- Find the total cost when \(b=0\) and when \(b=1\). - Which amount changes when one more badge is made?

1. The \(12\) is the one-time table fee, so it stays constant. 2. The \(3b\) represents the badge cost and grows when \(b\) increases.

Fixed cost: \(\$12\). Changing cost: \(\$3\) per badge, represented by \(3b\).

- Work out how many objects are in one bag before counting all six bags. - Compare what the two forms make easiest to see.

1. The parentheses in \(6(x+3)\) show one bag contains \(x\) stickers and \(3\) buttons. 2. The expanded form \(6x+18\) separates the sticker total and the \(18\) buttons.

\(6(x+3)\) shows the contents of one bag; \(6x+18\) shows there are \(18\) buttons altogether.

- Identify what happens to the same \(x\) beads at the beginning and end. - Which two terms can you combine first?

1. Regroup \(x+9-x\) as \((x-x)+9\). 2. The variable terms cancel: \(x-x=0\), so \(9\) beads remain, whatever \(x\) was.

\(x+9-x=9\). The original \(x\) beads are all removed, so \(9\) remain.

- One whole price is \(100\%\) of \(p\). - What portion stays after the discount?

1. Combine like terms: \(p-0.2p=(1-0.2)p=0.8p\). 2. The multiplier \(0.8\) means paying \(80\%\) of the original price.

\(0.8p\); you pay \(80\%\) of the original price.

- Convert \(\frac{5}{4}\) to a decimal for the first method. - Convert both decimals to fractions for the second method. - Look for reciprocal factors.

1. Decimal method: \(\frac{5}{4} = 1.25\), so \(0.8 \cdot 1.25 + 0.25 = 1 + 0.25 = 1.25\). 2. Fraction method: \(0.8 = \frac{4}{5}\) and \(0.25 = \frac{1}{4}\). Then \(\frac{4}{5} \cdot \frac{5}{4} + \frac{1}{4} = 1 + \frac{1}{4} = \frac{5}{4}\). 3. The fraction method is especially efficient because the reciprocal factors cancel immediately.

a) \(1.25\), or \(\frac{5}{4}\) b) The fraction method is more efficient because \(\frac{4}{5} \cdot \frac{5}{4} = 1\).

- Rewrite a factor or dividend using nearby multiples of \(10\) or \(100\). - The required response includes the decomposition, not only the result. - Keep the negative sign attached to the factor in part d).

1. a) \(15(100+1)=1500+15=1515\). 2. b) \(18(10-1)=180-18=162\). 3. c) \((300+15)\div3=100+5=105\). 4. d) \(-22(10+1)=-220-22=-242\).

a) \(15(100+1)=1515\) b) \(18(10-1)=162\) c) \((300+15)\div3=105\) d) \(-22(10+1)=-242\)

- Look for a factor shared by two products. - Rewrite subtraction as addition of the opposite before rearranging addends. - Factoring can turn two products into one simpler product.

1. For a), factor out \(17\): \(17[(-4) + (-6)] = 17(-10) = -170\). 2. For b), rewrite subtraction as addition of the opposite, then rearrange and regroup: \((-45 - 55) + 128 = -100 + 128 = 28\). 3. For c), factor out \(0.5\): \(0.5(19 - 9) = 0.5 \cdot 10 = 5\).

a) \(-170\) b) \(28\) c) \(5\)

- Compare the two fractional parts before multiplying the decimal twice. - What is the sum of the two fractions with denominator \(7\)? - Check the sign of the result using the signs of both original products.

1. Factor out the common factor \(-15.8\): \(-15.8\left(\frac{3}{7}+\frac{4}{7}\right)\). 2. Add inside the parentheses: \(\frac{3}{7}+\frac{4}{7}=1\). 3. Multiply: \(-15.8\cdot1=-15.8\).

\(-15.8\left(\frac{3}{7}+\frac{4}{7}\right)=-15.8\)

- Follow the grouping symbols when using the first method. - For the second method, view division by \(4\) as multiplication by \(\frac{1}{4}\). - Compare the number of arithmetic operations in the two methods.

1. Subtract first: \(12.4 - 8.8 = 3.6\), then \(3.6 \div 4 = 0.9\). 2. Distribute \(\frac{1}{4}\): \(12.4 \div 4 - 8.8 \div 4 = 3.1 - 2.2 = 0.9\). 3. Both methods are valid. Subtracting first requires one subtraction and one division, while distributing requires two divisions and one subtraction.

Both methods give \(0.9\). Subtracting first requires less computation.

- Can \(42\) be divided evenly by both denominators in a)? - In b), what do the two decimal values add up to? - Check that your expanded or factored form uses the same numbers and signs.

1. For a), distribute \(42\): \(42\cdot\frac{1}{7}+42\cdot\frac{1}{6}=6+7=13\). 2. For b), factor out \(17\): \(17(3.8+6.2)=17\cdot10=170\).

a) \(42\cdot\frac{1}{7}+42\cdot\frac{1}{6}=13\) b) \(17(3.8+6.2)=170\)

- In a), can you find a useful whole-number total without multiplying by \(15\) twice? - Compare the two small factors in b) before multiplying decimals. - In c), both products are negative. What sign should their sum have?

1. For a), factor out \(15\): \(15(48+52)=15\cdot100=1500\). 2. For b), factor out \(12.5\): \(12.5(7-3)=12.5\cdot4=50\). 3. For c), factor out \(-20\): \((-20)(14+36)=(-20)\cdot50=-1000\).

a) \(15(48+52)=1500\) b) \(12.5(7-3)=50\) c) \((-20)(14+36)=-1000\)

- In a), what pair of factors makes the multiplication easier? - In b), look for a factor that appears in both products. - In c), split one factor into two easy addends.

1. Regroup a): \((250\cdot4)\cdot13=1000\cdot13=13{,}000\). 2. Factor b): \(19(0.8+0.2)=19\cdot1=19\). 3. Rewrite c): \(45(10+1)=450+45=495\).

a) \((250\cdot4)\cdot13=13{,}000\) b) \(19(0.8+0.2)=19\) c) \(45(10+1)=495\)

- In a), try multiplying each part of the sum by the outside number. - In b), which fraction occurs twice as a factor? - In c), compare the work needed before and after distributing.

1. Distribute in a): \(24\cdot\frac{3}{8}+24\cdot\frac{1}{6}=9+4=13\). 2. Factor in b): \(\frac{5}{12}\left(\frac{4}{7}+\frac{3}{7}\right)=\frac{5}{12}\cdot1=\frac{5}{12}\). 3. For c), distributing directly gives \(21-10=11\).

a) \(24\cdot\frac{3}{8}+24\cdot\frac{1}{6}=13\) b) \(\frac{5}{12}\left(\frac{4}{7}+\frac{3}{7}\right)=\frac{5}{12}\) c) \(15\cdot\frac{7}{5}-15\cdot\frac{2}{3}=21-10=11\)

- How could you check the decimal addition in the first method? - In the second method, think of one rectangle split into two parts with the same height. - Why must the areas of the two parts add to the original area?

1. Add first: \(8(12.5+5)=8(17.5)=140\). 2. Distribute first: \(8\cdot12.5+8\cdot5=100+40=140\). 3. Both methods give the same value because the expressions are equivalent.

a) \(8\cdot17.5=140\) b) \(8\cdot12.5+8\cdot5=100+40=140\) Both results agree.

