \(10\); \(\square=10.5\) b) Greater than \(10\); \(\square=12\) c) Less than \(10\); \(\square=9\)

- Evaluate all numerical expressions before solving. - Identify whether multiplication or division is applied to \(x\). - Use the inverse operation to isolate \(x\).

1. a) Evaluate the grouped quotient: \(45\div5=9\). Then \(9x=54\), so \(x=54\div9=6\). 2. b) Multiply both sides by \(8\): \(x=3\times8=24\). 3. c) Evaluate the right side: \(60-[(7\times6)+(2\times6)]=60-(42+12)=6\). Then \(6x=6\), so \(x=1\).

a) \(x=6\) b) \(x=24\) c) \(x=1\)

- Evaluate any numerical expression that does not contain \(x\) first. - Then identify whether \(x\) is a factor or dividend. - Use multiplication and division as inverse operations.

1. For a), \(42\div6=7\), so \(7x=63\) and \(x=9\). 2. For b), multiply by \(8\): \(x=40\). 3. For c), \(64\div8=8\), so \(8x=56\) and \(x=7\). 4. For d), multiply by \(6\): \(x=30\).

a) \(x=9\) b) \(x=40\) c) \(x=7\) d) \(x=30\)

- Evaluate the side containing only numbers first. - Use division to find the missing factor. - Check that both sides have equal values.

1. a) The left side is \(48\). Solve \(12\times x=48\): \(x=4\). 2. b) The left side is \(75\). Solve \(25\times x=75\): \(x=3\). 3. c) The left side is \(40\). Solve \(x\times8=40\): \(x=5\). 4. d) The left side is \(20+10=30\). Solve \(6\times x=30\): \(x=5\).

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

- Think of division as making equal groups or sharing equally. - Use the relationship between multiplication and division. - Ask which number multiplied by \(12\) equals \(144\).

1. One possible situation is: “A teacher has \(144\) worksheets and gives \(12\) worksheets to each student. How many students receive worksheets?” 2. Rewrite the relationship as \(12x = 144\). 3. Divide by \(12\): \(x = 144 \div 12 = 12\).

Answers will vary for the situation. One example is distributing \(144\) items in groups of \(12\). \(x = 12\)

- First evaluate the side of each equation that has no unknown. - Then identify whether the missing value is a factor or dividend. - Use multiplication and division as inverse operations.

1. For a), \(25+35=60\), so \(60=\square\times4\) and \(\square=15\). 2. For b), \(81\div9=9\), so \(\square\div7=9\) and \(\square=63\). 3. For c), \(6\times4=24\), so \(\square\div5=24\) and \(\square=120\).

a) \(\square=15\) b) \(\square=63\) c) \(\square=120\)

- Find the missing factor first. - Use the relationship between multiplication and division. - Substitute the missing factor into the second expression.

1. Find the missing factor: \(12x = 72\), so \(x = 72 \div 12 = 6\). 2. Use \(6\) as the divisor: \(54 \div 6 = 9\).

The quotient is \(9\).

- Think about how multiplication and division are related. - Identify two operations that undo each other. - Focus on the structure rather than carrying out every calculation.

1. a) Multiplying by \(25\) and then dividing by \(25\) are inverse operations, so they cancel. Therefore, \(x = 82\). 2. b) To return to \(120\) after multiplying by \(10\), the first operation must divide by \(10\). Therefore, \(x = 10\).

a) \(x = 82\) b) \(x = 10\) In each case, multiplication and division by the same nonzero number undo each other.

- Identify whether \(x\) is a factor, dividend, or divisor. - Convert kilograms to grams before solving part b). - Use multiplication to undo division in parts c) and d).

1. Since \(15\div5=3\), \(x=5\). Here \(x\) is a unitless divisor. 2. \(1\,\text{kg}=1000\,\text{g}\). Then \(1000\div4=250\), so \(x=250\,\text{g}\). 3. Multiply \(250\,\text{mL}\) by \(4\): \(x=1000\,\text{mL}=1\,\text{L}\). 4. Write \(x\div5=12\). Multiply by \(5\): \(x=60\,\text{cents}\).

a) \(x=5\) b) \(x=250\,\text{g}\) c) \(x=1000\,\text{mL}=1\,\text{L}\) d) \(x=60\,\text{cents}\)

- Can you reorder the factors to make an easier multiplication? - Which two known factors could you multiply first without changing the product? - Then solve the remaining one-step equation.

