Free Math Worksheets

- 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) \(5\)

- For Expression A, evaluate the difference inside the parentheses first. - For Expression B, evaluate the two products first. - Which property relates multiplying a difference to subtracting two products?

1. For Expression A, subtract inside the parentheses: \(\frac{1}{4} - \frac{1}{3} = \frac{3}{12} - \frac{4}{12} = -\frac{1}{12}\). Then \(12\left(-\frac{1}{12}\right) = -1\). 2. For Expression B, evaluate each product: \(12 \cdot \frac{1}{4} = 3\) and \(12 \cdot \frac{1}{3} = 4\). Then \(3 - 4 = -1\). 3. Both expressions equal \(-1\). 4. The two equivalent forms illustrate the distributive property.

a) Expression A: \(-1\) Expression B: \(-1\) b) The results are equal. This illustrates the distributive property.

- First evaluate the expression inside the parentheses as a check. - When distributing, multiply the outside factor by each term, including its sign. - Recall the sign of a product of two negative numbers.

1. Evaluate inside the parentheses: \(1.2 - 4 = -2.8\). Then \(-5(-2.8) = 14\), so Sophie is correct. 2. When distributing, the correct expression is \(-5 \cdot 1.2 - (-5 \cdot 4)\). 3. Therefore, \(-6 - (-20) = -6 + 20 = 14\). Lucas failed to account for the product of two negative factors.

1. Sophie is correct; the value is \(14\). 2. Lucas mishandled the signs. The second product contributes \(+20\), not \(-20\).

- 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.

- Identify the main operation in each description. - “Increased by” indicates addition. - Addition and multiplication are commutative, but subtraction and division are not.

1. Description A matches Cards 1 and 5 because addition is commutative. 2. Description B matches Cards 2 and 4 because addition and multiplication are commutative. 3. Description C matches Card 3.

(A) Cards 1 and 5 (B) Cards 2 and 4 (C) Card 3

- Multiply \(20\) by each term inside the parentheses. - Evaluate \(20 \cdot \frac{1}{4}\) and \(20 \cdot 0.2\). - Compare the resulting expression with Expression B.

1. Distribute \(20\) in Expression A: \(20 \cdot \frac{1}{4} - 20 \cdot 0.2\). 2. Evaluate the first product: \(20 \cdot \frac{1}{4} = 5\). 3. Evaluate the second product: \(20 \cdot 0.2 = 4\). 4. The expanded form is \(5 - 4\), which is exactly Expression B. Both expressions equal \(1\).

Distributing \(20\) changes Expression A to \(5 - 4\), so the expressions are equivalent and both have value \(1\).

- In a), distribute the factor to both terms inside the parentheses. - In b), determine how a subtraction sign affects every term in the grouped sum. - Look for reciprocal factors whose product is \(1\).

1. For a), distribute \(\frac{4}{9}\): \(\frac{4}{9} \cdot 2.7 + \frac{4}{9} \cdot \frac{9}{4}\). The second product is \(1\), so the equation is true. 2. For b), subtracting a sum means subtracting both addends: \(15.3 - 4.2 - \frac{1}{3}\). The right side adds \(\frac{1}{3}\), so the equation is false.

a) True; distributing gives \(\frac{4}{9} \cdot 2.7 + 1\). b) False; the correct form is \((15.3 - 4.2) - \frac{1}{3}\).

- Look for factors that combine to make \(1000\). - In b), identify the common factor and add the decimal coefficients. - Rewrite \(11\) as a sum that makes multiplication easy.

1. For a), rearrange and regroup the factors: \((250 \cdot 4) \cdot 13 = 1000 \cdot 13 = 13{,}000\). 2. For b), factor out \(19\): \(19(0.8 + 0.2) = 19 \cdot 1 = 19\). 3. For c), write \(11 = 10 + 1\) and distribute: \(45(10 + 1) = 450 + 45 = 495\).

a) \(13{,}000\) b) \(19\) c) \(495\)

- In a), distributing avoids adding fractions with unlike denominators first. - In b), identify the common factor. - In c), multiply and simplify each product before subtracting.

1. For a), distribute \(24\): \(24 \cdot \frac{3}{8} + 24 \cdot \frac{1}{6} = 9 + 4 = 13\). 2. For b), factor out \(\frac{5}{12}\): \(\frac{5}{12}\left(\frac{4}{7} + \frac{3}{7}\right) = \frac{5}{12} \cdot 1 = \frac{5}{12}\). 3. For c), factor out \(15\): \(15\left(\frac{7}{5} - \frac{2}{3}\right)\). Distributing directly gives \(21 - 10 = 11\).

a) \(13\) b) \(\frac{5}{12}\) c) \(11\)

- Convert the fraction to a decimal. - Look for values whose decimal parts combine to whole numbers. - Rewrite the expression as a sum before reordering terms.

1. Write \(\frac{11}{4}=2.75\). 2. Rewrite subtraction as addition and regroup: \((7.25+2.75)+(-4.8-1.2)\). 3. Evaluate the grouped sums: \(10+(-6)=4\).

\(4\)

- For a), evaluate the sum inside the parentheses first. - For b), multiply \(8\) by each addend. - Equivalent methods must produce the same value.

1. Add first: \(8(12.5 + 5) = 8(17.5) = 140\). 2. Distribute first: \(8 \cdot 12.5 + 8 \cdot 5 = 100 + 40 = 140\). 3. Both methods give the same value, as expected for equivalent expressions.

a) \(140\) b) \(140\) The results are equal.

- Rewrite \(101\) using \(100\) and \(1\). - In b), identify the common factor. - Choose the equivalent form that creates the simplest arithmetic.

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

a) \(1212\) b) \(5\)

- For the direct method, evaluate multiplication before addition or subtraction. - For the factoring method, identify the factor shared by both terms. - Compare the two methods after completing each part.

1. For a), direct calculation gives \(154 + 126 = 280\). Factoring gives \(14(11 + 9) = 14 \cdot 20 = 280\). 2. For b), direct calculation gives \(52 - 39 = 13\). Factoring gives \(6.5(8 - 6) = 6.5 \cdot 2 = 13\). 3. For c), direct calculation gives \(-6 + (-9) = -15\). Factoring gives \((-1.5)(4 + 6) = -1.5 \cdot 10 = -15\).

a) \(280\) b) \(13\) c) \(-15\)

- Look for a common factor in each pair of products. - Factor out the common factor. - The values inside the parentheses combine to convenient numbers.

1. For a), \(16\cdot27-16\cdot17=16(27-17)=16\cdot10=160\). 2. For b), \(0.2\cdot35+0.2\cdot15=0.2(35+15)=0.2\cdot50=10\).

a) \(160\) b) \(10\)

- Use the distributive property to remove parentheses. - Combine like terms in each expression. - Equivalent expressions simplify to the same result.

1. For part a, simplify the first expression: \(4(a+2)+3a=4a+8+3a=7a+8\). It matches the second expression, so the pair is equivalent. 2. For part b, simplify the first expression: \(15y-6y+2=9y+2\). 3. Simplify the second expression: \(3(3y+2)-4=9y+6-4=9y+2\). 4. Both expressions in part b simplify to \(9y+2\), so the pair is equivalent.

a) Equivalent; both simplify to \(7a+8\). b) Equivalent; both simplify to \(9y+2\).

- Pay close attention to the sign before each set of parentheses. - Combine only terms with the same variable part. - When a negative factor is distributed, how does it affect each term inside the parentheses?

1. For a), use the distributive property: \(4(2x - 5) + 3x = 8x - 20 + 3x\). Use the commutative property of addition to reorder the terms: \(8x - 20 + 3x = 8x + 3x - 20\). Combine like terms: \(8x + 3x - 20 = 11x - 20\). 2. For b), use the distributive property: \(15y - 3(5y - 2) = 15y - 15y + 6\). Combine like terms: \(15y - 15y + 6 = 6\). 3. For c), use the distributive property: \(8z + 4 - 2(4z + 2) = 8z + 4 - 8z - 4\). Use the commutative property of addition to reorder the terms: \(8z + 4 - 8z - 4 = 8z - 8z + 4 - 4\). Combine like terms: \(8z - 8z + 4 - 4 = 0\).

a) \(11x - 20\) b) \(6\) c) \(0\)

- Combine variable terms and constant terms separately. - Use the distributive property for expressions with parentheses. - Compare each simplified expression with \(4x+10\).

1. For part a, \(2x+10+2x=4x+10\), so it is equivalent. 2. For part b, \(2(2x+5)=4x+10\), so it is equivalent. 3. For part c, \(15x-11x+12-2=4x+10\), so it is equivalent. 4. For part d, \(4(x+10)=4x+40\), so it is not equivalent. 5. For part e, \(x+x+x+x+7+3=4x+10\), so it is equivalent.

The equivalent expressions are a), b), c), and e). Expression d) simplifies to \(4x+40\), so it is not equivalent.

- Simplify each expression completely. - Use the distributive property for expressions with parentheses. - Combine variable terms and constant terms separately. - Apply multiplication and division before addition and subtraction.

1. \(3(2x-4)=6x-12\), so a) is equivalent. 2. \(6(x-2)=6x-12\), so b) is equivalent. 3. \(10x-12-4x=6x-12\), so c) is equivalent. 4. \(2x+4x-6\cdot2=6x-12\), so d) is equivalent. 5. Expression e) is identical to the given expression. 6. \(x(6-12)=-6x\), so f) is not equivalent. 7. \(12x\div2-12=6x-12\), so g) is equivalent.

a), b), c), d), e), and g)

- Translate each direction into an algebraic expression. - Use the distributive property when you multiply the sum by \(3\). - Make sure the multiplication in step \(3\) applies to the entire previous result.