- In a), compare the sizes of \(10\) and \(2.5\) before combining results. - In b), which denominators divide \(18\) evenly? - In c), does multiplying a positive fraction by a positive number change its sign?

1. For a), \(\frac{4}{5}\cdot10-\frac{4}{5}\cdot2.5=8-2=6\). 2. For b), \(18\cdot\frac{1}{2}+18\cdot\frac{1}{3}-18\cdot\frac{1}{6}=9+6-3=12\). 3. For c), \(\frac{7}{11}\cdot22-\frac{7}{11}\cdot5.5=14-3.5=10.5\).

a) \(\frac{4}{5}\cdot10-\frac{4}{5}\cdot2.5=6\) b) \(18\cdot\frac{1}{2}+18\cdot\frac{1}{3}-18\cdot\frac{1}{6}=12\) c) \(\frac{7}{11}\cdot22-\frac{7}{11}\cdot5.5=10.5\)

- For a), which nearby power of ten makes the calculation easier? - For b), what factor appears in both terms? - Write the equivalent expression before its value.

1. For a), write \(101=100+1\): \(12(100+1)=1200+12=1212\). 2. For b), factor out \(\frac{5}{9}\): \(\frac{5}{9}(14-5)=\frac{5}{9}\cdot9=5\).

a) \(12(100+1)=1212\) b) \(\frac{5}{9}(14-5)=5\)

- Which numbers change between the two products, and which stay the same? - Compare the signs of the products in c) before finding their sum. - If your two methods disagree, check whether subtraction or a negative factor changed sign.

1. For a), direct calculation gives \(154+126=280\). Factoring gives \(14(11+9)=14\cdot20=280\). 2. For b), direct calculation gives \(52-39=13\). Factoring gives \(6.5(8-6)=6.5\cdot2=13\). 3. For c), direct calculation gives \(-6+(-9)=-15\). Factoring gives \((-1.5)(4+6)=-1.5\cdot10=-15\).

a) Direct: \(154+126=280\); factored: \(14(11+9)=280\) b) Direct: \(52-39=13\); factored: \(6.5(8-6)=13\) c) Direct: \(-6+(-9)=-15\); factored: \((-1.5)(4+6)=-15\)

- Combine the \(a\)-terms and the \(b\)-terms. - Look for a common factor in the resulting expression. - For part b), use the two blocks \(37\) and \(12\).

1. Add: \((100a+b)+(100b+a)=101a+101b\). 2. Factor out \(101\): \(101a+101b=101(a+b)\). 3. For \(3712\), \(a=37\) and \(b=12\). Then \(101(37+12)=101\cdot49=4949\).

a) \((100a+b)+(100b+a)=101(a+b)\) b) \(4949\)

- Distribute the subtraction sign to both terms of the second expression. - Combine like terms. - Look for a common factor in the difference.

1. Subtract: \((100a+b)-(100b+a)=99a-99b\). 2. Factor out \(99\): \(99a-99b=99(a-b)\). 3. For \(8634\), \(a=86\) and \(b=34\). Then \(99(86-34)=99\cdot52=5148\).

a) \((100a+b)-(100b+a)=99(a-b)\) b) \(5148\)

- Look for pairs of factors that make \(100\) or \(1000\). - Use place value to simplify the product with trailing zeros. - In c), how can you change both numbers in a quotient without changing its value?

1. Regroup a): \((4\cdot25)\cdot19=100\cdot19=1900\). 2. Use place value in b): \(8\cdot6\cdot1000=48{,}000\). 3. Scale both numbers in c): \(5400\div9=600\). 4. Regroup d): \((2\cdot50)\cdot37=100\cdot37=3700\).

a) \((4\cdot25)\cdot19=1900\) b) \(8\cdot6\cdot1000=48{,}000\) c) \(5400\div9=600\) d) \((2\cdot50)\cdot37=3700\)

- Which factors in a) and b) can be multiplied without using the middle factor first? - Which calculation is easier: multiply left to right or change the grouping? - In d), what is repeated in the two products?

1. For a), \((25\cdot4)\cdot17=100\cdot17=1700\). 2. For b), \((125\cdot8)\cdot11=1000\cdot11=11{,}000\). 3. For c), \((5\cdot20)\cdot42=100\cdot42=4200\). 4. For d), \(63(8+2)=63\cdot10=630\).

a) \((25\cdot4)\cdot17=1700\) b) \((125\cdot8)\cdot11=11{,}000\) c) \((5\cdot20)\cdot42=4200\) d) \(63(8+2)=630\)

- Translate each phrase “the product of” into multiplication. - Identify the factor that appears in both products. - Factor it out before adding the remaining factors.

1. The expression is \(47 \cdot 19 + 47 \cdot 81\). 2. Factor out the common factor \(47\): \(47(19 + 81)\). 3. Evaluate: \(47 \cdot 100 = 4700\).

\(47 \cdot 19 + 47 \cdot 81 = 47(19 + 81) = 4700\)

- Check whether the order of terms changes. - Check whether the grouping changes. - Check whether a common factor is distributed or factored out.

1. In a), the order of \(57\) and \(72\) changes, so the commutative property of addition is used. 2. In b), the factors are reordered and regrouped, so the commutative and associative properties of multiplication are used. 3. In c), the common factor \(9\) is factored out, so the distributive property is used in reverse.

a) Commutative property of addition b) Commutative and associative properties of multiplication c) Distributive property

- What happens when you remove the parentheses from the sum? - What is the sum of \(-1\) and \(1\)? - How many copies of \(n\) remain? - Substitute each given value into both the three-number sum and the simplified expression.

1. The sum is \((n-1)+n+(n+1)\). Combining like terms gives \(3n\). 2. The expression shows that the sum of three numbers spaced \(1\) unit apart is three times the middle number. 3. For \(n=15\), the numbers are \(14\), \(15\), and \(16\). Their sum is \(14+15+16=45\), and \(3\cdot15=45\). 4. For \(n=2.5\), the numbers are \(1.5\), \(2.5\), and \(3.5\). Their sum is \(1.5+2.5+3.5=7.5\), and \(3\cdot2.5=7.5\).

1. \(3n\) 2. The sum is three times the middle number. 3. Both methods give \(45\) when \(n=15\) and \(7.5\) when \(n=2.5\).

- How do you represent the integers immediately before and after \(m\)? - Try a numerical example with middle number \(10\). - What is the sum of \(-2\), \(-1\), \(1\), and \(2\)?

1. The two integers before \(m\) are \(m-2\) and \(m-1\), and the two integers after it are \(m+1\) and \(m+2\). 2. The sum is \((m-2)+(m-1)+m+(m+1)+(m+2)\). 3. Combining like terms gives \(5m\), because the constants sum to \(0\).

a) \(m-2\), \(m-1\), \(m\), \(m+1\), and \(m+2\) b) \(5m\)

- Write \(p\) as \(1p\). - Combine the coefficients of the same variable. - Interpret a multiplier greater than \(1\) as the original amount plus an increase.

1. Combine the coefficients of \(p\): \(p+0.20p=1.20p\). 2. The multiplier \(1.20\) means the new price is \(120\%\) of the original price, which is a \(20\%\) increase.

\(1.20p\). The multiplier \(1.20\) reveals that the new price is \(120\%\) of the original, or \(20\%\) more.

- Think of \(p\) as \(1p\). - Find the amount that must be subtracted from \(1\) to get \(0.80\). - Convert the decimal coefficient you find into a percent.

