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Write expressions from verbal phrases

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5222696
Translate each verbal description into an algebraic expression. a) The sum of a number \(x\) and \(14\) b) The product of the variables \(c\) and \(d\) c) The difference between \(20\) and a variable \(y\) d) Eight times a number \(z\)

Hints

- Match the words sum, product, and difference with their operations. - In a difference, preserve the stated order. - “Eight times” means multiplication by \(8\).

Solution

1. The sum of \(x\) and \(14\) is \(x+14\). 2. The product of \(c\) and \(d\) is \(cd\). 3. Subtracting \(y\) from \(20\) gives \(20-y\). 4. Eight times \(z\) is \(8z\).

Answer

a) \(x+14\) b) \(cd\) c) \(20-y\) d) \(8z\)
5495886
A museum charges \(d\) dollars for admission and \(4\) dollars for an audio guide. Write an expression for the total cost of admission and one audio guide.

Hints

- Which quantity can vary in this situation? - How is the guide charge combined with the admission charge?

Solution

1. Admission contributes \(d\) dollars. 2. The audio guide adds \(4\) dollars. 3. The total-cost expression is \(d+4\).

Answer

\(d+4\)
5495896
A ribbon is \(r\) inches long. Kai cuts off \(7\) inches. Write an expression for the length of ribbon that remains.

Hints

- Identify the starting amount and the amount removed. - Check the order of the subtraction against the story.

Solution

1. Start with the original length \(r\). 2. Removing \(7\) inches is represented by subtraction. 3. The remaining length is \(r-7\).

Answer

\(r-7\)
5222616
Translate each verbal description into an algebraic expression using the given variables. a) The product of \(x\) and the sum of \(y\) and \(z\) b) The difference between five times \(a\) and \(12\) c) The quotient whose dividend is the difference of \(p\) and \(q\) and whose divisor is \(3\)

Hints

- Match the words sum, difference, product, and quotient with their operations. - Use parentheses when an entire sum or difference is part of another operation. - Identify the dividend and divisor in the quotient description.

Solution

1. The sum of \(y\) and \(z\) is \(y+z\). Multiplying that entire sum by \(x\) gives \(x(y+z)\). 2. Five times \(a\) is \(5a\). Subtracting \(12\) gives \(5a-12\). 3. The difference of \(p\) and \(q\) is \(p-q\). Dividing that entire difference by \(3\) gives \(\frac{p-q}{3}\).

Answer

a) \(x(y+z)\) b) \(5a-12\) c) \(\frac{p-q}{3}\)
5222706
Write an algebraic expression for each instruction. a) Five times the difference of \(x\) and \(y\) b) The sum of the product of \(3\) and \(a\) and the quotient of \(b\) and \(4\) c) Subtract twice \(z\) from \(50\)

Hints

- Use parentheses when an entire difference is multiplied. - “Twice” means multiplication by \(2\). - Pay attention to which quantity is subtracted from which.

Solution

1. The difference is \(x-y\), and five times that entire difference is \(5(x-y)\). 2. The product is \(3a\), and the quotient is \(\frac{b}{4}\). Their sum is \(3a+\frac{b}{4}\). 3. Twice \(z\) is \(2z\). Subtracting it from \(50\) gives \(50-2z\).

Answer

a) \(5(x-y)\) b) \(3a+\frac{b}{4}\) c) \(50-2z\)
5222716
Write an algebraic expression for each description. a) The difference between twice a number \(x\) and \(7\) b) The quotient of the sum of \(a\) and \(b\) and \(3\) c) The product of a number \(y\) and \(1\) more than that number

Hints

- Match sum, difference, product, and quotient with their operations. - “Twice” means multiplication by \(2\). - Write an expression for \(1\) more than \(y\) before forming the product. - Use parentheses when an entire sum is divided or multiplied.

Solution

1. Twice \(x\) is \(2x\), so the difference is \(2x-7\). 2. The sum is \(a+b\). Dividing the entire sum by \(3\) gives \(\frac{a+b}{3}\). 3. One more than \(y\) is \(y+1\), so the product is \(y(y+1)\).

Answer

a) \(2x-7\) b) \(\frac{a+b}{3}\) c) \(y(y+1)\)
5223216
A hiker travels at a rate of \(x\) miles per hour. a) Write an expression for the distance traveled in \(4\) hours, in \(30\) minutes, and in \(t\) hours. b) Find the distance traveled at \(4.5\) miles per hour for \(3\) hours. c) After a break, the hiker travels at half the original rate. Write and simplify an expression for the total distance when the hiker first travels \(2\) hours at rate \(x\) and then \(1\) hour at half that rate.

