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Write equivalent expressions

Learning goals

Produces expressions equivalent to a given expression.
Uses properties of operations to justify each rewrite.
Chooses a form that reveals useful structure.
Checks equivalence by reasoning or substitution.

Difficulty: 1 – 5

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Write equivalent expressions 1

Fill in the missing number so the expressions have the same value for every \(x\): \(4(x+2)=4x+\square\). Explain why the distributive property makes your answer work for every \(x\).

Hints

- What does the \(4\) multiply inside the parentheses? - Think about the part that does not contain \(x\).

Solution

1. Multiply \(4\) by both terms in the parentheses: \(4(x+2)=4x+4\cdot2\). 2. Since \(4\cdot2=8\), the missing number is \(8\). The distributive property makes the forms equal for every \(x\).

Answer

\(8\), because \(4(x+2)=4x+8\).

Write equivalent expressions 2

Write \(n+n+n+7\) using just one \(n\)-term. Why does your expression have the same value for every \(n\)?

Hints

- Count how many times \(n\) is added. - Does the number without a variable change?

Solution

1. Three copies of \(n\) make \(3n\). 2. The constant \(7\) does not change, so the expression becomes \(3n+7\).

Answer

\(3n+7\); adding three copies of \(n\) is the same as multiplying \(n\) by \(3\).

Write equivalent expressions 3

Are \(2(x+5)\) and \(2x+5\) equivalent? Choose a value of \(x\) that proves your answer.

Hints

- Equivalent expressions agree at every input. - Try a value of \(x\) that makes multiplication especially easy.

Solution

1. Use \(x=0\). The first expression is \(2(0+5)=10\). 2. The second expression is \(2\cdot0+5=5\). 3. The values differ, so the expressions are not equivalent.

Answer

No. For \(x=0\), the values are \(10\) and \(5\).

Write equivalent expressions 4

Theo writes \(9+2y\) and Priya writes \(2y+9\). Do the expressions always give the same value? Name the operation property that explains why.

Hints

- Are the same two terms present in both expressions? - Which addition property allows the order to change?

Solution

1. The terms in the two expressions are identical but appear in the opposite order. 2. Changing the order of terms being added does not change their sum. This is the commutative property of addition.

Answer

Yes. The commutative property of addition gives \(9+2y=2y+9\) for every \(y\).

Write equivalent expressions 5

Without fully evaluating, decide whether each equation is true. Justify your decision using properties of operations. a) \(\frac{4}{9}\left(2.7 + \frac{9}{4}\right) = \frac{4}{9} \cdot 2.7 + 1\) b) \(15.3 - \left(4.2 + \frac{1}{3}\right) = (15.3 - 4.2) + \frac{1}{3}\)

Hints

- 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 \(1\).

Solution

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.

Answer

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

Write equivalent expressions 6

Let \(T = \frac{3}{4} \cdot 1.2 - \frac{3}{4} \cdot 0.4\). Haruto claims, “I can find \(T\) more easily by first subtracting \(0.4\) from \(1.2\), then multiplying the difference by \(0.75\).” Use a property of operations to determine whether the claim is correct. Would the same kind of rule work if the original expression used addition instead of subtraction? Explain.

Hints

- Look for a feature shared by the two products before evaluating them. - Ask whether that shared structure lets you rewrite the expression without changing its value. - Then consider whether the same reasoning depends on the sign between the two products.

Solution

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 Haruto’s method, so the claim is correct. 4. The distributive property also applies to addition: \(ab + ac = a(b + c)\). Therefore, the corresponding rule also works with a plus sign.

Answer

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 \(ab + ac = a(b + c)\).

Write equivalent expressions 7

Write a simpler expression that is equivalent for every value of the variable. Explain your steps. Then evaluate for the given value. a) \(15x-4.5x+0.5x\) for \(x=1.2\) b) \(y(12-5.5)-1.5y\) for \(y=\frac{4}{5}\)

Hints

- Which terms can be combined without knowing the variable's value? - Can you rewrite the expression before substituting a number? - Check that your simpler expression gives the same value as the original.

Solution

1. In a), all terms contain \(x\), so combining their coefficients gives \(15x-4.5x+0.5x=11x\) for every \(x\). 2. Substitute \(x=1.2\): \(11\cdot1.2=13.2\). 3. Simplify b): \(y(12-5.5)-1.5y=6.5y-1.5y=5y\). 4. Substitute \(y=\frac{4}{5}\): \(5\cdot\frac{4}{5}=4\).

Answer

a) \(11x\); value \(13.2\). b) \(5y\); value \(4\).

Write equivalent expressions 8

Determine whether each pair of expressions is equivalent. Show your reasoning by simplifying both expressions. a) \(4(a+2)+3a\) and \(7a+8\) b) \(15y-6y+2\) and \(3(3y+2)-4\)

Hints

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

Solution

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.

Answer

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

Write equivalent expressions 9

Which expressions are equivalent to \(4x+10\)? Simplify each expression to justify your choices. a) \(2x+10+2x\) b) \(2(2x+5)\) c) \(15x-11x+12-2\) d) \(4(x+10)\) e) \(x+x+x+x+7+3\)

Hints

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

Solution

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.