1. For \(x=2\): \(2+5=7\), \(7\cdot3=21\), \(21-15=6\), and \(6-2\cdot2=2\). 2. For \(x=-4\): \(-4+5=1\), \(1\cdot3=3\), \(3-15=-12\), and \(-12-2(-4)=-4\). 3. In general, the expressions after the five steps are \(x\), \(x+5\), \(3(x+5)=3x+15\), \(3x\), and \(3x-2x=x\). 4. The final result is always the original number.

a) For example, starting with \(2\) gives \(2\), and starting with \(-4\) gives \(-4\). The trick returns the starting number. b) The expressions are \(x\), \(x+5\), \(3x+15\), \(3x\), and \(x\).

- Write an implied coefficient of \(1\) when needed. - For fractions with the same denominator, add the numerators. - Apply the distributive property to every term inside parentheses.

1. Part a is incorrect. Since \(x=1x\), \(7x-x=(7-1)x=6x\). The variable should not disappear. 2. Part b is correct. The fractions have the same denominator, so their numerators add: \(\frac{x+x}{3}=\frac{2x}{3}\). 3. Part c is incorrect. The factor \(3\) must multiply both terms: \(3(2a+4)=6a+12\). 4. Part d is correct because \(4y+3y=(4+3)y=7y\).

a) Incorrect; \(7x-x=6x\). b) Correct. c) Incorrect; \(3(2a+4)=6a+12\). d) Correct.

- Check the sign before each set of parentheses. - A negative sign before parentheses reverses every sign inside. - Combine only terms with identical variable parts.

1. For part a, remove the parentheses and combine like terms: \(12x-5y+3x-4y=15x-9y\). 2. For part b, distribute the negative sign and combine like terms: \(8a-2a+7b-10b=6a-3b\). 3. For part c, distribute the negative sign and combine like terms: \(-15m+4n+6m-4n=-9m\).

a) \(15x-9y\) b) \(6a-3b\) c) \(-9m\)

- Look for two cards whose values can combine to make \(24\). - Remember that subtracting a negative number is the same as adding a positive number. - The cards \(8\) and \(-8\) can combine to make zero.

1. One expression is \(16+8\), which equals \(24\). 2. A different expression is \(16-(-8)\), which also equals \(24\). 3. Each number card is used no more than once in either expression.

Possible answers are \(16+8\) and \(16-(-8)\).

- Look for pairs that make convenient tens or hundreds. - Keep each sign attached to its number when reordering. - Add the partial sums last.

1. Regroup as \((-135+35)+(48-18)\). 2. The partial sums are \(-100\) and \(30\). 3. Therefore, the value is \(-100+30=-70\).

\(-70\)

- A coefficient tells how many copies of a variable expression are being added. - Think of each power, such as \(a^2\), as one complete unit. - For a grouped expression, repeat the entire group.

1. Interpret multiplication as repeated addition: \(5x=x+x+x+x+x\). 2. Write three copies of \(a^2\) and two copies of \(b\): \(3a^2+2b=a^2+a^2+a^2+b+b\). 3. Write four copies of \(m\) and subtract three copies of \(n\): \(4m-3n=m+m+m+m-n-n-n\). 4. Treat \(p+q\) as one grouped expression and add two copies: \(2(p+q)=(p+q)+(p+q)\).

a) \(x+x+x+x+x\) b) \(a^2+a^2+a^2+b+b\) c) \(m+m+m+m-n-n-n\) d) \((p+q)+(p+q)\)

- Count the edges in the square base and the edges from the base to the apex. - A square pyramid has eight edges in all. - Translate “twice as long” into an equation relating \(s\) and \(a\). - Combine like terms after substitution.

1. The square base has four edges of length \(a\), and the pyramid has four lateral edges of length \(s\). Thus, \(L=4a+4s=4(a+s)\). 2. If \(s=2a\), then \(L=4a+4(2a)=4a+8a=12a\).

1. \(L=4a+4s\), or \(L=4(a+s)\) 2. \(L=12a\)

- Subtraction can be rewritten as addition of the opposite. - Addition can be rewritten as subtraction of the opposite. - Simplify a double negative before completing the requested form.

1. For a), rewrite each subtraction as addition of the opposite: \(p + (-q) + (-r) + (-10)\). 2. For b), adding a value is equivalent to subtracting its opposite: \(15 - (-a) - (-b)\). 3. For c), subtracting \(-y\) is adding \(y\), and subtracting \(z\) is adding \(-z\): \(x + y + (-z)\).

a) \(p + (-q) + (-r) + (-10)\) b) \(15 - (-a) - (-b)\) c) \(x + y + (-z)\)

- Simplify each expression separately. - Subtracting a negative term changes to addition. - Compare the simplified results.

1. \(A=-15a+6a=-9a\). 2. \(B=-4a-5a=-9a\). 3. \(C=-12a+3a=-9a\). 4. \(D=2a-11a=-9a\). 5. \(E=-a-10a=-11a\). 6. Therefore, A, B, C, and D are equivalent. E is not equivalent to them.

A, B, C, and D are equivalent because each simplifies to \(-9a\). Expression E simplifies to \(-11a\).

- Work from the innermost grouping symbols outward. - A negative sign before parentheses reverses every sign inside. - Simplify inside the brackets before removing them.

1. Work from the innermost parentheses: \(2x-[5x-3x-4]\). 2. Combine like terms inside the brackets: \(2x-[2x-4]\). 3. Distribute the negative sign: \(2x-2x+4\). 4. The variable terms cancel, so the result is \(4\).

\(4\)

- Identify which terms belong in each group. - A negative sign outside parentheses changes the sign of every term inside when the parentheses are removed. - Check your expression by distributing the negative sign.

1. The terms containing \(b\) and \(c\) are \(-7b\) and \(-5c\). 2. The remaining terms are \(14a\) and \(3\), so the first group is \((14a+3)\). 3. Because a subtraction sign comes before the second group, write \(7b+5c\) inside it. Distributing the negative sign gives \(-7b-5c\). 4. The equivalent expression is \((14a+3)-(7b+5c)\).

\((14a+3)-(7b+5c)\)

- Multiply before adding or subtracting. - Simplify each product first. - Only terms with exactly the same variable part can be combined. - Track the sign of each product carefully.

1. For part a, multiply first: \((-3x)\cdot(-2y)=6xy\). Then \(6xy-8xy=-2xy\). 2. For part b, evaluate each product: \(4a\cdot(-b)=-4ab\) and \((-2a)\cdot(-3b)=6ab\). Then \(-4ab+6ab=2ab\). 3. For part c, the two factors of \(-1\) have product \(1\), so the first term is \(mn\). Therefore, \(mn-mn=0\).

a) \(-2xy\) b) \(2ab\) c) \(0\)

- Track the sign of the difference inside the parentheses. - When distributing, multiply \(12\) by both decimal terms. - Equivalent expressions must have the same value.

1. Subtract first: \(0.75 - 1.25 = -0.5\), so \(12(-0.5) = -6\). 2. Distribute first: \(12 \cdot 0.75 - 12 \cdot 1.25 = 9 - 15 = -6\). 3. Both methods give the same result.

Both methods give \(-6\).

- A fraction bar groups everything in the numerator and denominator. - When changing a fraction to a division expression, use parentheses to preserve grouped sums or differences. - Identify the dividend and divisor in each expression.

1. In a fraction, the dividend becomes the numerator and the divisor becomes the denominator: \((4z+1)\div(a-3)=\frac{4z+1}{a-3}\). 2. The numerator is divided by the denominator. Parentheses preserve the grouped numerator: \(\frac{x+y}{9}=(x+y)\div9\). 3. The dividend \(15\) is the numerator and \(2u+5v\) is the denominator: \(15\div(2u+5v)=\frac{15}{2u+5v}\). 4. Use parentheses around both grouped expressions: \(\frac{7a-2b}{c+d}=(7a-2b)\div(c+d)\).

1) \(\frac{4z+1}{a-3}\) 2) \((x+y)\div9\) 3) \(\frac{15}{2u+5v}\) 4) \((7a-2b)\div(c+d)\)

- First follow the directions with the number from part a). - Use \(x\) to represent an unknown starting number. - Combine the terms containing \(x\) before dividing. - Interpret what an expression of the form \(x+\text{constant}\) means.

1. Starting with \(4\) gives \(\frac{3\cdot4+15-4}{2}=\frac{23}{2}=11.5\). 2. For a starting number \(x\), the expression is \(\frac{3x+15-x}{2}\). 3. Combine like terms in the numerator: \(\frac{2x+15}{2}=x+7.5\). 4. The final result is always \(7.5\) greater than the starting number.

a) \(11.5\) b) \(\frac{3x+15-x}{2}=x+7.5\) c) The final result is always \(7.5\) greater than the starting number.

- Represent the starting number with \(x\). - Translate the operations in order into one expression. - Distribute and combine like terms. - What factor remains in front of \(x\)?

1. Let the starting number be \(x\). The final result is \(5(4x+12)-60\). 2. Distribute: \(5(4x+12)-60=20x+60-60\). 3. Simplify to \(20x\). 4. Since the final result is \(20\) times the starting number, divide the final result by \(20\) to recover \(x\).

Divide the final result by \(20\). The expression \(5(4x+12)-60\) simplifies to \(20x\).

- Think of a negative sign before parentheses as multiplication by \(-1\). - Reverse every sign inside when you reverse the sign outside. - Expand your result to check that it matches the original expression.

1. Change subtraction to addition and reverse every sign inside the parentheses: \(12+(-4a+5b)\). 2. Change addition to subtraction and reverse every sign inside the parentheses: \(x-(-2y+3z-1)\). 3. Change subtraction to addition and reverse every sign inside the parentheses: \(3k+(-k+m)\).

