Free Math Worksheets

- Look for a factor that appears in both terms. - Use the distributive property to factor out the common factor. - Add the decimals before multiplying by the fraction.

1. Factor out the common factor: \((0.4 + 0.7) \times \frac{5}{11}\). 2. Add inside the parentheses: \(0.4 + 0.7 = 1.1\). 3. Write \(1.1\) as a fraction: \(1.1 = \frac{11}{10}\). 4. Multiply and simplify: \(\frac{11}{10} \times \frac{5}{11} = \frac{1}{2}\). 5. Maria’s method uses one multiplication instead of two, and the factors simplify immediately.

a) The value is \(\frac{1}{2}\), or \(0.5\). b) Maria’s method is more efficient because it uses one multiplication and allows immediate simplification.

- Find the factor shared by both products. - Write that common factor outside parentheses. - Look for a sum or difference inside the parentheses that is easy to calculate.

1. Use the distributive property in reverse: \(ab \pm ac = a(b \pm c)\). 2. a) \(7.2(14 + 6) = 7.2 \times 20 = 144\). 3. b) \(13(105 - 5) = 13 \times 100 = 1300\). 4. c) \((-4)(28 + 22) = (-4) \times 50 = -200\). 5. d) \(0.25(37 - 17) = 0.25 \times 20 = 5\).

a) \(144\) b) \(1300\) c) \(-200\) d) \(5\) Property: the distributive property

- Identify the factor that appears in both terms. - Add the other two factors inside parentheses before multiplying.

1. For a), factor out \(6\): \(13 \times 6 + 7 \times 6 = (13 + 7) \times 6 = 20 \times 6 = 120\). 2. For b), factor out \(5\): \(24 \times 5 + 6 \times 5 = (24 + 6) \times 5 = 30 \times 5 = 150\). 3. For c), factor out \(3\): \(18 \times 3 + 12 \times 3 = (18 + 12) \times 3 = 30 \times 3 = 90\).

a) \(120\) b) \(150\) c) \(90\)

- Identify the factor repeated in each pair of products. - Compare the factored form with the expanded form. - In part c, match each factor on the left with the corresponding product on the right.

1. For a), the common factor is \(14\), so \(\square = 14\). 2. For b), the common factor is \(5\), so \(\triangle = 5\). 3. For c), compare corresponding factors in \((20 + 12) \times 9 = 20 \times 9 + 12 \times 9\). Thus, \(\triangle = 20\) and \(\square = 9\). 4. For d), the second factor being subtracted is \(4\), so \(\square = 4\).

a) \(\square = 14\) b) \(\triangle = 5\) c) \(\triangle = 20\); \(\square = 9\) d) \(\square = 4\)

- Rewrite the factors so the repeated factor appears in the same position. - Factor out the common factor. - Add inside the grouping symbols before multiplying.

1. Use the commutative property to rewrite \(37 \times 38\) as \(38 \times 37\). 2. Factor out the common factor \(38\): \(38(63 + 37)\). 3. Evaluate: \(38 \times 100 = 3800\).

\(3800\); commutative and distributive properties

- Identify the factor present in all three terms. - Rewrite factors in the same order if needed. - Factor first, then evaluate inside the grouping symbols.

1. Rewrite \(5 \times 17\) as \(17 \times 5\). 2. Factor out \(17\): \(17(8 - 3 + 5)\). 3. Evaluate: \(17 \times 10 = 170\).

\(17(8 - 3 + 5) = 170\)

- Identify the factor present in both products. - Factor it out. - Add inside the grouping symbols before multiplying.

1. Factor out \(12\): \(12(17 + 3)\). 2. Evaluate: \(12 \times 20 = 240\).

\(12(17 + 3) = 240\); distributive property

- Notice that both terms have the same factor. - Use the distributive property in reverse to factor out the common factor. - Convert the measurements inside the parentheses to the same unit.

1. Factor out \(18\): \((125\,\text{mg} + 0.875\,\text{g}) \times 18\). 2. Convert: \(0.875\,\text{g} = 875\,\text{mg}\). 3. Add inside the parentheses: \(125\,\text{mg} + 875\,\text{mg} = 1000\,\text{mg} = 1\,\text{g}\). 4. Multiply: \(1\,\text{g} \times 18 = 18\,\text{g}\).

\(18\,\text{g}\)

- Find the common factor in the two products. - Combine the remaining factors inside parentheses. - Decide whether calculating inside the parentheses first makes the work easier.

1. a) Factor out \(-4.2\): \((-4.2) \times (17 + 3) = (-4.2) \times 20 = -84\). 2. b) Factor out \(\frac{2}{3}\): \(\frac{2}{3} \times (-5.8 - 4.2) = \frac{2}{3} \times (-10) = -\frac{20}{3}\).

a) \(-84\) b) \(-\frac{20}{3}\), or \(-6\frac{2}{3}\)

- Complete each method before comparing them. - In Method B, identify the common factor and add the remaining factors first. - Decide which intermediate numbers are easier to use mentally.

1. Method A gives \(128 + 32 = 160\). 2. Method B factors out the common factor \(8\): \((16 + 4) \times 8 = 20 \times 8 = 160\). 3. Method B is more efficient mentally because it creates the friendly factor \(20\) before multiplying.

Method A: \(160\) Method B: \(160\) Method B is more efficient for mental math because factoring out \(8\) creates the easier product \(20 \times 8\).