Hints

- Express all times in hours to match the rate unit. - Use distance equals rate times time. - Half the rate \(x\) is \(\frac{x}{2}\).

Solution

1. Distance equals rate times time. The expressions are \(4x\), \(0.5x\), and \(tx\), because \(30\) minutes is \(0.5\) hour. 2. At \(4.5\) miles per hour for \(3\) hours, the distance is \(4.5\times 3=13.5\) miles. 3. The first part is \(2x\), and the second part is \(\frac{x}{2}\). The total is \(2x+\frac{x}{2}=2.5x\).

Answer

a) \(4x\); \(0.5x\); \(tx\) b) \(13.5\) miles c) \(2.5x\) miles
5223276
A hiking trail is \(w\) miles long. A family hikes \(\frac{1}{5}\) of the trail on the first day and \(\frac{2}{5}\) of the trail on the second day. Write an expression in terms of \(w\) for the total number of miles they hike in the two days.

Hints

- Add the fractions of the trail completed on the two days. - The whole trail has length \(w\). - A fraction of an unknown quantity can be written as a product.

Solution

1. Add the fractions of the trail: \(\frac{1}{5}+\frac{2}{5}=\frac{3}{5}\). 2. Multiply the total fraction by the full trail length: \(\frac{3}{5}w\).

Answer

\(\frac{3}{5}w\) miles
5223356
The variables \(x\), \(y\), and \(z\) represent nonnegative measurement values. 1. How many milliliters are in \(x\) liters? 2. How many grams are in \(y\) kilograms? 3. How many seconds are in \(z\) minutes?

Hints

- Identify how many smaller units make one larger unit. - Try a specific value such as \(2\) before writing the expression with a variable. - Converting to a smaller unit requires multiplication.

Solution

1. Since \(1\,\text{L}=1000\,\text{mL}\), \(x\) liters is \(1000x\,\text{mL}\). 2. Since \(1\,\text{kg}=1000\,\text{g}\), \(y\) kilograms is \(1000y\,\text{g}\). 3. Since \(1\,\text{min}=60\,\text{s}\), \(z\) minutes is \(60z\,\text{s}\).

Answer

1. \(1000x\,\text{mL}\) 2. \(1000y\,\text{g}\) 3. \(60z\,\text{s}\)
5223366
The variables represent nonnegative measurement values. 1. How many centimeters are in a total of \(a\) meters and \(b\) centimeters? 2. How many cents are in an amount of \(d\) dollars and \(c\) cents?

Hints

- Convert each part to the same smaller unit. - Use \(100\,\text{cm}=1\,\text{m}\). - Use \(100\) cents for one dollar.

Solution

1. \(a\) meters is \(100a\) centimeters. Adding \(b\) centimeters gives \((100a+b)\,\text{cm}\). 2. \(d\) dollars is \(100d\) cents. Adding \(c\) cents gives \((100d+c)\) cents.

Answer

1. \((100a+b)\,\text{cm}\) 2. \((100d+c)\) cents
5223616
Write an algebraic expression for each described number. 1. A number made of \(x\) hundreds and \(y\) tens 2. A number made of \(a\) thousands, \(b\) hundreds, and \(c\) ones 3. Three times a number made of \(n\) tens and \(k\) ones

Hints

- Use the value of each place: hundreds, tens, and ones. - Think of a number such as \(345\) as \(300+40+5\). - Apply “three times” to the entire original number.

Solution

1. Each hundred contributes \(100\), and each ten contributes \(10\), so the expression is \(100x+10y\). 2. The place-value expression is \(1000a+100b+c\). 3. The original number is \(10n+k\). Three times that number is \(3(10n+k)\), which is equivalent to \(30n+3k\).

Answer

1. \(100x+10y\) 2. \(1000a+100b+c\) 3. \(3(10n+k)\), or \(30n+3k\)
5223996
Write an algebraic expression for each description. 1. Three times the sum of \(x\) and \(5\) 2. The difference between the square of \(a\) and twice \(b\) 3. The square of the difference of \(x\) and \(y\) 4. The sum of half of \(z\) and \(9\)

Hints

- Identify which operation must happen first. - Parentheses show that an entire expression is treated as one quantity. - Match sum, difference, product, and quotient with their operations. - “The square of” means raising the entire named quantity to the second power.