1) \(12+(-4a+5b)\) 2) \(x-(-2y+3z-1)\) 3) \(3k+(-k+m)\)

- Only the final three terms belong inside the parentheses. - Reverse each sign inside because a negative sign will be outside. - Expand the parentheses to verify your result.

1. The terms to group are \(-7b+4c-9\). 2. Reverse their signs when placing a minus sign before the parentheses: \(7b-4c+9\). 3. The equivalent expression is \(12a-(7b-4c+9)\). 4. Distributing the negative sign confirms that \(12a-(7b-4c+9)=12a-7b+4c-9\).

\(12a-(7b-4c+9)\)

- Treat the two yearly changes separately. - The second-year increase is based on the population after the first year. - Write the percent increase using the variable \(p\). - Leave the variable in the final expression.

1. For part a, the first-year increase is \(36{,}000 \cdot 0.005 = 180\). The population after the first year is \(36{,}000 + 180 = 36{,}180\). 2. For part b, the second-year increase is \(36{,}180 \cdot \frac{p}{100} = 361.8p\). 3. Therefore, the population after the second year is \(36{,}180 + 361.8p\), or equivalently \(36{,}180\left(1 + \frac{p}{100}\right)\).

a) The population after the first year is \(36{,}180\). b) The population after the second year is \(36{,}180\left(1 + \frac{p}{100}\right)\).

- Compare the signs inside the parentheses with the signs after the parentheses are removed. - Think about the difference between a plus sign and a minus sign before parentheses. - State the pattern in your own words.

1. \(A=-15+(-15)=-30\), and \(B=-15+30-45=-30\). 2. A plus sign before parentheses allows the parentheses to be removed without changing the signs inside. 3. With a minus sign before the parentheses, the signs inside change when the parentheses are removed: \(-15-30+45\).

1) \(A=-30\), \(B=-30\) 2) A plus sign before parentheses allows the parentheses to be removed without changing the signs inside. 3) \(-15-30+45\)

- In a), look for a factor shared by both products. - In b), which two factors multiply to make \(100\)? - In c), rewrite \(99\) using \(100\) and \(1\).

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) \(320\) b) \(3700\) c) \(4455\)

- Multiply the outside factor by every term inside the parentheses. - Keep each operation sign from inside the parentheses. - Each product simplifies to a whole number.

1. Distribute \(12\) to each term: \(12 \cdot \frac{3}{4} - 12 \cdot 0.5 + 12 \cdot \frac{1}{6}\). 2. Evaluate the products: \(12 \cdot \frac{3}{4} = 9\), \(12 \cdot 0.5 = 6\), and \(12 \cdot \frac{1}{6} = 2\). 3. Combine the results: \(9 - 6 + 2 = 5\).

\(5\)

- In a), multiply the outside factor by each addend. - In b), identify the common factor in both terms. - Choose the equivalent form that creates simpler arithmetic.

1. For a), distribute \(42\): \(42 \cdot \frac{1}{7} + 42 \cdot \frac{1}{6} = 6 + 7 = 13\). 2. For b), factor out \(17\): \(17(3.8 + 6.2) = 17 \cdot 10 = 170\).

a) \(13\) b) \(170\)

- Identify the operation performed last under the order of operations. - Look for a common factor inside the parentheses. - Determine the sign of the quotient before calculating.

1. The expression is a sum at its highest level because adding \(-4.5\) is the final operation. 2. Factor the common factor inside the parentheses: \(2.4\left(-1.5 - 3.5\right) = 2.4(-5) = -12\). 3. Divide: \(-12 \div (-1.2) = 10\). 4. Add: \(10 + (-4.5) = 5.5\).

It is a sum at its highest level. Its value is \(5.5\).

- 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 \(1\frac{2}{7}\).

- First determine the sign of the quantity inside the parentheses. - What does multiplying by \(-2\) do to a negative value? - What does absolute value do to a negative value? - Compare the resulting forms rather than calculating the final number.

1. Let \(x=-\frac{3}{4}+0.5\). Since \(0.5=\frac{1}{2}<\frac{3}{4}\), \(x<0\). 2. Expression A is \(x\cdot(-2)=(-x)\cdot2\). 3. Because \(x<0\), \(|x|=-x\). Therefore, Expression B is also \((-x)\cdot2\). 4. Both expressions have the same form, so they must have the same value.

Let \(x=-\frac{3}{4}+0.5\). Since \(x<0\), \(|x|=-x\). Both expressions therefore equal \((-x)\cdot2\), so they have the same value.

- What happens when \(-1\) is raised to an odd power? - Does an even or odd exponent change the sign? - Does changing the order of factors change a product?

1. Since \(3\) is odd, \((-1)^3=-1\). Therefore, Expression A is \(-1\cdot4.2=-4.2\). 2. Since \(7\) is odd, \((-1)^7=-1\). Therefore, Expression B is \(4.2\cdot(-1)=-4.2\). 3. Both expressions have the same value.

Yes. Both expressions equal \(-4.2\) because both odd powers of \(-1\) equal \(-1\), and multiplication can be written in either factor order.

- Complete both methods before comparing them. - Look for addends that combine to whole numbers. - Consider how reordering and regrouping affect the value of a sum.

1. Jordan’s method: \(2\frac{1}{4}=2.25\) and \(\frac{3}{4}=0.75\). Then \(4.8+2.25=7.05\), \(7.05+1.2=8.25\), and \(8.25+0.75=9\). 2. Maya’s method: \(4.8+1.2=6\) and \(2\frac{1}{4}+\frac{3}{4}=3\). Then \(6+3=9\). 3. Maya’s method is easier mentally because the commutative and associative properties create whole-number sums.

a) Both methods give \(9\). b) Maya’s method is generally easier mentally because the grouped addends make \(6\) and \(3\).

- In a), rearrange the factors to pair the two decimals. - Convert the decimals to fractions to check whether their product is \(1\). - In b), compare dividing by twice a number with dividing by the number and then doubling.

1. For a), rearrange and regroup the factors: \((0.8 \cdot 1.25)\frac{3}{7}\). Since \(0.8 = \frac{4}{5}\) and \(1.25 = \frac{5}{4}\), their product is \(1\). The equation is true. 2. For b), dividing by a product is equivalent to dividing successively by both factors: \(a \div (bc) = a \div b \div c\). The right side divides by \(\frac{1}{3}\) and then multiplies by \(2\), so the equation is false.

a) True, because \(0.8 \cdot 1.25 = 1\). b) False, because dividing by \(\frac{1}{3} \cdot 2\) is not the same as dividing by \(\frac{1}{3}\) and then multiplying by \(2\).

- Identify the factor shared by both terms. - Use the distributive property in reverse. - Compare \(\frac{3}{4}\) with \(0.75\). - Recall whether the distributive property applies to sums as well as differences.

1. Both terms have the common factor \(\frac{3}{4}\). 2. Factor using the distributive property: \(\frac{3}{4} \cdot 1.2 - \frac{3}{4} \cdot 0.4 = \frac{3}{4}(1.2 - 0.4)\). 3. Since \(\frac{3}{4} = 0.75\), this is exactly the student’s method, so the claim is correct. 4. The distributive property also applies to addition: \(ab + ac = a(b + c)\). Therefore, the corresponding rule would also work with a plus sign.

The claim is correct because \(\frac{3}{4}\) can be factored: \(\frac{3}{4}(1.2 - 0.4)\). The same rule works for addition because \(ab + ac = a(b + c)\).

- Evaluate each side separately. - Apply parentheses first. - What happens to every term inside parentheses when a negative sign is distributed?

1. For a), the left side is \(-12.5+7.5=-5\). The right side is \(-(12.5-7.5)=-5\). The statement is true. 2. For b), the left side is \(\frac{1}{2}-\left(\frac{1}{4}+\frac{1}{8}\right)=\frac{4}{8}-\frac{3}{8}=\frac{1}{8}\). 3. The right side is \(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}=\frac{4}{8}-\frac{2}{8}+\frac{1}{8}=\frac{3}{8}\). Since the two sides differ, the statement is false.

a) True; both sides equal \(-5\). b) False; the left side is \(\frac{1}{8}\) and the right side is \(\frac{3}{8}\).

- A subtraction sign before parentheses changes the sign of every term inside. - Convert a fraction to a decimal when that creates easy pairs. - Regroup terms to form whole numbers.

1. For a), remove the parentheses: \(-3.25-1.75+8=-5+8=3\). 2. For b), remove the parentheses and regroup: \(2\frac{1}{5}-4.8-\frac{1}{5}+0.8=(2.2-0.2)+(-4.8+0.8)=2-4=-2\). 3. For c), remove the parentheses: \(\frac{5}{8}+1.4-\frac{1}{8}+0.6\). Regroup: \(\left(\frac{5}{8}-\frac{1}{8}\right)+(1.4+0.6)=0.5+2=2.5\).

a) \(3\) b) \(-2\) c) \(2.5\)

- 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) \(13\)

- Multiply the outside factor by every term inside the parentheses. - Keep each addition or subtraction sign. - Simplify products by canceling common factors when possible.

1. For a), distribute: \(\frac{4}{5} \cdot 10 - \frac{4}{5} \cdot 2.5 = 8 - 2 = 6\). 2. For b), distribute: \(18 \cdot \frac{1}{2} + 18 \cdot \frac{1}{3} - 18 \cdot \frac{1}{6} = 9 + 6 - 3 = 12\). 3. For c), distribute: \(\frac{7}{11} \cdot 22 - \frac{7}{11} \cdot 5.5 = 14 - 3.5 = 10.5\).

a) \(6\) b) \(12\) c) \(10.5\)

- 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 \(-100\).

- Multiply the outside factor by each signed term inside the parentheses. - Pay close attention to the product of two negative numbers. - Check your result by evaluating the parentheses first.