Answer

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

Write equivalent expressions 10

Two expressions are equivalent if they have the same value for every value of the variable. Determine by simplifying whether \(T_1 = 4(k + 2) - 3k - 8\) and \(T_2 = k\) are equivalent. Then find the value of \(T_1\) when \(k = 12.5\).

Hints

- 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.

Solution

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 \(k = 12.5\), \(T_1 = k = 12.5\).

Answer

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

Write equivalent expressions 11

Which expressions have the same value as \(6x-12\) for every value of \(x\)? Select all that apply. a) \(3(2x-4)\) b) \(6(x-2)\) c) \(10x-12-4x\) d) \(2x+4x-6\cdot2\) e) \(6x-12\) f) \(x(6-12)\) g) \(12x\div2-12\)

Hints

- 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.

Solution

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.

Answer

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

Write equivalent expressions 12

Determine whether each statement is correct. For each incorrect statement, describe the error and give the correct result. a) \(7x-x=7\) b) \(\frac{x}{3}+\frac{x}{3}=\frac{2x}{3}\) c) \(3(2a+4)=6a+4\) d) \(4y+3y=7y\)

Hints

- 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.

Solution

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 \(4y+3y=(4+3)y=7y\).

Answer

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

Write equivalent expressions 13

Simplify each expression completely. a) \(12x-5y+(3x-4y)\) b) \(8a-(2a-7b)-10b\) c) \(-15m+4n-(-6m+4n)\)

Hints

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

Solution

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: \(-15m+4n+6m-4n=-9m\).

Answer

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

Write equivalent expressions 14

Rewrite each expression as a sum or difference so that no coefficient has an absolute value greater than \(1\). a) \(5x\) b) \(3a^2+2b\) c) \(4m-3n\) d) \(2(p+q)\)

Hints

- 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.

Solution

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: \(2(p+q)=(p+q)+(p+q)\).

Answer

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

Write equivalent expressions 15

A wire-frame square pyramid has a square base with side length \(a\) and four lateral edges of length \(s\). 1. Write an expression for the total wire length \(L\). 2. Simplify the expression when each lateral edge is twice as long as a base edge.

Hints

- 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.

Solution

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 \(L=4a+4(2a)=4a+8a=12a\).

Answer

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

Write equivalent expressions 16

Rewrite each expression in the requested equivalent form. a) Write \(p - q - r - 10\) as a sum of four terms. b) Write \(15 + a + b\) using subtraction signs between all terms. c) Write \(x - (-y) - z\) using only addition between terms.

Hints

- 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.

Solution

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 \(-z\): \(x + y + (-z)\).

Answer

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

Write equivalent expressions 17

For positive \(x\) and \(y\), Zofia claims that the square and triangle shown have equal perimeters. The drawings are not to scale. Check her claim by writing and simplifying an expression for each perimeter.

Figure for problem 522910

Hints

- Does a square need four different side expressions? - Can both perimeter expressions be written without parentheses? - Compare the coefficients of \(x\) and \(y\), not just one numerical example.

Solution

1. The square's perimeter is \(4(1.5x+2y)=6x+8y\). 2. The triangle's perimeter is \((2x+3y)+(3x+3y)+(x+2y)\). 3. Combining like terms gives \(6x+8y\). 4. The perimeters match for all positive \(x\) and \(y\), and the three triangle lengths satisfy the triangle inequalities for such inputs.

Answer

Square: \(4(1.5x+2y)=6x+8y\). Triangle: \((2x+3y)+(3x+3y)+(x+2y)=6x+8y\). The claim is correct for all positive \(x\) and \(y\).

Write equivalent expressions 18

For values of \(x\) and \(y\) that make the lengths positive and form a valid triangle, the side lengths are \(5x+2y\), \(3x-4y\), and \(2x+3y\). Write and simplify an expression for the perimeter \(P\).

Hints

- A triangle’s perimeter is the sum of its side lengths. - Combine all \(x\)-terms and all \(y\)-terms separately. - Keep the negative sign on \(-4y\).

Solution

1. Add the three side lengths: \(P=(5x+2y)+(3x-4y)+(2x+3y)\). 2. Combine the \(x\)-terms: \(5x+3x+2x=10x\). 3. Combine the \(y\)-terms: \(2y-4y+3y=y\). 4. Therefore, \(P=10x+y\).

Answer

\(P=10x+y\)

Write equivalent expressions 19

For values of \(x\) and \(y\) that make the lengths positive and form a valid quadrilateral, its side lengths are \(s_1=3x+2\) \(s_2=2y-1\) \(s_3=x+y+4\) \(s_4=2x-y+3\). Write and simplify an expression for the perimeter \(P\).

Hints

- Perimeter is the sum of all side lengths. - Combine the \(x\)-terms, \(y\)-terms, and constants separately. - Keep each term’s sign.

Solution

1. Add all four side lengths: \(P=(3x+2)+(2y-1)+(x+y+4)+(2x-y+3)\). 2. Combine the \(x\)-terms: \(3x+x+2x=6x\). 3. Combine the \(y\)-terms: \(2y+y-y=2y\). 4. Combine the constants: \(2-1+4+3=8\). 5. Therefore, \(P=6x+2y+8\).