1. Since \(0.80=1-0.20\), \(0.80p=p-0.20p\). 2. Thus \(r=0.20\). 3. The term \(0.20p\) is \(20\%\) of the original price, so the expression reveals a \(20\%\) discount.

\(0.80p=p-0.20p\), so \(r=0.20\). The discount is \(20\%\).

- Identify which terms depend on the number of participants. - Combine like terms that have the same variable. - Interpret the constant separately from the coefficient of \(n\).

1. Combine the two per-participant terms: \(4n+6n=10n\). 2. The equivalent expression is \(18+10n\). 3. The constant \(18\) is the one-time registration fee, and the coefficient \(10\) is the combined materials-and-meal cost for each participant.

\(18+10n\). The \(18\) is the fixed fee, and \(10\) is the cost per participant.

- What does each charge pay for? - How can you express the fixed fee using the hourly price? - Check whether the rewritten expression changes what actually happened.

1. The bike was rented for \(h\) hours, and the \(\$14\) fee pays for equipment, not extra rental time. 2. The fee is equal in cost to two \(\$7\) hourly charges. That is why \(7h+14=7(h+2)\). 3. The parentheses show a way to calculate the total money owed, not the number of hours actually rented.

No. The rental lasted \(h\) hours. The \(\$14\) fee costs the same as two \(\$7\) hourly charges, so \(7(h+2)\) gives the total cost, not the rental time.

- Can you take a common factor from the terms inside parentheses? - Can you group factors so one multiplication is especially easy? - Compare the number of calculations in your two forms.

1. Factor \(4\) inside parentheses: \(25(4q+8)=25\cdot4(q+2)\). 2. Multiply \(25\cdot4=100\), giving \(100(q+2)\). 3. At \(q=9\), the value is \(100(11)=1100\). Multiplying by \(100\) is simpler mentally.

Useful form: \(100(q+2)\). Value at \(q=9\): \(1100\). This form is easier mentally because multiplying by \(100\) is simpler than multiplying by \(25\) and then distributing.

- Which parts of each expression are powers rather than products or sums? - In b), how many negative factors are multiplied together? - In c), which operations are enclosed by the square brackets?

1. For a), evaluate the powers to get \(32-36+(-4)=-8\). 2. For b), \((-2)^3=-8\) because a product of three negative factors is negative. Also \(3^2=9\) and \(10-15=-5\), so \(-8+9-(-5)=6\). 3. For c), \(10^2=100\) and \(2^3=8\), giving \(100-[50-(8\cdot5)]=100-[50-40]=90\).

a) \(32-36+(-4)=-8\). b) \(-8+9-(-5)=6\). The power \((-2)^3\) is negative because three negative factors are multiplied. c) \(100-[50-(8\cdot5)]=90\).

- Write the expression you get after distributing \(12\). - Compare \(12\) with the denominators in the two fractions. - Carry out both methods, then compare the number and complexity of the steps.

1. Method 1: Convert \(0.75\) to \(\frac{3}{4}\). The least common denominator of \(3\) and \(4\) is \(12\). 2. Add inside the parentheses: \(\frac{2}{3} + \frac{3}{4} = \frac{8}{12} + \frac{9}{12} = \frac{17}{12}\). 3. Multiply: \(\frac{17}{12} \cdot 12 = 17\). 4. Method 2: Distribute \(12\): \(\frac{2}{3} \cdot 12 + 0.75 \cdot 12\). 5. Evaluate each product: \(\frac{2}{3} \cdot 12 = 8\) and \(0.75 \cdot 12 = \frac{3}{4} \cdot 12 = 9\). 6. Add: \(8 + 9 = 17\). 7. Method 2 requires less computation because \(12\) is divisible by both denominators, so each product becomes a whole number immediately.

Both methods give \(17\). Applying the distributive property first is more efficient because multiplying by \(12\) eliminates the denominators before any fraction addition is needed.

- In a), what factor appears in both products? - In b), which two factors are easiest to multiply first? - In c), can you replace \(99\) with a nearby round number and adjust?

1. For a), factor out the common factor \(16\): \(16(12 + 8) = 16 \cdot 20 = 320\). 2. For b), use the commutative and associative properties to group \(25\) and \(4\): \((25 \cdot 4) \cdot 37 = 100 \cdot 37 = 3700\). 3. For c), rewrite \(99\) as \(100 - 1\) and distribute: \(45(100 - 1) = 45 \cdot 100 - 45 \cdot 1 = 4500 - 45 = 4455\).

a) \(16(12+8)=320\) b) \((25\cdot4)\cdot37=3700\) c) \(45(100-1)=4455\)

- Look for factors that can make \(100\) or a divisor that can be split into simpler factors. - Decompose a factor or dividend rather than using a standard algorithm. - Show the equivalent expression that supports your mental method.

1. a) \((25\cdot4)\cdot7=700\). 2. b) \((126\div9)\div2=7\). 3. c) \(-15(10+2)=-180\). 4. d) \(-(70\div7+35\div7)=-15\).

a) \((25\cdot4)\cdot7=700\) b) \((126\div9)\div2=7\) c) \(-15(10+2)=-180\) d) \(-(70\div7+35\div7)=-15\)

- Evaluate both expressions separately and compare them. - What is the base of each power? - What is the sign of the product of two negative numbers?

1. For \(A\), the exponent applies only to \(6\): \(-6^2=-(6^2)=-36\). Then \(-36\div4+5=-9+5=-4\). 2. For \(B\), the parentheses make \(-6\) the base: \((-6)^2=(-6)\cdot(-6)=36\). Then \(36\div4+5=9+5=14\). 3. The parentheses determine whether the negative sign is part of the base being squared.

a) \(A=-4\); \(B=14\) b) In \(-6^2\), only \(6\) is squared. In \((-6)^2\), the entire negative number is squared, giving a positive power value.

- In a), what happens when two negative amounts are added? - In b), the two decimal numbers have different signs. What happens to their sum? - Check that your rewritten expression still has both original terms.

1. For a), factor out \(\frac{4}{7}\): \(\frac{4}{7}(-13.2-0.8)=\frac{4}{7}(-14)=-8\). 2. For b), factor out \(\frac{2}{3}\): \(\frac{2}{3}(-15.5+6.5)=\frac{2}{3}(-9)=-6\).

a) \(\frac{4}{7}(-13.2-0.8)=-8\) b) \(\frac{2}{3}(-15.5+6.5)=-6\)

- Is the quantity inside the parentheses greater or less than \(1\)? - Think about how multiplying \(12\) by a fraction changes its size. - The middle term is subtracted; check that its contribution remains negative.

1. Distribute \(12\) to each term: \(12\cdot\frac{3}{4}-12\cdot0.5+12\cdot\frac{1}{6}\). 2. Evaluate the products: \(12\cdot\frac{3}{4}=9\), \(12\cdot0.5=6\), and \(12\cdot\frac{1}{6}=2\). 3. Combine the results: \(9-6+2=5\).

\(12\cdot\frac{3}{4}-12\cdot0.5+12\cdot\frac{1}{6}=9-6+2=5\)

- Check whether a fraction has a terminating decimal representation. - Look for factors that simplify before multiplication. - Use decimals when all values have simple terminating decimal forms.