1. Group \(125\) and \(4\): \(125\times4=500\). 2. The equation becomes \(500\times\square=1500\). 3. Divide: \(\square=1500\div500=3\).

\(\square=3\)

- Which operation undoes multiplication? - If one factor stays fixed, what happens to the product when the other factor doubles? - Use the relationship among the two factors and the product.

1. a) Divide by \(13\): \(z = 182 \div 13 = 14\). 2. b) With one factor fixed, doubling the product requires doubling the other factor. Therefore, \(z\) doubles from \(14\) to \(28\). 3. Check: \(13 \times 28 = 364\).

a) \(z = 14\) b) \(z\) doubles to \(28\).

- Use multiplication to undo division in part 1. - With the divisor fixed, relate a change in the quotient to the dividend. - In part 3, find the divisor from the dividend and quotient.

1. Multiply the quotient by the divisor: \(y = 8 \times 25 = 200\). 2. If the quotient is halved to \(4\), then \(y = 4 \times 25 = 100\). Therefore, \(y\) is also halved. 3. Find the missing divisor by dividing the dividend by the quotient: \(m = 200 \div 8 = 25\).

1. \(y = 200\) 2. \(y\) is halved to \(100\). 3. \(m = 25\)

- Compare the answers to parts a and b. - Keep the dividend fixed while changing the quotient. - Rewrite the division equation as multiplication.

1. a) \(240 \div 10 = 24\), so the divisor is \(24\). 2. b) \(240 \div 30 = 8\), so the divisor is \(8\). 3. c) When the quotient is tripled from \(10\) to \(30\), the divisor is divided by \(3\), from \(24\) to \(8\). 4. d) The division equation is equivalent to \(\text{divisor} \times \text{quotient} = \text{dividend}\). To find one factor, divide the product by the other factor.

a) \(24\) b) \(8\) c) The divisor is divided by \(3\). d) Because \(\text{divisor} \times \text{quotient} = \text{dividend}\), divide the dividend by the quotient to find the divisor.

- Begin with “dividend divided by divisor equals quotient.” - Use multiplication to undo division. - Substitute \(d = 13\) into your expression.

1. The division equation is \(x \div d = 7\). Multiplying by \(d\) gives \(x = 7d\). 2. When \(d = 13\), \(x = 7 \times 13 = 91\). 3. Multiply the quotient by the divisor to find the dividend.

1. \(x = 7d\) 2. \(x = 91\) 3. \(\text{dividend} = \text{quotient} \times \text{divisor}\)

- How far is the shown top from the target? - Which bricks above the middle bottom brick change when it changes? - Write the multiplication equation before finding the new brick.

1. The current top is \(560\), so it must increase by \(600-560=40\). 2. A change of \(x\) in the middle bottom brick adds \(x\) to each brick directly above it. Thus \(2x=40\). 3. Solve: \(x=20\). The new middle bottom brick is \(140+20=160\). 4. Check: \((120+160)+(160+160)=600\).

\(2x=40\); \(x=20\); new middle bottom brick: \(160\)

- In a three-brick bottom row, the middle bottom brick contributes to both bricks above it. - Remove the contribution of the two outside bricks before writing the one-step equation. - Give the equation and its solution for each wall.

1. In wall a), the outside bottom bricks total \(200+250=450\), leaving \(950-450=500\) for the two contributions from the middle brick. 2. The required equation is \(2b=500\), so \(b=250\). 3. In wall b), the outside bottom bricks total \(111+111=222\), leaving \(666-222=444\). 4. The required equation is \(2b=444\), so \(b=222\).

a) \(2b=500\), so \(b=250\). b) \(2b=444\), so \(b=222\).

- Write the two bricks just below the top in terms of the missing middle brick. - How often does that missing brick appear when you add the two expressions? - Compare the outside bottom values across the walls while the top stays fixed.