Answer

\(P=6x+2y+8\)

Write equivalent expressions 20

Simplify each expression and identify which expressions are equivalent. A: \(-15a-(-6a)\) B: \(-4a+(-5a)\) C: \(-(12a-3a)\) D: \(2a-11a\) E: \(-a-(+10a)\)

Hints

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

Solution

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.

Answer

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

Write equivalent expressions 21

Simplify the expression. \(2x-[5x-(3x+4)]\)

Hints

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

Solution

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 \(4\).

Answer

\(4\)

Write equivalent expressions 22

A streaming service offers two monthly plans. Basic: a monthly fee of \(\$G\) plus \(\$k\) for each rented movie. Premium: a flat monthly fee of \(\$P\) with no additional charge per movie. Let \(K_P\) be the Premium plan's monthly cost. a) Write an expression for the Basic plan's monthly cost \(K_B\) when a customer rents \(n\) movies. b) Write an expression for the difference \(K_P-K_B\). c) What does it mean when the expression from part b) equals \(0\)?

Hints

- Add the fixed fee and the per-movie charges for the Basic plan. - Subtract the entire Basic expression from the Premium cost. - A difference of \(0\) means the two quantities are equal.

Solution

1. The Basic plan costs \(K_B=G+nk\). 2. The Premium plan costs \(K_P=P\). 3. The difference is \(K_P-K_B=P-(G+nk)=P-G-nk\). 4. If the difference is \(0\), then \(P=G+nk\), so the plans cost the same for that number of movies.

Answer

a) \(K_B=G+nk\) b) \(P-(G+nk)=P-G-nk\) c) Both plans have the same monthly cost for that number of movies.

Write equivalent expressions 23

Try this number puzzle: Choose a number, multiply it by \(3\), add \(15\), subtract the original number, and divide the result by \(2\). a) What result do you get when the starting number is \(4\)? b) Write and simplify an expression for the puzzle. c) Use the simplified expression to describe how the final result depends on the starting number.

Hints

- 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.

Solution

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.

Answer

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.

Write equivalent expressions 24

Choose a number. Multiply it by \(4\), add \(12\), multiply that sum by \(5\), and subtract \(60\). Once you report the final result, someone can immediately determine your starting number. Explain how to recover the starting number, and justify your rule with an algebraic expression.

Hints

- 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 \(x\)?

Solution

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 \(x\).

Answer

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

Write equivalent expressions 25

Change the sign before each set of parentheses to the opposite sign without changing the value of the expression. 1) \(12-(4a-5b)\) 2) \(x+(2y-3z+1)\) 3) \(3k-(k-m)\)

Hints

- 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.

Solution

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: \(3k+(-k+m)\).

Answer

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

Write equivalent expressions 26

Write the area of the whole rectangle two ways: as a sum of two areas and as one product. Explain why the expressions are equivalent.

Figure for problem 559557

Hints

- Which two smaller areas make up the whole rectangle? - How could you find the area without using the dividing line? - Why must these two ways of finding the area give the same result?

Solution

1. The left part has area \(3x\), and the right part has area \(3\cdot4=12\). 2. Adding the parts gives \(3x+12\). 3. The whole rectangle has width \(x+4\) and height \(3\), so its area is \(3(x+4)\). 4. The expressions are equivalent because \(3(x+4)=3x+12\) by the distributive property.

Answer

Sum: \(3x+12\). Product: \(3(x+4)\). Both give the same area; the distributive property shows that \(3(x+4)=3x+12\).

Write equivalent expressions 27

Three expressions are shown: \(A=6(t-1)\), \(B=6t-6\), and \(C=6t+6\). Which expressions are equivalent? Explain using algebra, not just one test value.

Hints

- Rewrite the expression with parentheses. - Why is a general rewrite stronger than matching at one input? - Compare the constant terms as well as the variable terms.

Solution

1. Expand \(A\): \(6(t-1)=6t-6\). 2. This matches \(B\) for every \(t\). 3. Expression \(C\) has a different constant term. For example, at \(t=0\), \(B=-6\) but \(C=6\).

Answer

\(A\) and \(B\) are equivalent because \(6(t-1)=6t-6\). Expression \(C\) is not equivalent.

Write equivalent expressions 28

Find the number \(c\) so that \(8x-(2x+c)\) is equivalent to \(6x-9\) for every \(x\). Explain why your value works.

Hints

- Simplify the expression with the parentheses without guessing \(c\). - Which terms must match if the expressions agree for every \(x\)?

Solution

1. Remove the parentheses: \(8x-(2x+c)=6x-c\). 2. To match \(6x-9\) for every \(x\), we need \(-c=-9\), so \(c=9\). 3. Substituting gives \(8x-(2x+9)=6x-9\).

Answer

\(c=9\). Both expressions then simplify to \(6x-9\).

Write equivalent expressions 29

The two rectangles could be joined along their equal-height sides. The drawings are not to scale. Write their total area as a sum and as one product. Explain why these forms are equivalent.

Figure for problem 559623

Hints

- Find the area of each smaller rectangle first. - If two rectangles have the same height, how can you describe their combined width? - Check your two expressions using the distributive property.