1. For a), fractions are efficient because \(\frac{1}{3}\) is a repeating decimal. Write \(0.5 = \frac{1}{2}\). Then \(\frac{1}{3} + \frac{1}{2} = \frac{5}{6}\). 2. For b), fractions are efficient because \(0.25 = \frac{1}{4}\). Then \(\frac{1}{4} \cdot \frac{4}{7} = \frac{1}{7}\). 3. For c), decimals are efficient because every value has a terminating decimal form. Write \(\frac{1}{5} = 0.2\). Then \(0.4 \cdot 1.2 - 0.2 = 0.48 - 0.2 = 0.28\).

a) Fractions: \(\frac{5}{6}\) b) Fractions: \(\frac{1}{7}\) c) Decimals: \(0.28\)

- Which operations have priority over addition? - What does an exponent mean? - Check the sign of the value inside the parentheses.

1. The first line contains two errors: \(1.5+2.5\) was added before the multiplication, and \(3^2\) was incorrectly treated as \(3\cdot2\). 2. Evaluate the parentheses: \(4.4-6.4=-2\). 3. Evaluate the power: \(3^2=9\). 4. Multiply before adding or subtracting: \(1.5+2.5\cdot(-2)-9=1.5-5-9\). 5. Finish: \(1.5-5-9=-12.5\).

The errors are in the first shown step: addition was done before multiplication, and \(3^2\) was treated as \(3\cdot2\). The correct value is \(-12.5\).

- Notice the expression that appears repeatedly. - Combine the decimal values before evaluating the repeated fraction. - Temporarily replace the repeated expression with a variable.

1. The expression is a difference at its highest level because the final operation subtracts \(-\frac{3}{7}\). 2. Let \(a = -\frac{3}{7}\). The expression becomes \([12.5-a+7.5]+[2.5-a-22.5]-a\). 3. Combine the decimal terms: \((20-a)+(-20-a)-a\). 4. Simplify: \(20-20-a-a-a=-3a\). 5. Substitute \(a=-\frac{3}{7}\): \(-3\left(-\frac{3}{7}\right)=\frac{9}{7}=1\frac{2}{7}\).

It is a difference at its highest level. Its value is \(\frac{9}{7}\), or \(1\frac{2}{7}\).

- Reverse the steps from last to first. - Use the inverse operation for each instruction. - Track decimal place value carefully.

1. Work backward from \(25\). Undo the subtraction: \(25+6.8=31.8\). 2. Undo multiplication by \(4\): \(31.8\div4=7.95\). 3. Undo addition of \(4.2\): \(7.95-4.2=3.75\). 4. Undo multiplication by \(0.5\): \(3.75\div0.5=7.5\).

\(7.5\)

- In a), the same divisor appears twice. What single quotient could replace the difference? - In b), each product shares a factor. - Display the factored form before evaluating.

1. In a), both parts divide by \(9\): \((720-180)\div9=540\div9=60\). 2. In b), factor out \(1.2\): \(1.2(15+5-10)=1.2\cdot10=12\).

a) \((720-180)\div9=60\) b) \(1.2(15+5-10)=12\)

- Compare the two numerical results before making a general claim. - Expand and collect the variable terms. - Which part of the simplified form changes when \(x\) increases?

1. At \(x=5\), the value is \(3(5+4)-2\cdot5=27-10=17\). 2. At \(x=10\), the value is \(3(10+4)-2\cdot10=42-20=22\). 3. Expand: \(3(x+4)-2x=3x+12-2x=x+12\). 4. Replacing \(x\) by \(x+5\) gives \(x+5+12\), which is \(5\) greater than \(x+12\) for every starting \(x\).

a) \(17\) and \(22\) b) \(x+12\) c) The result rises by \(5\) because the rewritten form is \(x+12\).

- Find the least common denominator of \(12\) and \(16\). - When distributing, simplify each product before subtracting. - Compare how much fraction arithmetic each method requires.

1. For a), the least common denominator is \(48\): \(\frac{7}{12} - \frac{5}{16} = \frac{28}{48} - \frac{15}{48} = \frac{13}{48}\). Then \(48 \cdot \frac{13}{48} = 13\). 2. For b), distribute: \(48 \cdot \frac{7}{12} - 48 \cdot \frac{5}{16} = 28 - 15 = 13\). 3. Both methods give \(13\).

a) \(13\) b) \(13\)

- Mentally preview both methods before choosing. - Distributing can avoid adding fractions with unlike denominators. - Evaluating parentheses first can be best when the result is \(10\) or \(100\).

1. For a), distributing is efficient because multiplying by \(100\) is easy: \(12 \cdot 100 + 12 \cdot 3 = 1200 + 36 = 1236\). 2. For b), distributing avoids adding unlike fractions first: \(15 \cdot \frac{1}{3} + 15 \cdot \frac{4}{5} = 5 + 12 = 17\). 3. For c), evaluate the parentheses first because \(13 - 3 = 10\): \(6.4 \cdot 10 = 64\).

a) \(1236\); distributing is efficient. b) \(17\); distributing is efficient. c) \(64\); evaluating the parentheses first is efficient.

- For a), subtract inside the parentheses before multiplying. - For b), multiply \(-8\) by both terms and keep track of the signs. - Compare the intermediate calculations, not only the number of steps.

1. For a), \(12.5 - 5 = 7.5\), so \((-8)(7.5) = -60\). 2. For b), distribute: \((-8)(12.5) - (-8)(5) = -100 - (-40) = -60\). 3. Both methods are valid. One reasonable choice is the distributive method because \((-8)(12.5) = -100\) creates an easy intermediate value.

a) \(-60\) b) \(-60\) c) Answers may vary. For example, distributing may be easier because it creates the product \(-100\).

- In a), which two numbers make an easy sum? - In c), think about multiplying a fraction by the difference between two whole numbers. - In d), is the shared number positive or negative? How does that affect the result?

1. For a), factor out \(27\): \(27(13+87)=2700\). 2. For b), factor out \(0.4\): \(0.4(35-15)=8\). 3. For c), factor out \(\frac{3}{8}\): \(\frac{3}{8}(17-9)=3\). 4. For d), factor out \(0.25\): \(0.25(-12-8)=-5\).

a) \(27(13+87)=2700\) b) \(0.4(35-15)=8\) c) \(\frac{3}{8}(17-9)=3\) d) \(0.25(-12-8)=-5\)

- Add all fixed time amounts first. - Increasing the total time used for other activities decreases the free time. - Write a new expression with the changed values, then compare it with the original expression.

1. The original free time is \(F=24-(8.5+6.5+1.5+t_H)\). 2. Combine the fixed times: \(8.5+6.5+1.5=16.5\). Thus, \(F=24-16.5-t_H=7.5-t_H\). 3. With the changes, \(F_{\text{new}}=24-(8.0+7.5+1.5+t_H)\). 4. Since \(8.0+7.5+1.5=17.0\), \(F_{\text{new}}=7.0-t_H\). This is \(0.5\) hour less than the original expression.

a) \(F=7.5-t_H\) b) The free time decreases by \(0.5\) hour because the fixed-time total increases from \(16.5\) hours to \(17.0\) hours.

- Carry out the subtraction inside the parentheses for one method. - For the other method, multiply \(-8\) by both terms and keep the subtraction sign. - Compare the kinds of intermediate values each method creates.

1. Lucas’s method: \(\frac{1}{4} - \frac{3}{8} = \frac{2}{8} - \frac{3}{8} = -\frac{1}{8}\). 2. Then \(-8\left(-\frac{1}{8}\right) = 1\). 3. Sarah’s method: \(-8 \cdot \frac{1}{4} - \left(-8 \cdot \frac{3}{8}\right) = -2 - (-3) = 1\). 4. Both methods are efficient. Distributing may be more convenient because \(-8\) cancels with both denominators and produces whole-number products.