1. The error occurs when distributing \(-4\). The second term should be \(-(-4 \cdot 6)\), not \(-4 \cdot 6\). 2. The correct expanded form is \(-4 \cdot 2.5 - (-4 \cdot 6)\). 3. Evaluate: \(-10 - (-24) = -10 + 24 = 14\). 4. As a check, evaluate inside the parentheses first: \((-4)(-3.5) = 14\).

The distribution step is incorrect. The correct value is \(14\).

- Check whether the distributive property has been applied correctly. - A subtraction sign before parentheses changes the sign of every term inside. - When uncertain, evaluate both sides separately.

1. For a), factoring out \(-0.2\) gives \((-0.2)(15 + 5)\), so the equation is true. 2. For b), subtracting the grouped difference changes both signs: \(\frac{3}{4} - \frac{1}{2} + \frac{1}{4}\). The given right side is false. 3. For c), factoring out \(-3\) gives \((-3)(7 - 4)\), so the equation is true.

a) True. b) False. Corrected right side: \(\frac{3}{4} - \frac{1}{2} + \frac{1}{4}\). c) True.

- When subtracting a grouped difference, change the signs inside. - Look for terms that cancel. - Use the distributive property in part b.

1. For a), the expression is \((10.2-12.4)-(7.8-12.4)\). Removing the second parentheses gives \(10.2-12.4-7.8+12.4=10.2-7.8=2.4\). 2. For b), \((25-2.5)\cdot4=25\cdot4-2.5\cdot4=100-10=90\).

a) \(2.4\) b) \(90\)

- In part a), distribute \(20\) to both fractions. - In part b), identify exactly what is subtracted from what. - Rewrite \(0.75\) as a fraction if that makes the multiplication easier.

1. For a), \(20\cdot(-\frac{2}{5}+\frac{1}{2})=20\cdot(-\frac{2}{5})+20\cdot(\frac{1}{2})=-8+10=2\). 2. For b), \((15+(-3))-(12\cdot0.75)=12-9=3\).

a) \(2\) b) \(3\)

- Combine the fixed costs first. - The variable \(x\) represents the shipping charge. - A larger shipping charge leaves a smaller balance.

1. Subtract all costs from the gift-card balance: \(100-(24.99+35.50+x)\). 2. Combine the fixed costs: \(24.99+35.50=60.49\). Therefore, \(100-(60.49+x)=39.51-x\). 3. For \(x=5.95\), \(39.51-5.95=33.56\).

a) \(100-(24.99+35.50+x)\), or \(39.51-x\) b) \(\$33.56\)

- 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.

- Choose fraction or decimal forms that make useful pairs. - Combine fractions with the same denominator directly. - Simplify the final fraction.

1. For a), write \(\frac{1}{8}=0.125\). Regroup: \((0.125+0.875)-(0.25+1.75)=1-2=-1\). 2. For b), remove the parentheses: \(\frac{5}{6}+1.2-\frac{1}{6}+0.8\). 3. Regroup: \(\left(\frac{5}{6}-\frac{1}{6}\right)+(1.2+0.8)=\frac{4}{6}+2=\frac{2}{3}+2=2\frac{2}{3}\).

a) \(-1\) b) \(2\frac{2}{3}\), or \(\frac{8}{3}\)

- Simplify each expression before substituting the variable value. - Combine coefficients of like terms. - In part b, simplify the expression inside parentheses first.

1. For part a, combine the coefficients of \(x\): \(15x-4.5x+0.5x=(15-4.5+0.5)x=11x\). 2. Substitute \(x=1.2\): \(11\cdot1.2=13.2\). 3. For part b, simplify inside the parentheses and combine like terms: \(y(12-5.5)-1.5y=6.5y-1.5y=5y\). 4. Substitute \(y=\frac{4}{5}\): \(5\cdot\frac{4}{5}=4\).

a) \(13.2\) b) \(4\)

- Evaluate the expression inside the parentheses before multiplying. - Compare the two results for each input. - Which property lets you rewrite a product involving parentheses?

1. For \(x = 10\), \(A = 5 \cdot 10 - 15 = 35\) and \(B = 5(10 - 3) = 35\). 2. For \(x = 0\), \(A = 5 \cdot 0 - 15 = -15\) and \(B = 5(0 - 3) = -15\). 3. For \(x = -2\), \(A = 5 \cdot (-2) - 15 = -25\) and \(B = 5(-2 - 3) = -25\). 4. By the distributive property, \(5(x - 3) = 5x - 15\). Therefore, the expressions are equivalent and have the same value for every \(x\).

a) At \(x = 10\), both values are \(35\); at \(x = 0\), both values are \(-15\); at \(x = -2\), both values are \(-25\). b) The expressions always have the same value because the distributive property gives \(5(x - 3) = 5x - 15\).

- Simplify each expression separately. - If both simplify to exactly the same expression, they are equivalent. - In \(T_1\), be especially careful when subtracting \((5z - 8)\).

1. Simplify \(T_1\): \(4(2z - 3) - (5z - 8) = 8z - 12 - 5z + 8 = 3z - 4\). 2. Simplify \(T_2\): \(3(z - 2) + 2 = 3z - 6 + 2 = 3z - 4\). 3. Both expressions simplify to \(3z - 4\), so they are equivalent.

Yes. The expressions are equivalent because both simplify to \(3z - 4\).

- Substitute each given value of \(x\) carefully. - Use the distributive property to remove parentheses. - Combine like terms and compare the simplified expressions.

1. For \(x=0\): \(A=3\cdot(0+4)=12\), \(B=3\cdot0+4=4\), and \(C=5\cdot0+12-2\cdot0=12\). 2. For \(x=2\): \(A=3\cdot(2+4)=18\), \(B=3\cdot2+4=10\), and \(C=5\cdot2+12-2\cdot2=18\). 3. For \(x=5\): \(A=3\cdot(5+4)=27\), \(B=3\cdot5+4=19\), and \(C=5\cdot5+12-2\cdot5=27\). 4. Simplify \(A\): \(3(x+4)=3x+12\). 5. Simplify \(C\): \(5x+12-2x=3x+12\). 6. Expressions \(A\) and \(C\) are equivalent because they simplify to the same expression. Expression \(B\) is not equivalent to them.

<table> <tr><td>\(x\)</td><td>\(0\)</td><td>\(2\)</td><td>\(5\)</td></tr> <tr><td>Expression \(A\)</td><td>\(12\)</td><td>\(18\)</td><td>\(27\)</td></tr> <tr><td>Expression \(B\)</td><td>\(4\)</td><td>\(10\)</td><td>\(19\)</td></tr> <tr><td>Expression \(C\)</td><td>\(12\)</td><td>\(18\)</td><td>\(27\)</td></tr> </table> Expressions \(A\) and \(C\) are equivalent because both simplify to \(3x+12\).

- Distribute the negative sign in the original expression carefully. - Rewrite each expression in the form \(ax+b\). - Use the distributive property and combine like terms.

1. Simplify the original expression: \(T=12x-(4x+8)=12x-4x-8=8x-8\). 2. \(T_1=8x+8\) is not equivalent because its constant term differs from \(8x-8\). 3. \(T_2=4(2x-2)=8x-8\), so it is equivalent. 4. \(T_3=16x-8-8x=8x-8\), so it is equivalent. 5. \(T_4=8(x-1)=8x-8\), so it is equivalent.

The simplified expression is \(T=8x-8\). The equivalent expressions are \(T_2\), \(T_3\), and \(T_4\).

- Equivalent expressions must have the same value for every input. - Substitute \(x=0\) into all three expressions. - Check how \(-2\) multiplies both terms inside the parentheses.

1. Substitute \(x=0\) into the original expression: \(6\cdot0-2\cdot(0-4)=0-2\cdot(-4)=8\). 2. Luke’s expression gives \(4\cdot0-8=-8\). 3. Maya’s expression gives \(4\cdot0+8=8\). 4. Maya’s result matches the original expression for \(x=0\), so Maya could be correct. 5. Luke did not distribute \(-2\) correctly to \(-4\). Since \((-2)\cdot(-4)=8\), the constant term should be \(+8\), not \(-8\).

a) The original expression equals \(8\) when \(x=0\). Luke’s result is \(-8\), and Maya’s result is \(8\). Maya could be correct. b) Luke made a sign error when distributing \(-2\). The product \((-2)\cdot(-4)\) is \(+8\).

- Focus on the negative sign before the parentheses. - Simplify the expression independently before comparing results. - Check the variable terms and constants separately.

1. Distribute the negative sign: \(12b-(5b+7)+3=12b-5b-7+3\). 2. Combine the variable terms: \(12b-5b=7b\). 3. Combine the constants: \(-7+3=-4\). 4. The correct simplified expression is \(7b-4\). 5. The student failed to change the sign of \(+7\) when removing the parentheses and incorrectly combined the constants as \(7+3\).

The result is incorrect. The \(+7\) inside the subtracted parentheses must become \(-7\). The correct simplification is \(7b-4\).

- Expand the known product first. - Treat the box as an unknown expression. - Subtract the target expression from the expanded expression. - Substitute your result back into the original statement to check it.

1. Expand the known expression: \(5(2n+1)=10n+5\). 2. Let the missing expression be \(M\): \(10n+5-M=3n+8\). 3. Solve for \(M\): \(M=(10n+5)-(3n+8)\). 4. Distribute the negative sign and combine like terms: \(M=10n+5-3n-8=7n-3\). 5. Check: \(10n+5-(7n-3)=10n+5-7n+3=3n+8\).

\(7n-3\)

- Combine the constant terms in each expression. - Pay attention to the subtraction sign before the parentheses in Expression B. - When subtracting the expressions, distribute the subtraction sign to both terms in the second expression.