1. Wall 1: subtract the outside contributions, \(800-200-200=400\). Hence \(2b=400\), so \(b=200\). 2. Wall 2: \(800-100-100=600\). Hence \(2b=600\), so \(b=300\). 3. Wall 3: \(800-50-50=700\). Hence \(2b=700\), so \(b=350\). 4. As the sum of the outside bricks decreases, the middle brick increases by half that decrease because it counts twice in the top.

a) Wall 1: \(2b=400\), \(b=200\); Wall 2: \(2b=600\), \(b=300\); Wall 3: \(2b=700\), \(b=350\) b) If the outside bottom bricks decrease in total by a certain amount, the middle brick increases by half that amount.

- How much must the top brick increase? - Trace how a change in each bottom brick affects the top. - Your answer should contain the equation and its solution.

1. The current top is \(25\), so it must increase by \(45-25=20\). 2. Adding \(x\) to every bottom brick raises each middle brick by \(2x\), raising the top by \(4x\). 3. Therefore \(4x=20\) and \(x=5\). 4. Check: bottom bricks \(10,12,11\) give middle bricks \(22,23\) and top \(45\).

\(4x=20\), so \(x=5\).

- Find the difference between the old and new top values. - How much do the three bricks in the row above the bottom increase? - Use the increase at the top to write an equation.

1. The original top is \(320\), so the increase needed is \(1000-320=680\). 2. Adding \(x\) to all four bottom bricks increases the top by \(8x\). 3. Hence \(8x=680\) and \(x=85\).

\(8x=680\), so \(x=85\).

- What do two equal bottom bricks add up to? - Express the middle bricks and top in terms of \(x\). - Use the number at the top to solve the equation.

1. Both bricks in the middle row are \(2x\), so the top is \(4x\). 2. The equation is \(4x=44\), giving \(x=11\). 3. The bottom row is \(11,11,11\), and the middle row is \(22,22\).

Top: \(4x\); equation: \(4x=44\); \(x=11\). Bottom row: \(11,11,11\); middle row: \(22,22\).

- Translate “two-thirds of” as multiplication by a fraction. - Think about what operation will undo multiplication by \(\frac{2}{3}\). - Check the solution by finding two-thirds of the full capacity you obtained.

1. “Two-thirds of the full capacity” is \(\frac{2}{3}c\), so the equation is \(\frac{2}{3}c=14\). 2. Divide \(14\) by \(\frac{2}{3}\), or multiply by its reciprocal: \(c=14\times\frac{3}{2}=21\). 3. Check: \(\frac{2}{3}\times21=14\).

\(\frac{2}{3}c=14\), so \(c=21\,\text{gal}\).

- Represent the whole numbers immediately before and after the unknown middle number in terms of \(m\). - Add the two outer expressions and simplify before writing the equation. - Your answer must include both the multiplication equation and its solution.

1. The three consecutive whole numbers are \(m-1\), \(m\), and \(m+1\). 2. The smallest and largest numbers have sum \((m-1)+(m+1)=2m\). 3. The required one-step multiplication equation is \(2m=2{,}000{,}000\). 4. Divide by \(2\): \(m=1{,}000{,}000\). 5. Check: \(999{,}999+1{,}000{,}001=2{,}000{,}000\).

\(2m=2{,}000{,}000\), so \(m=1{,}000{,}000\).

- Complete the original wall from bottom to top first. - Track how a common change in the four bottom positions contributes to the top. - For part b), your answer must include the one-step multiplication equation for the common increase.

1. The second row is \(130\), \(140\), and \(100\). The third row is \(270\) and \(240\), so the original top is \(510\). 2. Increasing every bottom brick by \(x\) increases the top by \(8x\), because the four bottom positions contribute with coefficients \(1\), \(3\), \(3\), and \(1\). 3. The top must increase by \(590-510=80\), so the required equation is \(8x=80\). Thus \(x=10\). 4. If every bottom brick increases by \(20\), the top increases by \(8\times20=160\). The new top is \(510+160=670\).

a) \(510\) b) \(8x=80\), so \(x=10\). Increase each bottom brick by \(10\). c) \(670\)

- Use the same variable for all four equal bottom bricks. - Track how many copies of \(x\) appear in each higher row. - Write the equation using the number at the top.

1. The next row contains \(2x,2x,2x\). 2. The following row contains \(4x,4x\), so the top is \(8x\). 3. The top is \(240\), giving \(8x=240\). Thus \(x=30\).

Top: \(8x\); equation: \(8x=240\); \(x=30\)