Both methods give \(1\). Distributing is especially efficient because the factor \(-8\) cancels with both denominators.

- What changes when you move one place left or right in a row? - What changes when you move up or down one row? - Which numbers cancel when you add the five expressions?

1. The left and right neighbors are \(x-1\) and \(x+1\). The numbers above and below are \(x-10\) and \(x+10\). 2. Adding the cross gives \(x+(x-1)+(x+1)+(x-10)+(x+10)\). 3. The constants cancel in opposite pairs, leaving \(5x\). 4. Therefore the sum depends only on the center number, and is always five times it.

a) Above: \(x-10\); below: \(x+10\); left: \(x-1\); right: \(x+1\). b) \(x+(x-10)+(x+10)+(x-1)+(x+1)=5x\).

- Write an expression that follows the operations in order. - Can you divide each term in the numerator by \(2\)? - Once you know how the final result compares with the starting number, reverse that change.

1. The final result is \(\frac{2x+10}{2}\). 2. Dividing both terms in the numerator by \(2\) gives \(x+5\). 3. For a final result of \(42\), solve \(x+5=42\). Subtracting \(5\) gives \(x=37\). 4. Since the final result is always \(5\) more than the starting number, the performer subtracts \(5\) from the reported result.

a) \(\frac{2x+10}{2}=x+5\) b) \(37\) c) Subtract \(5\) from the final result.

- Look for combinations that make easy whole numbers. - Decide whether regrouping or evaluating parentheses first is more efficient. - Simplify fractions before multiplying when possible.

1. For a), \(18-50=-32\), so \(42+(-32)=10\). 2. For b), \(2.3+1.7=4\). Then regroup as \(6.7+3.3-4=10-4=6\). 3. For c), \(\frac{17}{8}-\frac{5}{8}=\frac{12}{8}=\frac{3}{2}\). Then \(\frac{4}{9}\cdot\frac{3}{2}=\frac{2}{3}\).

a) \(10\) b) \(6\) c) \(\frac{2}{3}\)

- Represent both four-digit numbers using their two-digit blocks. - Add the expressions and look for a common factor. - Use the total sum to recover the sum of the two blocks before finding \(b\).

1. The two numbers are \(100a+b\) and \(100b+a\). 2. Add and regroup: \((100a+b)+(100b+a)=101a+101b=101(a+b)\). 3. From \(101(a+b)=9191\), divide by \(101\): \(a+b=91\). 4. Since \(a=50\), \(b=41\), and the original number is \(100\cdot50+41=5041\).

The sum is \(101(a+b)\), so it is divisible by \(101\). Thus \(b=41\), and the original number is \(5041\).

- Break the expression into parenthetical values. - Decide whether the changed parentheses are added or subtracted. - Increasing a subtracted quantity decreases the total by the same amount.

1. An estimate is \(10-100-(-20)=-70\). 2. Exactly, \(38+62=100\) and \(24-44=-20\). 3. Therefore, \(12-100-(-20)=-68\). 4. Replacing \(38\) with \(48\) increases the first parenthetical value by \(10\). Because that value is subtracted, the entire expression decreases by \(10\).

a) Estimate: \(\approx -70\); exact value: \(-68\) b) The value decreases by \(10\).

- Determine the role of \(42\) inside its parentheses. - What happens to a difference when the number being subtracted becomes smaller? - Then account for the subtraction sign before the parentheses.

1. In a), \(92-42=50\) and \(16+34=50\). 2. Therefore, \(25-50-50=-75\). 3. Replacing \(42\) with \(22\) increases the first parenthetical value by \(20\), because less is subtracted from \(92\). 4. Since that parenthetical value is subtracted from the total, the final value decreases by \(20\).

a) \(-75\) b) The value decreases by \(20\).

- Pay attention to the subtraction sign before the second parentheses. - Determine how changing \(12\) changes the value inside the parentheses. - Subtracting a quantity that is \(10\) greater makes the total \(10\) less.

1. In a), \(17+33=50\) and \(12-42=-30\). 2. Therefore, \(-10-50-(-30)=-30\). 3. Replacing \(12\) with \(22\) increases the second parenthetical value by \(10\). 4. Since that parenthetical value is subtracted, the total decreases by \(10\).

a) \(-30\) b) The value decreases by \(10\).

- Compare the sums of the first and last terms and of the second and next-to-last terms. - Divide the number of terms by \(2\) to find the number of pairs. - Multiply the number of pairs by the sum of each pair.

1. The commutative property allows the terms to be reordered, and the associative property allows them to be regrouped. 2. There are \(80 \div 2 = 40\) pairs. 3. Each pair has a sum of \(81\), since \(1 + 80 = 81\), \(2 + 79 = 81\), and so on. 4. Therefore, the total is \(40 \cdot 81 = 3240\).

1) Commutative property and associative property 2) \(40\) pairs 3) \(81\) 4) \(3240\)

- Write the first few and last few terms of the sum. - Pair terms from opposite ends of the list. - Determine how many pairs are formed and the sum of each pair.

1. The total is represented by \(1 + 2 + 3 + \dots + 40\). 2. Pair the first and last terms: \(1 + 40 = 41\), \(2 + 39 = 41\), and so on. Since \(40 \div 2 = 20\), there are \(20\) pairs. 3. Each pair has a sum of \(41\), so the total is \(20 \cdot 41 = 820\).

1) \(1 + 2 + 3 + \dots + 40\) 2) \(820\) apples 3) There are \(20\) pairs, and each pair has a sum of \(41\).

- Divide \(60\) by \(2\) to count the even numbers. - Pair the smallest term with the largest term. - Multiply the number of pairs by the sum of each pair.

1. There are \(60 \div 2 = 30\) even numbers from \(2\) through \(60\). 2. Pair terms from opposite ends: \(2 + 60 = 62\), \(4 + 58 = 62\), and so on. The \(30\) terms form \(30 \div 2 = 15\) pairs. 3. The total is \(15 \cdot 62 = 930\).

1) \(30\) terms 2) \(15\) pairs; each pair has a sum of \(62\) 3) \(930\)

- Identify the factor that appears in both products. - Factor that number out of the sum. - Add the remaining factors before multiplying.

1. Both products have a common factor of \(14\). 2. Factor out \(14\): \(14 \cdot (63 + 37)\). 3. Evaluate: \(14 \cdot 100 = 1400\).

\(14 \cdot 63 + 14 \cdot 37 = 14 \cdot (63 + 37) = 1400\)

- Count how many times each addend appears. - Write each repeated addend as a product. - Look for a common factor in the two products.

1. The number \(18\) appears five times, and \(12\) appears five times, so the expression is \(5 \cdot 18 + 5 \cdot 12\). 2. Factor out the common factor \(5\): \(5(18 + 12)\). 3. Evaluate: \(5 \cdot 30 = 150\).

\(5 \cdot 18 + 5 \cdot 12 = 5(18 + 12) = 150\)

- Would the products be easier if you multiplied the factors in a different order? - Look for a short multiplication you already know among the factors in b). - How could you check that a rearrangement has kept the value the same?