1. Simplify Expression A by combining the constants: \(30-x-10-5=15-x\). 2. Simplify Expression B by evaluating the parentheses: \(30-x-(10-5)=30-x-5=25-x\). 3. Subtract the expressions: \((25-x)-(15-x)=25-x-15+x=10\). 4. The \(x\)-terms cancel, so the difference is always \(10\), regardless of the value of \(x\).

a) Expression A: \(15-x\); Expression B: \(25-x\) b) The difference is \(10\). It does not depend on \(x\) because the \(x\)-terms cancel.

- Simplify every expression before grouping them. - Distribute a factor to every term inside parentheses. - Combine like terms after distributing.

1. Expression 1 simplifies to \(6a-12+2a=8a-12\). 2. Expression 2 simplifies to \(3(a+4)-a=3a+12-a=2a+12\). 3. Expression 3 simplifies to \(4(2a-3)=8a-12\). 4. Expression 4 is already \(2a+12\). 5. Expression 5 simplifies to \(10a-12-2a=8a-12\). 6. Expression 6 simplifies to \(2(a+6)=2a+12\). 7. Therefore, expressions 1, 3, and 5 form one group, and expressions 2, 4, and 6 form the other.

Group 1, equivalent to \(8a-12\): expressions 1, 3, and 5 Group 2, equivalent to \(2a+12\): expressions 2, 4, and 6

- Use the distributive property to remove the parentheses. - If two expressions simplify to exactly the same form, they are equivalent. - For part b), using the original form of \(T_1\) can help check your simplification.

1. Simplify \(T_1\): \(5(z - 3) + 10 = 5z - 15 + 10 = 5z - 5\). 2. For \(z = -2\), each expression has the value \(5 \cdot (-2) - 5 = -15\). 3. For \(z = 6\), each expression has the value \(5 \cdot 6 - 5 = 25\). 4. Since \(T_1\) simplifies exactly to \(T_2\), the expressions are equivalent for every value of \(z\).

a) \(T_1 = 5z - 5\) b) For \(z = -2\), both values are \(-15\). For \(z = 6\), both values are \(25\). c) The expressions are equivalent because they simplify to the same expression.

- If one expression simplifies to exactly the other expression, what does that show? - For the evaluation, can you use the simpler equivalent expression? - Look for terms that cancel.

1. Expand and combine like terms: \(T_1 = 4k + 8 - 3k - 8 = k\). 2. Since \(T_1\) simplifies to \(T_2\), the expressions are equivalent. 3. When \(k = 12.5\), \(T_1 = k = 12.5\).

The expressions are equivalent because \(T_1\) simplifies to \(k\). For \(k = 12.5\), the value of \(T_1\) is \(12.5\).

- Treat each term as representing one or more side lengths. - How many side lengths are represented by the expression? - Use the distributive property before combining like terms.

1. One possible figure is an irregular hexagon with four sides of length \(a\) and two sides of length \(a+5\). 2. For \(a=3.5\,\text{cm}\), \(P=4\cdot3.5+2(3.5+5)=14+17=31\). The perimeter is \(31\,\text{cm}\). 3. Distribute and combine like terms: \(P=4a+2a+10=6a+10\).

a) Answers will vary. One example is a hexagon with side lengths \(a\), \(a\), \(a\), \(a\), \(a+5\), and \(a+5\). b) \(31\,\text{cm}\) c) \(P=6a+10\)

- How many equal sides does an equilateral triangle have? How many sides does a square have? - Write a perimeter expression for each figure. - Use the distributive property to simplify both expressions. - What must be true of the simplified expressions for the perimeters always to match?

1. The triangle's perimeter is \(P_A = 3(2s + 4) = 6s + 12\). 2. The square's perimeter is \(P_B = 4(1.5s + 3) = 6s + 12\). 3. The perimeter expressions are identical, so \(P_A = P_B\) for every allowable value of \(s\). Both side lengths are positive when \(s > -2\).

Both perimeter expressions simplify to \(6s + 12\), so the figures have equal perimeters for every \(s > -2\).

- Write a separate algebraic expression for each rule. - In Rule A, the phrase “the entire result” means the sum should be inside parentheses. - Expand any parentheses and combine like terms. - How can simplified expressions show that two rules always agree?

1. Rule A is represented by \(T_A = 3(x + 4)\). Using the distributive property, \(T_A = 3x + 12\). 2. Rule B is represented by \(T_B = 2x + x + 12\). Combining like terms gives \(T_B = 3x + 12\). 3. Since both rules simplify to \(3x + 12\), they always produce the same result.

Yes. Rule A gives \(3(x + 4) = 3x + 12\), and Rule B gives \(2x + x + 12 = 3x + 12\). The expressions are equivalent.

- The delivery charge is a one-time fixed cost. - When adding expressions, combine only like terms. - Check the combined expression by substituting a value for \(x\).

1. Vendor A's cost is \(A(x)=15+0.80x\). Vendor B's cost is \(B(x)=5+1.20x\). 2. For \(x=30\), \(A(30)=15+0.80\cdot 30=39\), and \(B(30)=5+1.20\cdot 30=41\). 3. Add the expressions: \((15+0.80x)+(5+1.20x)=20+2x\). The student's expression is correct.

a) Vendor A: \(15+0.80x\); Vendor B: \(5+1.20x\) b) Vendor A: \(\$39.00\); Vendor B: \(\$41.00\) c) Yes. \((15+0.80x)+(5+1.20x)=20+2x\).

- Substitute \(n=2\) into both expressions. - Distribute the factor outside the parentheses to every term inside. - Find the number that becomes \(8\) after multiplication by \(4\).

1. For \(n=2\), \(T_1=4\cdot2+8=16\), while \(T_2=4\cdot(2+8)=40\). Since \(16\ne40\), the claim is false. 2. By the distributive property, \(4(n+8)=4n+32\), not \(4n+8\). The factor \(4\) must multiply both terms inside the parentheses. 3. The number inside the parentheses must produce \(8\) when multiplied by \(4\). Since \(8\div4=2\), the equivalent expression is \(4(n+2)\).

1. \(T_1=16\) and \(T_2=40\), so the claim is false. 2. \(4(n+8)=4n+32\), not \(4n+8\). 3. Change \(T_2\) to \(4(n+2)\).

- Distribute the negative sign before combining like terms. - Compare the simplified form of \(A\) with \(B\). - Substitute the value of \(x\) after simplifying.

1. Distribute the negative sign: \(A=5x-2x-3\). 2. Combine like terms: \(A=3x-3\), which is identical to \(B\). Therefore, the expressions are equivalent. 3. For \(x=4.5\), \(3\cdot4.5-3=13.5-3=10.5\). Both expressions have this value.

1. \(A=5x-2x-3=3x-3=B\). 2. For \(x=4.5\), both expressions equal \(10.5\).

- Write a separate expression for each rule. - In Sophie's rule, which operation happens first? - Use the distributive property to make the expressions easier to compare. - When do two expressions represent the same rule for every input?

1. Leon's rule is represented by \(3x + 6\). 2. Sophie's rule is represented by \(3(x + 2)\). 3. Using the distributive property, \(3(x + 2) = 3x + 6\). 4. The expressions are equivalent, so the rules produce the same result for every \(x\).

Yes. Leon's rule is \(3x + 6\), and Sophie's rule is \(3(x + 2) = 3x + 6\). Therefore, the rules are equivalent.

- Expand the parentheses first. - Combine the terms containing \(x\) separately from the constants. - What does it mean for two expressions to be equivalent?

1. Simplify Expression 1: \(3(4x - 5) + 8 = 12x - 15 + 8 = 12x - 7\). 2. Simplify Expression 2: \(2x + 5(2x - 1) - 2 = 2x + 10x - 5 - 2 = 12x - 7\). 3. Since both simplify to \(12x - 7\), the expressions are equivalent.

Yes. The expressions are equivalent because both simplify to \(12x - 7\).

- Substitute the given value into each expression separately. - What must be true for two expressions to be equivalent? - Use the distributive property to expand \(T_1\).

1. For \(x = 4\), \(T_1 = 3(4 + 2) = 18\). 2. For \(x = 4\), \(T_2 = 3 \cdot 4 + 2 = 14\). 3. Since the expressions have different values for the same input, they are not equivalent. 4. The distributive property confirms this: \(3(x + 2) = 3x + 6\), which is not \(3x + 2\).

a) \(T_1 = 18\) and \(T_2 = 14\) b) The expressions are not equivalent. The distributive property gives \(3(x + 2) = 3x + 6\), not \(3x + 2\).

- Evaluate both expressions for each input. - Compare the two outputs in each row. - Distribute the \(2\) and combine like terms in \(B\).

1. Evaluating both expressions gives matching values for every tested input. 2. Expanding and combining like terms in \(B\) gives \(B=2x+8+3x-5=5x+3\). 3. Therefore, \(A\) and \(B\) are equivalent expressions.

<table> <tr><th>\(x\)</th><th>\(A\)</th><th>\(B\)</th></tr> <tr><td>\(-2\)</td><td>\(-7\)</td><td>\(-7\)</td></tr> <tr><td>\(-1\)</td><td>\(-2\)</td><td>\(-2\)</td></tr> <tr><td>\(0\)</td><td>\(3\)</td><td>\(3\)</td></tr> <tr><td>\(1\)</td><td>\(8\)</td><td>\(8\)</td></tr> <tr><td>\(2\)</td><td>\(13\)</td><td>\(13\)</td></tr> </table> The expressions are equivalent because \(2(x+4)+3x-5=5x+3\).

- In Step 1, identify the property that distributes a factor across a sum. - Compare the order of the addends in Steps 1 and 2. - In Step 3, notice that \(x\) is factored out of two terms.