Solution

1. Both rectangles have height \(3\). Their areas are \(3x\) and \(3\cdot4=12\). 2. Adding separate areas gives \(3x+12\). 3. Together they have width \(x+4\) and height \(3\), giving \(3(x+4)\). 4. Distributing confirms \(3(x+4)=3x+12\).

Answer

Sum: \(3x+12\). Product: \(3(x+4)\). They are equivalent by the distributive property.

Write equivalent expressions 30

Determine whether each equation is true or false. For each false equation, correct the right side so that it is equivalent to the original expression on the left. a) \((-0.2) \cdot 15 + (-0.2) \cdot 5 = (-0.2)(15 + 5)\) b) \(\frac{3}{4} - \left(\frac{1}{2} - \frac{1}{4}\right) = \frac{3}{4} - \frac{1}{2} - \frac{1}{4}\) c) \((-3) \cdot 7 - (-3) \cdot 4 = (-3)(7 - 4)\)

Hints

- 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.

Solution

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.

Answer

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

Write equivalent expressions 31

A shopper uses a \(\$100\) gift card for an online order. The cart contains a game that costs \(\$24.99\), headphones that cost \(\$35.50\), and shipping that costs \(\$x\). a) Write and simplify an expression for the amount left on the gift card. b) Find the amount left when shipping costs \(\$5.95\).

Hints

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

Solution

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 \(x=5.95\), \(39.51-5.95=33.56\).

Answer

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

Write equivalent expressions 32

Consider the expressions \(A = 5x - 15\) and \(B = 5(x - 3)\). a) Evaluate both expressions for \(x = 10\), \(x = 0\), and \(x = -2\). b) What do you notice about the results? Justify your observation using a property of operations.

Hints

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

Solution

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 \(x\).

Answer

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 \(5(x - 3) = 5x - 15\).

Write equivalent expressions 33

Two expressions are equivalent if they have the same value for every value of the variable. Simplify the expressions to determine whether \(T_1\) and \(T_2\) are equivalent. \(T_1 = 4(2z - 3) - (5z - 8)\) \(T_2 = 3(z - 2) + 2\)

Hints

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

Solution

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.

Answer

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

Write equivalent expressions 34

Complete the table, then simplify the expressions to determine which are equivalent. Expression \(A=3(x+4)\) Expression \(B=3x+4\) Expression \(C=5x+12-2x\) <table> <tr> <td>\(x\)</td> <td>\(0\)</td> <td>\(2\)</td> <td>\(5\)</td> </tr> <tr> <td>Expression \(A\)</td> <td></td> <td></td> <td></td> </tr> <tr> <td>Expression \(B\)</td> <td></td> <td></td> <td></td> </tr> <tr> <td>Expression \(C\)</td> <td></td> <td></td> <td></td> </tr> </table> Which expressions are equivalent? Justify your answer by simplifying them.

Hints

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

Solution

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.

Answer

<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 \(3x+12\).

Write equivalent expressions 35

Let \(T=12x-(4x+8)\). Simplify \(T\). Then determine which expressions are equivalent to \(T\). Justify each choice by rewriting the expression. \(T_1=8x+8\) \(T_2=4(2x-2)\) \(T_3=16x-8-8x\) \(T_4=8(x-1)\)

Hints

- 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.

Solution

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.

Answer

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

Write equivalent expressions 36

Luke and Maya simplify \(6x-2(x-4)\). Luke writes: \(6x-2(x-4)=6x-2x-8=4x-8\) Maya writes: \(6x-2(x-4)=6x-2x+8=4x+8\) a) Substitute \(x=0\) into the original expression and into each student’s result. Which student could be correct? b) Explain the error in the incorrect work.

Hints

- 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.

Solution

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 \(-8\).

Answer

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 \(+8\).

Write equivalent expressions 37

Ingrid wrote: \(12b-(5b+7)+3=7b+10\) Determine whether the result is correct. If it is not, identify the error and give the correct simplified expression.

Hints

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

Solution

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. Ingrid failed to change the sign of \(+7\) when removing the parentheses and incorrectly combined the constants as \(7+3\).

Answer

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

Write equivalent expressions 38

Consider these two expressions: Expression A: \(30-x-10-5\) Expression B: \(30-x-(10-5)\) a) Simplify each expression. b) Find \(\text{Expression B}-\text{Expression A}\). Does the result depend on the value of \(x\)? Explain.

Hints

- 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.

Solution

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 \(x\).

Answer

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.

Write equivalent expressions 39

Sort the expressions into two groups of three equivalent expressions. 1. \(6a-12+2a\) 2. \(3(a+4)-a\) 3. \(4(2a-3)\) 4. \(2a+12\) 5. \(10a-12-2a\) 6. \(2(a+6)\)

Hints

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

Solution

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.

Answer

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

Write equivalent expressions 40

A rectangular garden bed has width \(x\) meters. Its length is \(3\,\text{m}\) more than twice its width. a) Write an expression for the perimeter \(P\). b) Simplify the expression. c) Find the total fence length needed when the width is \(4.5\,\text{m}\).

Hints

- Express the length in terms of \(x\). - Use the perimeter formula for a rectangle. - Distribute before combining like terms.