1. For a), \((4\cdot5\cdot5)\cdot19=100\cdot19=1900\). 2. For b), \((8\cdot125)\cdot15=1000\cdot15=15{,}000\). 3. For c), \(((2\cdot5)\cdot10)\cdot13=100\cdot13=1300\). 4. For d), \((250\cdot4)\cdot7=1000\cdot7=7000\).

a) \((4\cdot5\cdot5)\cdot19=1900\) b) \((8\cdot125)\cdot15=15{,}000\) c) \(((2\cdot5)\cdot10)\cdot13=1300\) d) \((250\cdot4)\cdot7=7000\)

- For a), imagine the total as eight groups made from two smaller groups. - For b), compare the two factors beside \(17\). - In d), which pair of numbers can be compared before multiplying by \(15\)?

1. Expand a): \(8\cdot100+8\cdot4=800+32=832\). 2. Factor b): \(17(12+8)=17\cdot20=340\). 3. Expand c): \(50\cdot6-2\cdot6=300-12=288\). 4. Factor d): \((44-34)15=10\cdot15=150\).

a) Expanded: \(8\cdot100+8\cdot4=832\) b) Factored: \(17(12+8)=340\) c) Expanded: \(50\cdot6-2\cdot6=288\) d) Factored: \((44-34)15=150\)

- In a), are both parts multiples of \(8\)? - What calculation can check either quotient without repeating the division? - In b), a subtraction is involved. How do you check the sign of the smaller part?

1. For a), Method 1: \(456\div8=57\); Method 2: \(400\div8+56\div8=50+7=57\). 2. For b), Method 1: \(558\div6=93\); Method 2: \(600\div6-42\div6=100-7=93\). 3. Method 2 is often easier because the numbers in its smaller quotients are convenient multiples of the divisor. Either justified preference is acceptable.

a) Method 1: \(456\div8=57\); Method 2: \(50+7=57\). For example, Method 2 is easier because both parts divide evenly by \(8\). b) Method 1: \(558\div6=93\); Method 2: \(100-7=93\). For example, Method 2 is easier because both parts divide evenly by \(6\). A different method preference is acceptable if supported by a comparison of the work.

- Compare the divisions needed in the two methods. - Can both methods give the same answer while differing in how easy the steps are? - What would have to be true for the student's claim to hold for every example?

1. Subtract first: \((1100-55)\div11=1045\div11=95\). 2. Divide first: \(1100\div11-55\div11=100-5=95\). 3. The claim is false. Here, dividing each term first gives two easy quotients, \(100\) and \(5\), instead of \(1045\div11\).

Subtract first: \(1045\div11=95\). Divide first: \(100-5=95\). No. Dividing first gives \(1100\div11-55\div11=100-5=95\), which is easier here than \(1045\div11\). This disproves “always.”

- For a), compare the two numbers multiplied by \(13\). - For b), how close is \(99\) to a number that is easy to multiply by \(14\)? - For c), does \(62\) times \(11\) seem closer to ten groups or twenty groups of \(62\)?

1. Factor a): \(13(57+43)=13\cdot100=1300\). 2. Rewrite b): \((100-1)14=1400-14=1386\). 3. Rewrite c): \(62(10+1)=620+62=682\).

a) \(13(57+43)=1300\) b) \((100-1)14=1386\) c) \(62(10+1)=682\)

- For a), which nearby multiples of \(5\) are easy to divide? - In b), is the result positive, and should it be smaller than \(144\div12\)? - In c), could one check the final answer by multiplying by \(11\)?

1. Split a): \(200\div5+45\div5=40+9=49\). 2. Divide b) by parts: \(144\div12-36\div12=12-3=9\). 3. Combine c): \((132+88)\div11=220\div11=20\).

a) \(200\div5+45\div5=49\) b) \(144\div12-36\div12=9\) c) \((132+88)\div11=20\)

- Translate each phrase “the product of” into multiplication with grouping symbols. - Look for factors that multiply to \(1000\). - Use the commutative and associative properties to reorder and regroup.

1. The expression is \((125 \cdot 9)(8 \cdot 3)\). 2. Reorder and regroup the factors: \((125 \cdot 8)(9 \cdot 3)\). 3. Evaluate: \(1000 \cdot 27 = 27{,}000\).

\((125 \cdot 9)(8 \cdot 3) = (125 \cdot 8)(9 \cdot 3) = 27{,}000\)

- Pay attention to which quotient is being subtracted from which. - Notice that both quotients have the same divisor. - Subtract the dividends before dividing.

1. The expression is \(1024 \div 8 - 224 \div 8\). 2. Since both quotients have the same divisor, combine the dividends: \((1024 - 224) \div 8\). 3. Evaluate: \(800 \div 8 = 100\).

\(1024 \div 8 - 224 \div 8 = (1024 - 224) \div 8 = 100\)

- Look for a factor close to a multiple of \(10\) or \(100\). - Rewrite that factor as a sum or difference. - Distribute the other factor to both terms.

1. For a), write \(102\) as \(100 + 2\): \(7(100 + 2) = 700 + 14 = 714\). 2. For b), write \(98\) as \(100 - 2\): \(9(100 - 2) = 900 - 18 = 882\). 3. For c), write \(21\) as \(20 + 1\): \(15(20 + 1) = 300 + 15 = 315\).

a) \(7(100 + 2) = 714\) b) \(9(100 - 2) = 882\) c) \(15(20 + 1) = 315\)

- Identify the factor shared by both terms. - Place the remaining factors inside grouping symbols. - Add or subtract inside the grouping symbols before multiplying.

1. For a), factor out \(8\): \((14 + 6)8 = 20 \cdot 8 = 160\). 2. For b), factor out \(12\): \((37 - 27)12 = 10 \cdot 12 = 120\). 3. For c), factor out \(5\): \(5(43 + 57) = 5 \cdot 100 = 500\).

a) \(8(14 + 6) = 160\) b) \(12(37 - 27) = 120\) c) \(5(43 + 57) = 500\)

- Compare each expanded expression with its factored form. - Identify the common factor or divisor. - In part d, determine which number added to \(8\) gives \(20\).

1. For a), factor out \(4\): \((36 + 14) \cdot 4\). Thus, \(\square = 14\) and \(\triangle = 4\). 2. For b), compare with \((25 - 15) \cdot 7\). Thus, \(\triangle = 25\). 3. For c), both quotients have divisor \(8\), so \(\square = 8\). 4. For d), the factors inside the implied sum must satisfy \(\square + 8 = 20\), so \(\square = 12\).

a) \(\square = 14\); \(\triangle = 4\) b) \(\triangle = 25\) c) \(\square = 8\) d) \(\square = 12\)

- Combine the two multiplication steps. - Then account for the division by \(2\). - Determine how many times the original number appears in the final result.

1. Combine the operations: multiplying by \(4\), then by \(5\), and dividing by \(2\) is equivalent to multiplying by \((4 \cdot 5) \div 2 = 10\). 2. If the original number is \(x\), then \(10x = 70\). 3. Divide by \(10\): \(x = 7\).

\(7\)

- Carry out the steps for \(6\). - Combine the constant factors \(25\), \(8\), and \(100\). - Represent the starting number with a variable.

1. Starting with \(6\): \(6 \cdot 25 \cdot 8 \div 100 = 12\). Half of \(12\) is \(6\), so the claim works in this case. 2. For any starting number \(n\), the steps give \(n \cdot 25 \cdot 8 \div 100\). 3. Since \(25 \cdot 8 = 200\) and \(200 \div 100 = 2\), the final result is \(2n\). Taking half returns \(n\).

a) The final result is \(12\), and half of \(12\) is \(6\). b) The operations are equivalent to multiplying the starting number by \(2\), so halving the result recovers the starting number.