1. Step 1 uses the distributive property: \(4(x+5)=4x+4\cdot5\). 2. Step 2 uses the commutative property of addition to place the like terms \(4x\) and \(6x\) next to each other. 3. Step 3 uses the distributive property in reverse to factor out \(x\): \(4x+6x=(4+6)x\). 4. Step 2 groups the like terms together, making it easier to combine them in the next step.

Step 1: distributive property Step 2: commutative property of addition Step 3: distributive property Step 2 places the like terms next to each other so they can be combined easily.

- Represent the original number with a variable. - Group a number with its opposite. - Remember that zero is not positive.

1. For any number \(n\), \(n+12+(-n)=12+[n+(-n)]=12\), so a) is true. 2. A number and its opposite sum to \(0\), which is neither positive nor negative. Therefore, b) is false.

a) True; \(n+12+(-n)=12\). b) False; a number and its opposite sum to \(0\).

- Look for pairs that make convenient hundreds. - Addition allows addends to be reordered and regrouped. - Combine the negative and positive partial sums last.

1. Use the commutative property to reorder and the associative property to regroup: \([-240+(-60)]+[115+85]\). 2. The partial sums are \(-300\) and \(200\). 3. Therefore, the value is \(-300+200=-100\).

The value is \(-100\). The commutative and associative properties of addition are used.

- Look for pairs that make convenient hundreds. - Keep each sign attached to its number when reordering. - Use the commutative and associative properties.

1. Reorder and regroup as \((-123-77)+(57+43)+(265-65)\). 2. The partial sums are \(-200\), \(100\), and \(200\). 3. Therefore, the value is \(-200+100+200=100\).

\(100\)

- Write the percent increase as a decimal. - Express each class's amount separately before adding. - Combine like terms when simplifying the total. - Distinguish variable terms from fixed amounts.

1. A \(20\%\) increase gives \(x+0.20x=1.20x\). 2. Class C raises \(1.20x-50\). 3. Add the three amounts: \(x+1.20x+(1.20x-50)=3.40x-50\).

1. \(1.20x\), or \(x+0.20x\) 2. \(1.20x-50\) 3. \(3.40x-50\)

- “\(k\) dollars less” means subtract \(k\) from the other price. - Add the two individual prices to find the total. - Combine terms with the same variable. - In part d), substitute the same value for both variables.

1. The eraser costs \(f-k\) dollars. 2. The total price is \(f+(f-k)=2f-k\). 3. Substitute the values: \(2\cdot 8.50-5.65=17.00-5.65=11.35\). 4. If \(k=f\), then the eraser price is \(f-f=0\), so the eraser would be free.

a) \(f-k\) dollars b) \(2f-k\) dollars c) \(\$11.35\) d) The eraser would cost \(\$0\), so it would be free.

- Multiply each lunch price by the number of lunches sold. - Add \(0.50\) to each original price. - The increase applies to all \(135\) lunches. - Substitute the original prices only after writing the new expression.

1. The original revenue is \(15x+120y\). 2. The new prices are \(x+0.50\) and \(y+0.50\), so the new revenue is \(15(x+0.50)+120(y+0.50)\). 3. Distribute and combine constants: \(15x+7.50+120y+60=15x+120y+67.50\). 4. With original prices \(x=5.50\) and \(y=3.50\), the new prices are \(6.00\) and \(4.00\). The revenue is \(15\cdot 6+120\cdot 4=90+480=570\).

a) \(15x+120y\) b) \(15(x+0.50)+120(y+0.50)\), or \(15x+120y+67.50\) c) \(\$570.00\)

- Expand parentheses and combine like terms before comparing expressions. - Follow the order of operations. - To show that two expressions are not equivalent, one input that gives different values is enough.

1. Pair A: \(2(3x + 4) - 2x = 6x + 8 - 2x = 4x + 8\), so the pair is equivalent. 2. Pair B: \(8a \div 2 = 4a\), so the pair is equivalent. 3. Pair C: \(5\) and \(3b\) are not like terms, so \(5 + 3b\) cannot simplify to \(8b\). For example, when \(b = 0\), the values are \(5\) and \(0\). The pair is not equivalent.

Pair A: Equivalent; both expressions simplify to \(4x + 8\). Pair B: Equivalent; \(8a \div 2 = 4a\). Pair C: Not equivalent; for example, when \(b = 0\), the values are \(5\) and \(0\).

- The base fee is added once and is not multiplied by usage. - Translate an increase as addition and a decrease as subtraction. - Keep the new unit price in parentheses before multiplying by \(k\).

1. The usage charge is \(kp\), so the original monthly cost is \(G+kp\). 2. The new base fee is \(G+5\), and the new unit price is \(p-0.02\). The new cost is \((G+5)+k(p-0.02)\). 3. Distribute and simplify: \((G+5)+k(p-0.02)=G+5+kp-0.02k\).

a) \(G+kp\) dollars b) \((G+5)+k(p-0.02)\), or \(G+5+kp-0.02k\), dollars

- 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.

- Reversing a difference gives its opposite. - In b), remove the parentheses and combine the constants. - Use the result from a) to evaluate c).

1. Since \(y - x = -(x - y)\), part a) equals \(-7\). 2. Remove the parentheses in part b): \(x + 10 - y - 10 = x - y = 7\). 3. Use part a) in part c): \(2(y - x) = 2(-7) = -14\).

a) \(-7\) b) \(7\) c) \(-14\)

- Substitute each pair of values into all three expressions. - Use the distributive property on \(T_1\). - In \(T_3\), both variables are multiplied by \(2\). - Compare the three results for each part.

1. For \(a = 5\) and \(b = 3\): \(T_1 = 4\), \(T_2 = 7\), and \(T_3 = 4\). 2. For \(a = 1.5\) and \(b = 2\): \(T_1 = -1\), \(T_2 = 1\), and \(T_3 = -1\). 3. For \(a = \frac{1}{2}\) and \(b = \frac{1}{4}\): \(T_1 = \frac{1}{2}\), \(T_2 = \frac{3}{4}\), and \(T_3 = \frac{1}{2}\). 4. The results suggest that \(T_1\) and \(T_3\) are equivalent. The distributive property proves it: \(2(a - b) = 2a - 2b\). The values above provide counterexamples showing that \(T_2\) is not equivalent to them.

a) \(T_1 = 4\), \(T_2 = 7\), \(T_3 = 4\) b) \(T_1 = -1\), \(T_2 = 1\), \(T_3 = -1\) c) \(T_1 = \frac{1}{2}\), \(T_2 = \frac{3}{4}\), \(T_3 = \frac{1}{2}\) \(T_1\) and \(T_3\) are equivalent because \(2(a - b) = 2a - 2b\). \(T_2\) is not equivalent to them.

- Treat the missing expression as an unknown. - Subtract the target expression from the first expression. - Combine each variable separately.

1. Let the missing expression be \(T\): \((12m-5n+2k)-T=7m+n-4k\). 2. Subtract the result from the first expression: \(T=(12m-5n+2k)-(7m+n-4k)\). 3. Distribute the negative sign and combine like terms: \(T=12m-5n+2k-7m-n+4k=5m-6n+6k\).

\(5m-6n+6k\)

- Distribute each factor to every term inside its parentheses. - Reverse the signs in a subtracted group. - Group terms by variable before combining coefficients.

1. Distribute both factors: \(\frac{2}{3}(6a-9b)=4a-6b\) and \(\frac{1}{4}(8a+12b)=2a+3b\). 2. Subtract and combine like terms: \((4a-6b)-(2a+3b)=2a-9b\). 3. For part 2, remove the parentheses: \(x-\frac{1}{2}y+\frac{1}{3}z-\frac{1}{2}x-y+\frac{2}{3}z+\frac{1}{4}x-\frac{1}{2}y-z\). 4. Combine coefficients: \(1-\frac{1}{2}+\frac{1}{4}=\frac{3}{4}\) for \(x\), \(-\frac{1}{2}-1-\frac{1}{2}=-2\) for \(y\), and \(\frac{1}{3}+\frac{2}{3}-1=0\) for \(z\). 5. The result for part 2 is \(\frac{3}{4}x-2y\).

1) \(2a-9b\) 2) \(\frac{3}{4}x-2y\)

- Work from the innermost grouping symbols outward. - Reverse every sign when subtracting a grouped expression. - Rewrite the entire expression after removing each set of grouping symbols. - Combine only terms with the same variable part.

1. Remove the innermost parentheses: \(10a-\{3b-[4a-5b+2c+a-c]\}\). 2. Combine like terms inside the brackets: \(10a-\{3b-[5a-5b+c]\}\). 3. Subtract the bracketed expression: \(10a-\{3b-5a+5b-c\}\). 4. Combine like terms inside the braces: \(10a-\{8b-5a-c\}\). 5. Subtract the expression in braces and combine like terms: \(10a-8b+5a+c=15a-8b+c\).

\(15a-8b+c\)

- Work from the innermost grouping symbols outward. - A negative sign before parentheses or brackets reverses every sign inside. - Simplify inside the brackets before removing them. - Combine only terms with the same variable.

1. Simplify inside the brackets: \(3y-2x-5y+7x=5x-2y\). 2. Substitute this result into the full expression: \(4x-[5x-2y]-(x-2y)\). 3. Remove the brackets and parentheses: \(4x-5x+2y-x+2y\). 4. Combine like terms to obtain \(-2x+4y\).

\(-2x+4y\)

- Decide how the terms not in the first group must appear after a negative sign is distributed. - Expand your rewritten expression to check whether it matches the original. - Remember that a negative sign outside parentheses changes every sign inside.