Solution

1. The width is \(x\), and the length is \(2x+3\). 2. The perimeter is \(P=2x+2(2x+3)\). 3. Distribute and combine like terms: \(P=2x+4x+6=6x+6\). 4. For \(x=4.5\), \(P=6\cdot4.5+6=27+6=33\,\text{m}\).

Answer

a) \(P=2x+2(2x+3)\) b) \(P=6x+6\) c) \(33\,\text{m}\)

Write equivalent expressions 41

The expression \(P=4a+2(a+5)\) represents the perimeter of a geometric figure. a) Describe one possible figure and list its side lengths so that its perimeter matches the expression. b) Find the perimeter when \(a=3.5\,\text{cm}\). c) Simplify the expression.

Hints

- 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.

Solution

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: \(P=4a+2a+10=6a+10\).

Answer

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) \(P=6a+10\)

Write equivalent expressions 42

Figure A is an equilateral triangle with side length \(2s + 4\). Figure B is a square with side length \(1.5s + 3\). Use algebra to show that the two figures have the same perimeter for every value of \(s\) that gives positive side lengths.

Hints

- 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?

Solution

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 \(s > -2\).

Answer

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

Write equivalent expressions 43

Determine whether these two rules always produce the same result for any number \(x\). Justify your answer by writing and simplifying expressions. Rule A: Add \(4\) to a number, then multiply the entire result by \(3\). Rule B: Add twice the number to the number itself, then add \(12\).

Hints

- 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?

Solution

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.

Answer

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

Write equivalent expressions 44

For a school carnival, a seventh-grade class compares two drink suppliers. Lakeview Drinks charges \(\$15.00\) for delivery and \(\$0.80\) per bottle. Riverbend Bottling charges \(\$5.00\) for delivery and \(\$1.20\) per bottle. a) Write an expression for the total cost from each supplier for \(x\) bottles. b) Find the cost of \(30\) bottles from each supplier. c) Two classes order the same number, \(x\), of bottles. One orders from Lakeview Drinks and the other from Riverbend Bottling. Jiho claims their combined cost is \(20+2x\) dollars. Verify the claim by combining the expressions.

Hints

- 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 \(x\).

Solution

1. Lakeview Drinks' cost is \(15+0.80x\). Riverbend Bottling's cost is \(5+1.20x\). 2. For \(x=30\), Lakeview Drinks costs \(15+0.80\cdot30=39\), and Riverbend Bottling costs \(5+1.20\cdot30=41\). 3. Add the expressions: \((15+0.80x)+(5+1.20x)=20+2x\). Jiho's expression is correct.

Answer

a) Lakeview Drinks: \(15+0.80x\); Riverbend Bottling: \(5+1.20x\) b) Lakeview Drinks: \(\$39.00\); Riverbend Bottling: \(\$41.00\) c) Yes. \((15+0.80x)+(5+1.20x)=20+2x\).

Write equivalent expressions 45

Let \(T_1=4n+8\). Ximena claims that \(T_2=4(n+8)\) is equivalent to \(T_1\). 1. Test the claim by evaluating both expressions for \(n=2\). 2. Explain the error using the distributive property. 3. How should the expression inside the parentheses in \(T_2\) be changed so that it is equivalent to \(T_1\)?

Hints

- Substitute \(n=2\) into both expressions. - Use the distributive property to compare the structures of the two expressions. - Compare the factored form with \(4n+8\) and think about what the expression inside the parentheses must contribute after distribution.

Solution

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 \(4(n+2)\).

Answer

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 \(4(n+2)\).

Write equivalent expressions 46

Two expressions are given: \(A=5x-(2x+3)\) \(B=3x-3\) 1. Simplify \(A\) step by step to show that the expressions are equivalent for every value of \(x\). 2. Evaluate both expressions for \(x=4.5\).

Hints

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

Solution

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.

Answer

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

Write equivalent expressions 47

There are three piles of counters. The left and right piles each start with exactly \(12\) counters. The middle pile starts with \(z\) counters. Follow these steps: 1. Move \(5\) counters from the left pile to the middle pile. 2. Move \(3\) counters from the right pile to the middle pile. 3. Count the counters remaining in the left pile. Remove that same number of counters from the middle pile and set them aside. a) How many counters remain in the left pile after step \(1\)? b) Write and simplify an expression for the number of counters in the middle pile after step \(3\). c) How many counters are in the middle pile at the end if \(z=10\)?

Hints

- Track the number of counters in each pile after each step. - First find the fixed number left in the left pile. - In step \(3\), subtract that fixed number from the middle-pile expression.

Solution

1. After step \(1\), the left pile has \(12-5=7\) counters. 2. After steps \(1\) and \(2\), the middle pile has \(z+5+3=z+8\) counters. 3. Step \(3\) removes \(7\) counters from the middle pile, so the final amount is \((z+8)-7=z+1\). 4. When \(z=10\), the final amount is \(10+1=11\).

Answer

a) \(7\) counters b) \(z+1\) c) \(11\) counters

Write equivalent expressions 48

Try this number trick: 1. Choose any number \(x\). 2. Add \(5\). 3. Multiply the result by \(3\). 4. Subtract \(15\). 5. Subtract twice your original number \(x\). a) Try the trick with two different starting numbers. What do you notice? b) Write an expression for the result after each step. Simplify the final expression to explain why the trick works.