- Write a multiplication expression for the cost of each color first. - Both groups have the same price per jump rope, so that common factor can be used once. - Add the numbers of jump ropes before multiplying in the second expression.

1. For a), calculate each box separately: \(14 \times 4 + 16 \times 4\). 2. For b), factor out the common cost of \(4\): \((14 + 16) \times 4\). 3. Evaluate the factored expression: \((14 + 16) \times 4 = 30 \times 4 = 120\).

a) \(14 \times 4 + 16 \times 4\) b) \((14 + 16) \times 4\) c) \(\$120\)

- One method is to find each class total before subtracting. - Another method is to find the price difference for one student first. - Look for the common factor shared by both class totals.

1. Method 1: Find each total and subtract. Class 4A spends \(22 \times \$6 = \$132\), and Class 4B spends \(22 \times \$4 = \$88\). The difference is \(\$132 - \$88 = \$44\). 2. Method 2: Find the difference in price per student and use the common factor \(22\): \(22 \times (\$6 - \$4) = 22 \times \$2 = \$44\).

Class 4A spends \(\$44\) more than Class 4B.

- For one method, combine the adult and child seats in one paired row before multiplying. - For the other method, find the two seating totals separately and add. - Use only the number of adult seats when finding the ticket revenue.

1. For a), Method 1 combines the seats in one adult-and-child row: \((15 + 25) \times 12 = 40 \times 12 = 480\). 2. Method 2 finds the two groups separately: \(12 \times 15 + 12 \times 25 = 180 + 300 = 480\). 3. For b), there are \(12 \times 15 = 180\) adult seats. The ticket revenue is \(180 \times \$8 = \$1440\).

a) There are \(480\) seats altogether. b) The school will collect \(\$1440\).

- Evaluate any side with no variable first. - In parts a and c, factor out the common numerical factor. - Then use inverse operations to find the missing value.

1. a) Factor out \(17\): \(17(5 + x) = 170\). Since \(170 \div 17 = 10\), \(5 + x = 10\), so \(x = 5\). 2. b) Evaluate the right side: \(100 - 60 = 40\). Then \((y - 15) \times 4 = 40\), so \(y - 15 = 10\) and \(y = 25\). 3. c) Factor out \(12\): \((13 - x) \times 12 = 60\). Since \(60 \div 12 = 5\), \(13 - x = 5\), so \(x = 8\).

a) \(x = 5\) b) \(y = 25\) c) \(x = 8\)

- Write the single \(17\) as a product with factor \(1\). - Identify the common factor. - Factor before calculating.

1. Rewrite \(17\) as \(1 \times 17\): \(99 \times 17 + 1 \times 17\). 2. Factor out \(17\): \((99 + 1) \times 17\). 3. Evaluate: \(100 \times 17 = 1700\).

\(1700\)

- A factor outside parentheses applies to the entire sum. - Try evaluating the sum before multiplying. - When distributing, multiply every addend by the outside factor.

1. The factor \(4\) was applied to \(12\) but not to \(9\). 2. Evaluate the parentheses first: \((12 + 9) \times 4 = 21 \times 4 = 84\). 3. Use the distributive property: \((12 + 9) \times 4 = 12 \times 4 + 9 \times 4 = 48 + 36 = 84\).

The factor \(4\) must multiply both addends. Parentheses first: \((12 + 9) \times 4 = 84\). Distributive property: \(12 \times 4 + 9 \times 4 = 84\).

- One method begins by finding the total number of notebooks. - Another method finds each class cost before adding. - The price per notebook is the common factor in all three class costs.

1. Method 1: Add the numbers of notebooks first: \(28 + 32 + 30 = 90\). Then \(90 \times \$0.85 = \$76.50\). 2. Method 2: Find each class cost separately: \(28 \times \$0.85 = \$23.80\), \(32 \times \$0.85 = \$27.20\), and \(30 \times \$0.85 = \$25.50\). Then \(\$23.80 + \$27.20 + \$25.50 = \$76.50\).

The notebooks cost \(\$76.50\) altogether.

- Identify the factor shared by every term. - A term such as \(13\) can be written as \(13 \times 1\). - Factor first, then evaluate inside the grouping symbols.

1. For a), factor out \(12\): \(12(17 + 11 + 2) = 12 \times 30 = 360\). 2. For b), rewrite \(16 \times 18\) as \(18 \times 16\), then factor: \(18(24 + 16) = 18 \times 40 = 720\). 3. For c), factor out \(15\): \(15(27 - 7) = 15 \times 20 = 300\). 4. For d), write \(13\) as \(13 \times 1\), then factor: \(13(29 - 18 - 1) = 13 \times 10 = 130\). 5. For e), factor out \(19\): \(19(14 + 23 + 13) = 19 \times 50 = 950\). 6. For f), write \(22\) as \(22 \times 1\), then factor: \(22(35 - 14 - 1) = 22 \times 20 = 440\).

a) \(360\) b) \(720\) c) \(300\) d) \(130\) e) \(950\) f) \(440\)

- Simplify \(5 \times 4\) first. - Identify the factor shared by all three terms. - Factor it out, then evaluate inside the grouping symbols.

1. Simplify the coefficient in the last term: \(5 \times 4 = 20\). The expression becomes \(36 \times 12 + 14 \times 12 - 20 \times 12\). 2. Factor out the common factor \(12\): \((36 + 14 - 20) \times 12\). 3. Evaluate: \(30 \times 12 = 360\).

\(360\); distributive property