1. Group \(20u\) and \(10w\) in the first set of parentheses: \((20u+10w)\). 2. The remaining terms are \(-15v\) and \(-5\). 3. To place a subtraction sign before the second group, write \(-(15v+5)\). Distributing the negative sign reproduces \(-15v-5\). 4. Therefore, the expression is \((20u+10w)-(15v+5)\). 5. The \(5\) is positive inside the second group because the negative sign outside the parentheses changes it to \(-5\) when distributed.

a) \((20u+10w)-(15v+5)\) b) The negative sign outside the parentheses changes \(+5\) to \(-5\) when the parentheses are removed, preserving the original expression.

- Represent the unknown addend with a variable or placeholder. - Subtract the known addend from the complete expression. - Distribute the negative sign before combining like terms.

1. Let the missing addend be \(M\). Then \((x-3y)+M=4x-7y+2\). 2. Subtract the given addend from the target expression: \(M=(4x-7y+2)-(x-3y)\). 3. Distribute the negative sign and combine like terms: \(M=4x-7y+2-x+3y=3x-4y+2\). 4. Therefore, \(4x-7y+2=(x-3y)+(3x-4y+2)\).

\((x-3y)+(3x-4y+2)\)

- Consider simplifying the expression inside parentheses before applying the negative sign. - Opposite terms, such as \(t\) and \(-t\), cancel. - Rewrite the full expression after each step to keep track of signs.

1. Substitute the expressions: \((3r-5s+2t)-(-r+4s-3t)+(2r-s+t)\). Remove the parentheses and combine like terms: \(3r-5s+2t+r-4s+3t+2r-s+t=6r-10s+6t\). 2. First find \(u+v=(3r-5s+2t)+(-r+4s-3t)=2r-s-t\). Then \(-(u+v)-w=-(2r-s-t)-(2r-s+t)\). Remove the parentheses and combine like terms: \(-2r+s+t-2r+s-t=-4r+2s\).

1) \(6r-10s+6t\) 2) \(-4r+2s\)

- Work from the innermost grouping symbols outward. - Simplify the expression in \(a\) and \(b\) before substituting. - After substituting, use the distributive property. - Reverse every sign when subtracting a grouped expression.

1. Simplify the inner grouping: \(3a-[2b-a+b]=3a-[3b-a]\). 2. Remove the brackets: \(3a-3b+a=4a-3b\). 3. Substitute \(a=2x+y\) and \(b=x-y\): \(4(2x+y)-3(x-y)\). 4. Distribute: \(8x+4y-3x+3y\). 5. Combine like terms: \(5x+7y\).

\(5x+7y\)

- Write the condition in part b as an equation involving \(W\). - Simplify \(X+Y\) before finding \(W\). - When subtracting an entire expression, use parentheses and reverse every sign inside.

1. For part a, substitute the expressions: \((3a-4b)-(5a+2b)+(-a+6b)\). 2. Remove the parentheses and combine like terms: \(3a-5a-a-4b-2b+6b=-3a\). 3. For part b, solve the condition \((X+Y)+W=Z\) for \(W\): \(W=Z-(X+Y)\). 4. Find \(X+Y=(3a-4b)+(5a+2b)=8a-2b\). 5. Substitute and simplify: \(W=(-a+6b)-(8a-2b)=-a-8a+6b+2b=-9a+8b\).

a) \(-3a\) b) \(W=-9a+8b\)

- After removing the grouping symbols, check what happens to the \(x\)-terms. - Look at which variables remain in the simplified expression. - Decide whether you need to substitute \(x=100\) after simplifying.

1. Remove the inner parentheses: \(T=5x+3y-[3x-y+2x]\). 2. Combine like terms inside the brackets: \(T=5x+3y-[5x-y]\). 3. Remove the brackets and combine like terms: \(T=5x+3y-5x+y=4y\). 4. The \(x\)-terms cancel, so changing \(x\) does not affect the value of \(T\). 5. For \(y=-2.5\), \(T=4\cdot(-2.5)=-10\).

a) \(4y\) b) The \(x\)-terms cancel, so the simplified expression contains no \(x\). c) \(-10\)

- 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 \(130\,\text{m}\).

- Write the first three side lengths one at a time. - The perimeter is the sum of all four sides. - Distribute the subtraction sign carefully when removing the sum of the known sides.

1. The second side is \(s_2=(2a+b)+a=3a+b\). 2. The third side is \(s_3=(3a+b)-2b=3a-b\). 3. The sum of the first three sides is \((2a+b)+(3a+b)+(3a-b)=8a+b\). 4. Subtract from the perimeter: \(s_4=(10a+6b)-(8a+b)=2a+5b\).

\(s_4=2a+5b\)

- Multiply the first-side expression by \(\frac{1}{2}\) to find the second side. - Add the three known side expressions before subtracting from the total length. - Evaluate both side expressions at \(y=10\) before comparing.

1. The second side is \(\frac{1}{2}(5y+4)=2.5y+2\). 2. The sum of the three known sides is \((5y+4)+(2.5y+2)+(3y+6)=10.5y+12\). 3. The fourth side is \(s_4=(16y+20)-(10.5y+12)=5.5y+8\). 4. For \(y=10\), the first side is \(5\cdot 10+4=54\,\text{cm}\), and the fourth side is \(5.5\cdot 10+8=63\,\text{cm}\). Therefore, the fourth side is longer.

a) \(s_4=5.5y+8\) centimeters b) Yes. When \(y=10\), \(s_4=63\,\text{cm}\) and \(s_1=54\,\text{cm}\).

- Equivalent expressions have the same value for every input. - Simplify the more complicated expression first. - Use the distributive property on both products. - Pay close attention to the negative factor \(-4\).

1. Expand \(T_1\): \(3(2a + b) - 4(a - b) = 6a + 3b - 4a + 4b\). 2. Combine like terms: \(6a - 4a + 3b + 4b = 2a + 7b\). 3. This is exactly \(T_2\), so the expressions are equivalent.

Yes. The expressions are equivalent because \(T_1\) simplifies to \(2a + 7b\).

- Use parentheses when substituting an expression for a variable. - A minus sign before parentheses changes every sign inside. - Combine the \(a\)-terms and constants separately.

1. \(5(2a + 3) + 2(a - 5) = 10a + 15 + 2a - 10 = 12a + 5\). 2. \(3(2a + 3) - 4(a - 5) = 6a + 9 - 4a + 20 = 2a + 29\). 3. \(2\left((2a + 3) - (a - 5)\right) = 2(a + 8) = 2a + 16\).

1) \(12a + 5\) 2) \(2a + 29\) 3) \(2a + 16\)

- Find the interest as a percent of the original balance. - The ending balance must be greater than the beginning balance. - A growth factor can combine the original balance and the increase. - Substitute each set of values into your expression.

1. The balance after one year is \(K\left(1 + \frac{p}{100}\right)\). An equivalent expression is \(K + \frac{Kp}{100}\). 2. For case 1, \(\$2500\left(1 + \frac{2}{100}\right) = \$2500 \cdot 1.02 = \$2550\). 3. For case 2, \(\$12{,}000\left(1 + \frac{1.4}{100}\right) = \$12{,}000 \cdot 1.014 = \$12{,}168\).

Expression: \(K\left(1 + \frac{p}{100}\right)\) 1) \(\$2550\) 2) \(\$12{,}168\)

- A discount subtracts part of the original price. - After a discount of \(r\%\), what percent of the original price remains? - Follow the order of operations when substituting values.

1. A discount of \(r\%\) leaves \(1 - \frac{r}{100}\) of the original price, so the sale price is \(x\left(1 - \frac{r}{100}\right)\). An equivalent expression is \(x - \frac{xr}{100}\). 2. For case 1, \(\$45\left(1 - \frac{20}{100}\right) = \$45 \cdot 0.80 = \$36\). 3. For case 2, \(\$129\left(1 - \frac{15}{100}\right) = \$129 \cdot 0.85 = \$109.65\).

Expression: \(x\left(1 - \frac{r}{100}\right)\) 1) \(\$36\) 2) \(\$109.65\)

- What must be true of two expressions if the rules always give the same result? - Use parentheses around each quantity that is multiplied. - Apply the distributive property to both expressions. - Compare the simplified expressions.

1. Lukas’s rule gives \(5(n+4)=5n+20\). 2. Sophie’s rule gives \(2.5(2n+8)=5n+20\). 3. Both rules simplify to the same expression, \(5n+20\). Therefore, they give the same result for every starting number.

Lukas: \(5(n+4)=5n+20\) Sophie: \(2.5(2n+8)=5n+20\) The rules are equivalent, so they always give the same result.

- Express the downstream and upstream speeds first. - Subtract the entire upstream expression. - Distribute the negative sign carefully. - Check whether one variable cancels.

1. The downstream speed is \(v+c\), and the upstream speed is \(v-c\). Their difference is \((v+c)-(v-c)\). 2. Simplifying gives \(v+c-v+c=2c\). The difference is twice the current speed and does not depend on the boat's still-water speed. 3. The downstream speed is \(22+4=26\) miles per hour, and the upstream speed is \(22-4=18\) miles per hour. Their difference is \(26-18=8\) miles per hour, which equals \(2\cdot 4=8\) miles per hour.

1) \((v+c)-(v-c)\) 2) \(2c\); the difference is twice the current speed. 3) \(8\) miles per hour

- Add the wind speed for a tailwind and subtract it for a headwind. - Look for terms that cancel when the two expressions are added. - Simplify the sum before dividing by \(2\). - Compare either ground speed with the still-air speed to find \(w\).

1. With a tailwind, \(v_T=v+w\). With a headwind, \(v_H=v-w\). 2. Their mean is \(\frac{(v+w)+(v-w)}{2}=\frac{2v}{2}=v\). It equals the airplane's still-air speed. 3. The still-air speed is \(\frac{520+440}{2}=480\) miles per hour. 4. The wind speed is \(520-480=40\) miles per hour.

a) \(v_T=v+w\); \(v_H=v-w\) b) \(v\), the airplane's still-air speed c) Still-air speed: \(480\) miles per hour; wind speed: \(40\) miles per hour

- Begin with the innermost grouping symbols. - Reverse every sign when subtracting a grouped expression. - Combine only terms with the same variable. - Rewrite the entire expression after each step.