Hints

- 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.

Solution

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.

Answer

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 equivalent expressions 49

Determine whether the expressions are equivalent. Justify your conclusion by simplifying both. Expression 1: \(3(4x - 5) + 8\) Expression 2: \(2x + 5(2x - 1) - 2\)

Hints

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

Solution

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.

Answer

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

Write equivalent expressions 50

The expression \(4(x+5)+6x\) is simplified below. Step 1: \(4x+4\cdot5+6x\) Step 2: \(4x+6x+4\cdot5\) Step 3: \((4+6)x+4\cdot5\) Step 4: \(10x+20\) Name the mathematical property used in Steps 1, 2, and 3. Briefly explain how Step 2 helps with the remaining simplification.

Hints

- 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.

Solution

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.

Answer

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.

Write equivalent expressions 51

At a school-supply store, a mechanical pencil costs \(\$f\). An eraser costs \(\$k\) less than the mechanical pencil. a) Write an expression for the price of the eraser. b) Write and simplify an expression for the total price of one mechanical pencil and one eraser. c) Find the total price when \(f=8.50\) and \(k=5.65\). d) What would \(k=f\) mean for the price of the eraser?

Hints

- “\(k\) 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.

Solution

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.

Answer

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

Write equivalent expressions 52

In a school cafeteria, a staff lunch costs \(\$x\), and a student lunch costs \(\$y\). a) Write an expression for the total revenue when \(15\) staff members and \(120\) students buy lunch. b) Every lunch price increases by \(\$0.50\). Write and simplify a new expression for the revenue if the same numbers of people buy lunch. c) Find the revenue after the increase if the original staff lunch cost \(\$5.50\) and the original student lunch cost \(\$3.50\).

Hints

- Multiply each lunch price by the number of lunches sold. - Add the same price increase to each original price. - Consider how that increase affects every lunch sold. - Substitute the original prices only after writing the new expression.

Solution

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 \(15\cdot6+120\cdot4=90+480=570\).

Answer

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

Write equivalent expressions 53

Determine whether each pair of expressions is equivalent. Justify each decision by simplifying. Pair A: \(T_1 = 2(3x + 4) - 2x\) and \(T_2 = 4x + 8\) Pair B: \(T_3 = 8a \div 2\) and \(T_4 = 4a\) Pair C: \(T_5 = 5 + 3b\) and \(T_6 = 8b\)

Hints

- 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.

Solution

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.

Answer

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 \(0\).

Write equivalent expressions 54

An electric company models a monthly bill as \( 18.50+0.16k, \) where \(k\) is the number of kilowatt-hours used. A billing analyst splits the usage charge into two equal components and writes \( 18.50+0.08k+0.08k. \) a) Show that the two expressions are equivalent for every value of \(k\). b) Name the property or like-term reasoning that justifies your rewrite. c) Explain what the coefficient \(0.16\) represents in the simplified form.

Hints

- Focus on the two terms containing \(k\). - Factor out \(k\) mentally from \(0.08k+0.08k\). - Add the numerical coefficients. - Equivalent expressions must have the same value for every allowed \(k\).

Solution

1. Combine the like terms in the analyst's expression: \( 0.08k+0.08k=(0.08+0.08)k=0.16k. \) 2. Therefore, \( 18.50+0.08k+0.08k=18.50+0.16k. \) 3. The rewrite uses the distributive property in reverse, together with addition of the numerical coefficients of like terms. 4. The coefficient \(0.16\) is the total charge in dollars for each kilowatt-hour used.

Answer

a) \(18.50+0.08k+0.08k=18.50+0.16k\) b) Combine like terms; equivalently, use the distributive property in reverse. c) \(0.16\) is the per-kilowatt-hour charge in dollars.

Write equivalent expressions 55

Suppose \(x - y = 7\). Determine the value of each expression. a) \(y - x\) b) \((x + 10) - (y + 10)\) c) \(2(y - x)\)

Hints

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

Solution

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): \(2(y - x) = 2(-7) = -14\).

Answer

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

Write equivalent expressions 56

Three sections of temporary fencing are installed at a construction site. The first section is \(4a+3b\) feet long. The second section is \(a-b+2\) feet longer than the first section. The third section is \(2a+4b-5\) feet long. For values of \(a\) and \(b\) that make all three lengths positive, write and simplify an expression for the total length of the fencing.

Hints

- Write an expression for the second section first. - “Longer than” indicates addition. - Combine the \(a\)-terms, \(b\)-terms, and constants separately.

Solution

1. The second section has length \((4a+3b)+(a-b+2)=5a+2b+2\). 2. Add all three section lengths: \((4a+3b)+(5a+2b+2)+(2a+4b-5)\). 3. Combine like terms: \(4a+5a+2a=11a\), \(3b+2b+4b=9b\), and \(2-5=-3\). 4. The total length is \(11a+9b-3\) feet.

Answer

The total length is \(11a+9b-3\) feet.

Write equivalent expressions 57

For values of \(a\) and \(b\) that make all original and new dimensions positive, a rectangle has length \(L=4a+3b\) and width \(W=2a+b\). A second rectangle is built with a length that is \(a-b\) greater than \(L\) and a width that is \(b\) less than \(W\). Write and simplify an expression for the perimeter of the second rectangle.