- Carry out every step for each test number. - Represent the starting number with a variable. - Distribute the division by \(3\) across both terms in the numerator.

1. For \(4\): \((3 \cdot 4 + 15) \div 3 - 5 = 9 - 5 = 4\). 2. For \(10\): \((3 \cdot 10 + 15) \div 3 - 5 = 15 - 5 = 10\). 3. For \(25\): \((3 \cdot 25 + 15) \div 3 - 5 = 30 - 5 = 25\). 4. For any starting number \(n\), \((3n + 15) \div 3 - 5 = n + 5 - 5 = n\). The process always returns the starting number.

The final result always equals the starting number because \((3n + 15) \div 3 - 5 = n\).

- Rewrite a factor as a convenient sum or difference. - In part c, identify the common factor. - Choose a rewrite that creates multiples of \(10\) or \(100\).

1. For a), write \(104\) as \(100 + 4\): \(7(100 + 4) = 700 + 28 = 728\). 2. For b), write \(9\) as \(10 - 1\): \(18(10 - 1) = 180 - 18 = 162\). 3. For c), factor out \(35\): \(35(16 + 4) = 35 \cdot 20 = 700\).

a) \(7(100 + 4) = 728\) b) \(18(10 - 1) = 162\) c) \(35(16 + 4) = 700\)

- Write \(14\) as both a sum and a product. - Apply the named property in each method. - Compare the number and difficulty of the steps.

1. Distributive method: \((10 + 4) \cdot 5 = 10 \cdot 5 + 4 \cdot 5 = 50 + 20 = 70\). 2. Associative method: \((7 \cdot 2) \cdot 5 = 7 \cdot (2 \cdot 5) = 7 \cdot 10 = 70\). 3. Both methods give \(70\). A reasonable comparison is that the second method is shorter because it creates the factor \(10\).

1) \((10 + 4) \cdot 5 = 70\) 2) \(7 \cdot (2 \cdot 5) = 70\) 3) The associative method is easier because it creates \(2 \cdot 5 = 10\) and uses fewer steps.

- Expand the sum in Strategy A. - Regroup the factors in Strategy B. - Match each change to the property that permits it.

1. Strategy A: \(25(10 + 2) = 250 + 50 = 300\). This uses the distributive property. 2. Strategy B: \(25(4 \cdot 3) = (25 \cdot 4) \cdot 3 = 100 \cdot 3 = 300\). This uses the associative property.

a) Strategy A: \(25(10 + 2) = 300\). Strategy B: \((25 \cdot 4) \cdot 3 = 300\). b) Strategy A uses the distributive property; Strategy B uses the associative property.

- A factor outside grouping symbols must multiply every term inside. - Identify the missing product in Sarah’s expansion. - Add the two correct partial products.

1. Sarah multiplied \(18\) by \(100\) but did not multiply \(18\) by \(2\). 2. The correct expansion is \(18(100 + 2) = 18 \cdot 100 + 18 \cdot 2\). 3. Therefore, the product is \(1800 + 36 = 1836\).

a) Sarah failed to distribute \(18\) to the second term. b) \(18(100 + 2) = 1800 + 36 = 1836\)

- Compare the structure on the two sides of the first equation. - Look for a decomposition of \(105\) that makes mental multiplication convenient. - Check the rewritten expression by evaluating it.

1. Each addend inside the grouping symbols is multiplied by \(6\), so the missing number is \(7\). 2. Write \(105\) as \(100+5\): \(8(100+5)=8\cdot100+8\cdot5=800+40=840\).

a) \(7\) b) \(8(100+5)=800+40=840\)

- Substitute \(s=5\) into each expression. - Apply the distributive property to Expression 2. - Count the noncorner squares along the four sides, then count the corners.

1. Using Expression 1, \(4\cdot5+4=24\). 2. Using Expression 2, \(4(5+1)=24\). 3. Apply the distributive property: \(4(s+1)=4s+4\). Therefore, the expressions are equivalent. 4. The term \(4s\) counts \(s\) border squares along each of the four sides, excluding the corners. The additional \(4\) counts the four corner squares.

a) Both expressions give \(24\) border squares. b) Yes. \(4(s+1)=4s+4\). c) \(4s\) counts the noncorner border squares along the four sides, and \(+4\) counts the four corners.

- First find each boat's speed relative to the shore. - Speeds add when objects move directly toward each other. - Remove parentheses and combine like terms. - Identify which variables remain after simplification.

1. Boat A's shore speed is \(v_A+v_s\). Boat B's shore speed is \(v_B-v_s\). 2. Because they move toward each other, add their shore speeds: \(v_c=(v_A+v_s)+(v_B-v_s)\). 3. Simplify: \(v_c=v_A+v_s+v_B-v_s=v_A+v_B\). 4. The current cancels from the closing-speed expression. Therefore, under this model, the meeting time is \(t=\frac{d}{v_A+v_B}\) and does not depend on \(v_s\).

a) Boat A: \(v_A+v_s\); Boat B: \(v_B-v_s\) b) \(v_c=v_A+v_B\) c) The current speed does not affect the meeting time because its effects cancel in the closing speed.

- Start with the perimeter formula for a rectangle. - Replace the original side lengths with \(a+x\) and \(b-x\). - Simplify the expression inside the parentheses first. - Observe what happens to the \(x\)-terms.

1. The original perimeter is \(P_{\text{old}}=2(a+b)\). 2. The new side lengths are \(a+x\) and \(b-x\). Thus, \(P_{\text{new}}=2[(a+x)+(b-x)]\). 3. Inside the brackets, \(a+x+b-x=a+b\), so \(P_{\text{new}}=2(a+b)=P_{\text{old}}\). The perimeter is unchanged. 4. \(P_{\text{old}}=2\cdot(40+25)=130\,\text{m}\). The new sides are \(45\,\text{m}\) and \(20\,\text{m}\), so \(P_{\text{new}}=2\cdot(45+20)=130\,\text{m}\).

1. \(P_{\text{old}}=2(a+b)\) 2. \(P_{\text{new}}=2(a+b)\) 3. The perimeters are equal. 4. Both perimeters are \(130\,\text{m}\).

- List all six expressions before forming the two sums. - Use parentheses around each entire sum when subtracting. - After simplifying, check whether the variable remains.

1. The six numbers are \(n+5\), \(n+6\), \(n+7\), \(n+8\), \(n+9\), and \(n+10\). 2. The sum of the three smallest is \((n+5)+(n+6)+(n+7)=3n+18\). 3. The sum of the three largest is \((n+8)+(n+9)+(n+10)=3n+27\). 4. Their difference is \((3n+27)-(3n+18)=9\). 5. The result is constant; it does not depend on \(n\).

a) \(n+5\), \(n+6\), \(n+7\), \(n+8\), \(n+9\), and \(n+10\) b) Smallest three: \(3n+18\); largest three: \(3n+27\) c) \(9\). The difference is the same for every value of \(n\).

- Express the downstream and upstream speeds separately. - Subtract the entire expression for upstream speed. - What does the simplified difference depend on?

1. Downstream speed is \(v+c\), upstream speed is \(v-c\), so the difference is \((v+c)-(v-c)\). 2. Simplify: \(v+c-v+c=2c\). The difference depends on the current speed, not on \(v\). 3. Downstream is \(26\,\text{mi/h}\), upstream is \(18\,\text{mi/h}\), and the difference is \(8\,\text{mi/h}\), equal to \(2\cdot4\).