1. Simplify the innermost grouping: \(2a-(5b-3a)=2a-5b+3a=5a-5b\). 2. Simplify inside the braces: \(4b+(5a-5b)-2b=5a-3b\). 3. Subtract the expression in braces: \(15a-(5a-3b)=15a-5a+3b\). 4. Combine like terms to obtain \(10a+3b\).

\(10a+3b\)

- Work from the innermost grouping symbols outward. - Remember the negative sign before the first parentheses. - Group terms with the same variable before combining them. - Check every sign after removing the grouping symbols.

1. Simplify the innermost grouping: \(5v-(u+3v)=5v-u-3v=-u+2v\). 2. Continue inside the braces: \(3u-(-u+2v)=4u-2v\). 3. Subtract the remaining expression: \((4u-2v)-(2u-v)=4u-2v-2u+v=2u-v\). 4. Simplify the first term: \(-(6u-2v)=-6u+2v\). 5. Combine the two results: \((-6u+2v)+(2u-v)=-4u+v\).

\(-4u+v\)

- Reversing the sign before parentheses requires reversing every sign inside. - Remove the parentheses from both sides in part b and compare the resulting terms. - Use the commutative property of addition when comparing term order.

1. For part a, reverse every sign inside when changing subtraction to addition: \(20+(-x+10)\). 2. For part b, the left side becomes \(a+b-c\). The right side becomes \(a-c+b\). By the commutative property of addition, these are equivalent. 3. For part c, reverse every sign inside when changing addition to subtraction: \(m^2-(n-1)\). Expanding gives \(m^2-n+1\), the same as the original expression.

a) \(20+(-x+10)\) b) Left side: \(a+b-c\); right side: \(a-c+b\). The terms are the same, so the expressions are equivalent. c) \(m^2-(n-1)\)

- 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: \(2 \cdot 1.2 = 2.4\) meters of long strips and \(2 \cdot 0.85 = 1.7\) meters of short strips. It adds these to get \(4.1\) meters per frame, then multiplies by \(4\). 2. Tree a) gives \(4(2.4 + 1.7) = 4 \cdot 4.1 = 16.4\,\text{m}\). 3. Tree b) uses \(8\) long strips and \(8\) short strips for all four frames. The partial lengths are \(8 \cdot 1.2 = 9.6\,\text{m}\) and \(8 \cdot 0.85 = 6.8\,\text{m}\). 4. Tree b) gives \(9.6 + 6.8 = 16.4\,\text{m}\). 5. The distributive property shows the methods are equivalent. The expressions are \(4(2 \cdot 1.2 + 2 \cdot 0.85)\) and \(8 \cdot 1.2 + 8 \cdot 0.85\).

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; \(4(2 \cdot 1.2 + 2 \cdot 0.85)\) and \(8 \cdot 1.2 + 8 \cdot 0.85\).

- What does the initial \(-3\) in each expression represent on the graph? - Find the temperature change from one hour to the next. - Identify how Clara grouped increases and decreases. - Identify the longer intervals in which Jonas combined changes.

1. Lucas's expression gives \(-3-1+2+3+4+2+0-1-3-4-2=-3\). 2. Clara's expression gives \(-3+(2+3+4+2)-(1+1+3+4+2)=-3+11-11=-3\). 3. Jonas's expression gives \(-3-1+11-10=-3\). Therefore, all three expressions give a final temperature of \(-3\,^\circ\text{C}\). 4. Lucas lists every hourly change in order. Clara groups all increases together and all decreases together. Jonas groups consecutive phases: an initial \(1\)-degree decrease, an \(11\)-degree increase, no change, and a final \(10\)-degree decrease.

a) Each expression equals \(-3\), so the temperature at \(4{:}00\) p.m. is \(-3\,^\circ\text{C}\). b) Lucas uses each hourly change. Clara groups positive changes and negative changes. Jonas combines consecutive changes into longer warming and cooling phases.

- Distribute the factor in \(A\) and compare the resulting terms with \(B\). - Pay close attention to the product of two negative numbers. - Find the value of the first two terms in \(B\), then determine its additive inverse.

1. Distributing \(-\frac{3}{4}\) in \(A\) gives \(-\frac{3}{4} \cdot 2.4 - \left(-\frac{3}{4} \cdot 6.4\right) + 1.5\), which simplifies to \(B\). 2. Evaluate \(A\): \(2.4 - 6.4 = -4\), so \(-\frac{3}{4}(-4) + 1.5 = 3 + 1.5 = 4.5\). 3. The part of \(B\) before the final number equals \(3\). Solve \(3 + x = 0\), giving \(x = -3\).

1. The distributive property shows that the expressions are equivalent. 2. \(A = 4.5\) 3. Replace \(1.5\) with \(-3\).

- Identify the operation performed last under the order of operations. - Look for a common factor in the two products inside the brackets. - Evaluate the exponent before the final subtraction. - To divide by a fraction, multiply by its reciprocal.

1. The expression is a difference at its highest level because subtracting \((-2)^3\) is the final operation. 2. Factor \(\frac{2}{3}\) inside the brackets: \(\left(-\frac{3}{4} - \frac{1}{4}\right)\frac{2}{3} = (-1) \cdot \frac{2}{3} = -\frac{2}{3}\). 3. Divide: \(\frac{7}{2} \div \left(-\frac{2}{3}\right) = \frac{7}{2} \cdot \left(-\frac{3}{2}\right) = -\frac{21}{4}\). 4. Evaluate the exponent: \((-2)^3 = -8\). 5. Subtract: \(-\frac{21}{4} - (-8) = -\frac{21}{4} + \frac{32}{4} = \frac{11}{4} = 2.75\).

It is a difference at its highest level. Its value is \(\frac{11}{4}\), or \(2.75\).

- Equivalent expressions simplify to the same expression. - A minus sign before parentheses changes the sign of every term inside. - For part b), substitute \(x = 2\) into \(11x\), then determine the constant needed to make the total \(0\).

1. Simplify \(T_1\): \(2(3x - 4) + 5x = 6x - 8 + 5x = 11x - 8\). 2. Simplify \(T_2\): \(12x - (x + 8) = 12x - x - 8 = 11x - 8\). 3. Therefore, \(T_1\) and \(T_2\) are equivalent. Since \(T_3 = 11x - 4\), it is not equivalent to them. 4. At \(x = 2\), the variable term is \(11 \cdot 2 = 22\). To make the value \(0\), the constant term must be \(-22\). The revised expression is \(11x - 22\).

a) \(T_1\) and \(T_2\) are equivalent; both simplify to \(11x - 8\). \(T_3\) is not equivalent to them. b) \(11x - 22\)

- Simplify each expression separately. - Compare both the variable coefficient and the constant term. - Identify which number can be changed to correct the variable coefficient without changing the constant term.

1. Simplify \(T_1\): \(0.5(12y-8)+3y=6y-4+3y=9y-4\). 2. Simplify \(T_2\): \(3(3y-2)-y+2=9y-6-y+2=8y-4\). 3. The expressions are not equivalent because their variable coefficients differ. 4. One valid change is to replace the \(3\) multiplying \(y\) inside the parentheses with \(\frac{10}{3}\): \(T_3=3\left(\frac{10}{3}y-2\right)-y+2\). 5. Simplify to check: \(T_3=10y-6-y+2=9y-4\), so \(T_3\) is equivalent to \(T_1\).

a) The claim is false. \(T_1=9y-4\), while \(T_2=8y-4\). b) One possible answer is \(T_3=3\left(\frac{10}{3}y-2\right)-y+2\).

- Simplify the first expression completely. - Expand the second expression so the two forms can be compared. - What value of \(a\) makes the constant terms equal?

1. Simplify \(T_1\): \(5k - 2(k - 6) = 5k - 2k + 12 = 3k + 12\). 2. Expand \(T_2\): \(3(k + a) = 3k + 3a\). 3. For the expressions to be equivalent, the constant terms must match: \(3a = 12\). 4. Divide by \(3\): \(a = 4\).

\(a = 4\)

- Represent the original number with a variable. - Pair one copy of the number with one copy of its opposite. - Think of opposites as reflections across zero.

1. Let the number be \(n\). Then \(n+2(-n)=n-2n=-n\), so a) is true. 2. Every integer has exactly one opposite. If two integers had the same opposite, taking the opposite again would show that the integers are equal. Therefore, b) is false.

a) True; \(n+2(-n)=-n\). b) False; each integer has exactly one opposite.

- Find the active-ingredient volume in one liter of concentrate. - Find the total volume after adding water. - Write the active-ingredient portion as a percent of the total volume. - Substitute the given values to test the claim.

1. One liter of concentrate contains \(\frac{k}{100}\,\text{L}\) of active ingredient. 2. The total volume is \((1+n)\,\text{L}\). 3. The concentration as a percent is \(\frac{\frac{k}{100}}{1+n} \cdot 100\% = \frac{k}{1+n}\%\). 4. For \(k = 20\) and \(n = 1\), the concentration is \(\frac{20}{2}\% = 10\%\). 5. For \(k = 20\) and \(n = 2\), the concentration is \(\frac{20}{3}\% \approx 6.67\%\). 6. Since \(6.67\%\) is not half of \(10\%\), the claim is false.

a) The concentration is \(\frac{k}{1+n}\%\). b) The claim is false. The concentrations are \(10\%\) when \(n = 1\) and approximately \(6.67\%\) when \(n = 2\).