Hints

- Find the new length and width separately. - “Greater than” indicates addition, and “less than” indicates subtraction. - Use \(P=2(L+W)\) for a rectangle.

Solution

1. The new length is \(L_2=(4a+3b)+(a-b)=5a+2b\). 2. The new width is \(W_2=(2a+b)-b=2a\). 3. Use the perimeter formula: \(P=2(L_2+W_2)\). 4. Substitute and simplify: \(P=2[(5a+2b)+2a]=2(7a+2b)=14a+4b\).

Answer

\(P=14a+4b\)

Write equivalent expressions 58

For values of \(a\) and \(b\) that make all side lengths positive and form a valid triangle, the first two sides are Side 1: \(a+4b\) Side 2: \(3a-2b+5\). The third side is \(2a+b+3\) less than the sum of the first two sides. Write and simplify an expression for the perimeter \(P\).

Hints

- Add the first two side lengths first. - “Less than” means subtract the entire given expression. - Add all three side lengths to find the perimeter.

Solution

1. Add the first two sides: \((a+4b)+(3a-2b+5)=4a+2b+5\). 2. Find the third side: \((4a+2b+5)-(2a+b+3)=2a+b+2\). 3. Add all three sides: \(P=(4a+2b+5)+(2a+b+2)=6a+3b+7\).

Answer

\(P=6a+3b+7\)

Write equivalent expressions 59

Consider \(T_1 = 2(a - b)\), \(T_2 = 2a - b\), and \(T_3 = 2a - 2b\). First evaluate all three expressions for each set of values and make a conjecture. Then prove by simplifying which expressions are equivalent. a) \(a = 5\), \(b = 3\) b) \(a = 1.5\), \(b = 2\) c) \(a = \frac{1}{2}\), \(b = \frac{1}{4}\)

Hints

- 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.

Solution

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.

Answer

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.

Write equivalent expressions 60

Simplify each expression. 1) \(\frac{2}{3}(6a-9b)-\frac{1}{4}(8a+12b)\) 2) \(\left(x-\frac{1}{2}y+\frac{1}{3}z\right)-\left(\frac{1}{2}x+y-\frac{2}{3}z\right)+\left(\frac{1}{4}x-\frac{1}{2}y-z\right)\)

Hints

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

Solution

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 \(\frac{3}{4}x-2y\).

Answer

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

Write equivalent expressions 61

Simplify the expression. \(4x-[3y-(2x+5y)+7x]-(x-2y)\)

Hints

- 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.

Solution

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 \(-2x+4y\).

Answer

\(-2x+4y\)

Write equivalent expressions 62

Consider the expression \(20u-15v+10w-5\). a) Rewrite it in the form \((\square)-(\square)\), with \(20u\) and \(10w\) in the first set of parentheses. b) Explain why the sign of \(5\) inside the second set of parentheses differs from its sign in the original expression.

Hints

- 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.

Solution

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.

Answer

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.

Write equivalent expressions 63

Let \(u=3r-5s+2t\), \(v=-r+4s-3t\), and \(w=2r-s+t\). Simplify each expression. 1) \(u-v+w\) 2) \(-(u+v)-w\)

Hints

- 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.

Solution

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: \(-2r+s+t-2r+s-t=-4r+2s\).

Answer

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

Write equivalent expressions 64

Consider \(T=5x+3y-[3x-(y-2x)]\). a) Simplify \(T\). b) Use the simplified expression to explain why the value of \(T\) does not depend on \(x\). c) Evaluate \(T\) for \(x=100\) and \(y=-2.5\).

Hints

- 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.

Solution

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 \(y=-2.5\), \(T=4\cdot(-2.5)=-10\).

Answer

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

Write equivalent expressions 65

For \(a>0\), \(b>0\), and \(2b<3a\), a quadrilateral has perimeter \(P=10a+6b\). Three side lengths are described below. <ul><li>\(s_1=2a+b\)</li><li>\(s_2\) is \(a\) longer than \(s_1\).</li><li>\(s_3\) is \(2b\) shorter than \(s_2\).</li></ul> Write and simplify an expression for the fourth side length \(s_4\).

Hints

- 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.

Solution

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: \(s_4=(10a+6b)-(8a+b)=2a+5b\).

Answer

\(s_4=2a+5b\)

Write equivalent expressions 66

A wire with total length \((16y+20)\,\text{cm}\) is bent into a quadrilateral. The value of \(y\) is chosen so that all side lengths are positive and form a quadrilateral. <ul><li>The first side is \((5y+4)\,\text{cm}\).</li><li>The second side is half as long as the first side.</li><li>The third side is \((3y+6)\,\text{cm}\).</li></ul> a) Write and simplify an expression for the fourth side \(s_4\). b) When \(y=10\), is the fourth side longer than the first side? Show the calculation.

Hints

- 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.

Solution

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.

Answer

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

Write equivalent expressions 67

Consider \(T_1 = 3(2a + b) - 4(a - b)\) and \(T_2 = 2a + 7b\). Determine whether the expressions are equivalent. Justify your answer by simplifying \(T_1\).