1) \((v+c)-(v-c)\) 2) \(2c\), twice the current's speed. The difference does not depend on the boat's speed \(v\) in still water. 3) \(8\,\text{mi/h}\)

- A tailwind increases ground speed, while a headwind decreases it. - Add the two speed expressions and look for cancellation. - Use the average before finding the wind speed.

1. With a tailwind, \(v_T=v+w\). With a headwind, \(v_H=v-w\). 2. Their average is \(\frac{(v+w)+(v-w)}{2}=\frac{2v}{2}=v\), the still-air speed. 3. The still-air speed is \(\frac{520+440}{2}=480\,\text{mi/h}\). 4. The wind speed is \(520-480=40\,\text{mi/h}\).

a) \(v_T=v+w\); \(v_H=v-w\). b) \(v\), the still-air speed. c) \(v=480\,\text{mi/h}\), \(w=40\,\text{mi/h}\).

- Determine how many long and short strips are needed for all four frames. - Work from the leaves toward the root of each tree. - One tree calculates one frame first; the other calculates each strip type for all frames. - Which property distributes multiplication across a sum?

1. Tree a) finds the wood needed for one frame: \(2\cdot1.2=2.4\) meters of long strips and \(2\cdot0.85=1.7\) meters of short strips. It adds these to get \(4.1\) meters per frame, then multiplies by \(4\). 2. Tree a) gives \(4(2.4+1.7)=4\cdot4.1=16.4\,\text{m}\). 3. Tree b) uses \(8\) long strips and \(8\) short strips for all four frames. The partial lengths are \(8\cdot1.2=9.6\,\text{m}\) and \(8\cdot0.85=6.8\,\text{m}\). 4. Tree b) gives \(9.6+6.8=16.4\,\text{m}\). 5. The distributive property shows the methods are equivalent. The expressions are \(4(2\cdot1.2+2\cdot0.85)\) and \(8\cdot1.2+8\cdot0.85\).

a) Tree a) finds the wood for one frame and then multiplies by \(4\). Tree b) finds the total lengths of all long and all short strips separately, then adds them. b) Tree a) has missing values \(2.4\), \(1.7\), \(4.1\), and \(16.4\). Tree b) has missing values \(9.6\), \(6.8\), and \(16.4\). The total length is \(16.4\,\text{m}\). c) Distributive property; \(4(2\cdot1.2+2\cdot0.85)\) and \(8\cdot1.2+8\cdot0.85\).

- Look for the factor shared by both terms. - Ask what \(h+k\) represents in the situation. - Connect the shared numerical factor to the unit rate.

1. Factor the common rate: \(2.4h+2.4k=2.4(h+k)\). 2. The quantity \(h+k\) is the total irrigation time over the two days. 3. The factor \(2.4\) is the water-use rate in gallons per hour.

\(2.4(h+k)\). The factor \(2.4\) is the gallons-per-hour rate, and \(h+k\) is the total irrigation time.

- Evaluate the expression inside the parentheses first. - For part 2, find how much the first number changes. - Track how that change is affected by the division that follows.

1. Write \(\frac{3}{4}=0.75\). Then \(1.25-0.75=0.5\), \(0.5\div0.5=1\), and \(1+0.1=1.1\). 2. Replacing \(1.25\) with \(1.5\) increases the value inside the parentheses by \(0.25\). Dividing that increase by \(0.5\) doubles it, so the final value increases by \(0.5\). The new value is \(1.6\).

1. \(1.1\) 2. The value increases by \(0.5\), from \(1.1\) to \(1.6\).

- Evaluate each numerical expression carefully. - What happens to the signs inside parentheses when a minus sign is in front? - Use \(x\) for the first number and \(y\) for the second, then combine like terms.

1. Expression A is \((15.2 + 6.4) - (15.2 - 6.4)\). Its value is \(21.6 - 8.8 = 12.8\). 2. Expression B is \(2 \cdot 6.4 = 12.8\). 3. In general, \((x + y) - (x - y) = x + y - x + y = 2y\). 4. Therefore, subtracting the difference from the sum always gives twice the second number.

a) Expression A: \((15.2 + 6.4) - (15.2 - 6.4) = 12.8\); Expression B: \(2 \cdot 6.4 = 12.8\) b) \((x + y) - (x - y) = 2y\), so the result depends only on the second number and equals twice that number.

- Evaluate both parenthetical expressions first. - Determine how far the original value is from \(20\). - Increasing a value in an added group increases the total, while decreasing a value in a subtracted group increases the total.

1. In a), \(12-45=-33\) and \(-18+5=-13\), so \(36-33-(-13)=16\). 2. To increase the total from \(16\) to \(20\), increase the first parenthetical value by \(4\). Replace \(12\) with \(16\). 3. The second parentheses are subtracted. To increase the total by \(4\), decrease that parenthetical value by \(4\). Replace \(-18\) with \(-22\).

a) \(16\) b) Replace \(12\) with \(16\). c) Replace \(-18\) with \(-22\).

- Evaluate the left and right sides separately. - Note whether the grouped expression is being divided or is doing the dividing. - One counterexample disproves an “always” claim.

1. In Example A, the left side is \(120 \div 10 = 12\), and the right side is \(8 + 4 = 12\). This equality is true. 2. In Example B, the left side is \(120 \div 6 = 20\), and the right side is \(30 + 60 = 90\). This equality is false. 3. Division can be distributed over a sum or difference in the dividend, but not over a sum or difference in the divisor.

Paul is incorrect. Example A is true, but Example B is false because \(20 \ne 90\). Division can distribute over a sum or difference in the dividend, not in the divisor.

- Write one expression for the original rule. - Simplify to see why the result does not depend on \(x\). - After doubling, compare \(2x+10\) with the desired result \(x\). - What terms must be removed so that only \(x\) remains?

1. The expression for the claim is \(2(x+5)-2x\). 2. Distributing and combining like terms gives \(2x+10-2x=10\), so the claim is true. 3. After doubling, the expression is \(2x+10\). To leave only \(x\), subtract \(10\) and subtract one copy of \(x\). 4. One possible revised ending is: subtract \(10\), then subtract the original number. Algebraically, \(2x+10-10-x=x\).

a) \(2(x+5)-2x=10\), so the claim is true. b) One possible rule is: after doubling, subtract \(10\), then subtract the original number. This gives \(2x+10-10-x=x\).

- Interpret \(0.75\) as the fraction of the original bill the club still pays. - Distribute the multiplier to compare the two forms. - For the change in \(n\), look for the form that makes the per-hiker contribution visible.

1. The form \(0.75(120+8n)\) shows that the club pays \(75\%\) of the entire original bill after the sponsor pays \(25\%\). 2. Distribute \(0.75\): \(0.75\cdot120+0.75\cdot8n=90+6n\), so the forms are equivalent. 3. The form \(90+6n\) shows a discounted fixed cost of \(\$90\) and a discounted cost of \(\$6\) per hiker. 4. Five additional hikers increase the club's share by \(6\cdot5=\$30\).

a) \(0.75(120+8n)\) reveals the \(25\%\) sponsorship of the whole bill; \(90+6n\) reveals a \(\$90\) fixed share plus \(\$6\) per hiker. b) \(0.75(120+8n)=90+6n\) c) The club's share increases by \(\$30\).