Hints

- 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 \(-4\).

Solution

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.

Answer

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

Write equivalent expressions 68

Let \(x = 2a + 3\) and \(y = a - 5\). Substitute these expressions and simplify. 1) \(5x + 2y\) 2) \(3x - 4y\) 3) \(2(x - y)\)

Hints

- 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.

Solution

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. \(2\left((2a + 3) - (a - 5)\right) = 2(a + 8) = 2a + 16\).

Answer

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

Write equivalent expressions 69

Lukas and Sophie use two different calculation rules. Lukas says, “Choose a number, add \(4\), and multiply the result by \(5\).” Sophie says, “Choose the same number, double it, add \(8\), and multiply the new result by \(2.5\).” Write and simplify an expression in \(n\) for each rule. Determine whether the two rules give the same result for every starting number.

Hints

- 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.

Solution

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.

Answer

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

Write equivalent expressions 70

Simplify the expression. \(15a-\{4b+[2a-(5b-3a)]-2b\}\)

Hints

- 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.

Solution

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 \(10a+3b\).

Answer

\(10a+3b\)

Write equivalent expressions 71

Find the expression that belongs in the box so the two sides are equivalent. \(5(2n+1)-\square=3n+8\)

Hints

- First compare the expanded structure on the left with the target expression on the right. - Treat the missing part as one unknown expression rather than guessing individual terms. - Substitute your result back into the original statement to verify equivalence.

Solution

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: \(10n+5-(7n-3)=10n+5-7n+3=3n+8\).

Answer

\(7n-3\)

Write equivalent expressions 72

Consider the expressions \(T_1=2(3x-4)+5x\), \(T_2=12x-(x+8)\), and \(T_3=11x-4\). a) Simplify the expressions and determine which are equivalent. b) Change only the constant term in \(T_3\) so that its value is \(0\) when \(x=2\).

Hints

- Which expressions become identical after you simplify them? - How does the minus sign affect both terms in \(T_2\)? - In part b), how must the constant balance the variable term at the chosen input?

Solution

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. Thus \(T_1\) and \(T_2\) are equivalent, while \(T_3=11x-4\) is not. 4. When \(x=2\), the variable term in \(T_3\) is \(11\cdot2=22\). To make the total zero, change its constant term to \(-22\). The new expression is \(11x-22\), and \(11\cdot2-22=0\).

Answer

a) \(T_1\) and \(T_2\) are equivalent; both simplify to \(11x-8\). \(T_3\) is not equivalent. b) Change \(-4\) to \(-22\), giving \(11x-22\). At \(x=2\), its value is \(0\).

Write equivalent expressions 73

Chinedu claims that \(T_1=0.5(12y-8)+3y\) is equivalent to \(T_2=3(3y-2)-y+2\). a) Check the claim by simplifying both expressions. b) Change exactly one number in \(T_2\) so the resulting expression \(T_3\) is equivalent to \(T_1\). State \(T_3\).

Hints

- 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.

Solution

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 \(T_1\).

Answer

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

Write equivalent expressions 74

Find the value of \(a\) that makes the expressions equivalent: \(T_1 = 5k - 2(k - 6)\) \(T_2 = 3(k + a)\)

Hints

- 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?

Solution

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 \(3\): \(a = 4\).

Answer

\(a = 4\)

Write equivalent expressions 75

What expression belongs in the parentheses? \((12m-5n+2k)-(\square)=7m+n-4k\)

Hints

- Treat the missing parentheses as one unknown expression. - Compare like-variable coefficients on the two sides after accounting for the outside subtraction. - Expand the completed expression to verify that it matches the target.

Solution

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: \(T=12m-5n+2k-7m-n+4k=5m-6n+6k\).

Answer

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

Write equivalent expressions 76

Simplify the expression step by step. \(10a-\{3b-[4a-(5b-2c)+(a-c)]\}\)

Hints

- 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.

Solution

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: \(10a-8b+5a+c=15a-8b+c\).

Answer

\(15a-8b+c\)

Write equivalent expressions 77

Write \(4x-7y+2\) as a sum of two expressions when the first addend must be \(x-3y\).

Hints

- 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.

Solution

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, \(4x-7y+2=(x-3y)+(3x-4y+2)\).

Answer

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

Write equivalent expressions 78

First simplify \(T=3a-[2b-(a-b)]\). Then substitute \(a=2x+y\) and \(b=x-y\), and simplify the result.

Hints

- 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.

Solution

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: \(5x+7y\).

Answer

\(5x+7y\)

Write equivalent expressions 79

Let \(X=3a-4b\), \(Y=5a+2b\), and \(Z=-a+6b\). a) Simplify \(X-Y+Z\). b) Find an expression \(W\) such that \((X+Y)+W=Z\).

Hints

- 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.

Solution

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: \(W=(-a+6b)-(8a-2b)=-a-8a+6b+2b=-9a+8b\).

Answer

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

Write equivalent expressions 80

Simplify the expression. \(-(6u-2v)+\{3u-[5v-(u+3v)]-(2u-v)\}\)

Hints

- 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.

Solution

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: \((-6u+2v)+(2u-v)=-4u+v\).

Answer

\(-4u+v\)

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