Solution

1. The sum is \(x+5\), so three times the sum is \(3(x+5)\). 2. The square of \(a\) is \(a^2\), and twice \(b\) is \(2b\), so the expression is \(a^2-2b\). 3. The difference is \(x-y\), and squaring the entire difference gives \((x-y)^2\). 4. Half of \(z\) is \(\frac{z}{2}\), so the sum is \(\frac{z}{2}+9\).

Answer

1. \(3(x+5)\) 2. \(a^2-2b\) 3. \((x-y)^2\) 4. \(\frac{z}{2}+9\)
5224176
Write an algebraic expression for each description. 1) The sum of the square of \(x\) and four times \(y\) 2) The product of the difference of \(a\) and \(b\) and \(7\) 3) The quotient of the product of \(p\) and \(q\) and \(6\)

Hints

- Identify the final operation in each description. - Put a compound sum or difference in parentheses. - Form the product before writing the quotient in part 3.

Solution

1. The square of \(x\) is \(x^2\), and four times \(y\) is \(4y\), so the expression is \(x^2+4y\). 2. The difference is \(a-b\), so the product with \(7\) is \(7(a-b)\). 3. The product is \(pq\). Dividing that product by \(6\) gives \(\frac{pq}{6}\).

Answer

1) \(x^2+4y\) 2) \(7(a-b)\) 3) \(\frac{pq}{6}\)
5279206
A hiking group travels in two sections. 1) In the first section, the group hikes for \(t_1\) hours at \(v_1\) miles per hour. In the second section, the group hikes for \(t_2\) hours at \(v_2\) miles per hour. Write an expression for the total distance \(s\). 2) Find the total distance when \(t_1=2\) hours at \(v_1=3\) miles per hour and \(t_2=3\) hours at \(v_2=2.5\) miles per hour. 3) Write a new expression for \(s\) if the group hikes for the entire time \(t_1+t_2\) at one constant speed \(v\).

Hints

- Find each section's distance by multiplying speed by time. - Add the two section distances. - For part 3, multiply one speed by the total time.

Solution

1. The section distances are \(v_1t_1\) and \(v_2t_2\), so \(s=v_1t_1+v_2t_2\). 2. Substituting the values gives \(s=3\times 2+2.5\times 3=6+7.5=13.5\) miles. 3. At one constant speed for the full time, \(s=v(t_1+t_2)\).

Answer

1) \(s=v_1t_1+v_2t_2\) 2) \(13.5\) miles 3) \(s=v(t_1+t_2)\)
5495906
Write an expression for “\(9\) less than three times \(n\).”

Hints

- Translate the multiplication phrase before handling “less than.” - Read your expression back in words to check the subtraction order.

Solution

1. Three times \(n\) is \(3n\). 2. Nine less than that quantity means subtract \(9\). 3. The expression is \(3n-9\).

Answer

\(3n-9\)
5495916
Write an expression for “the quotient of \(42\) and the sum of \(x\) and \(5\).”

Hints

- Decide which words describe a grouped quantity. - Make sure the whole sum, not just one term, is used in the quotient.

Solution

1. The sum is \(x+5\). 2. The entire sum belongs in the denominator. 3. The expression is \(\frac{42}{x+5}\).

Answer

\(\frac{42}{x+5}\)
5495926
Write an expression for “four times the sum of \(m\) and \(6\).”

Hints

- Identify the grouped quantity named by “the sum of.” - Decide which quantity is multiplied by \(4\).

Solution

1. The sum of \(m\) and \(6\) is \(m+6\). 2. Four times that entire sum means multiply the grouped quantity by \(4\). 3. The expression is \(4(m+6)\).

Answer

\(4(m+6)\)
5495936
A square has side length \(s\) centimeters. A larger square is made by increasing the side length by \(2\) centimeters. Write an expression for the perimeter of the larger square.

Hints

- First express the new side length. - Then use the number of equal sides in a square.

Solution

1. The larger side length is \(s+2\) centimeters. 2. A square has four equal sides. 3. The perimeter is \(4(s+2)\,\text{cm}\).

Answer

\(4(s+2)\,\text{cm}\)
5495946
A theater row has \(p\) seats. Three identical rows are reserved, but \(5\) seats in the reserved section are blocked. Write an expression for the number of usable reserved seats.

Hints

- Separate the repeated equal groups from the later adjustment. - Ask whether the blocked seats are removed from each row or from the total section.

Solution

1. Three identical rows contain \(3p\) seats. 2. Five blocked seats are removed from that total. 3. The expression is \(3p-5\).

Answer

\(3p-5\)
5495966
A school club pays a one-time printing fee of \(\$18\) and \(\$0.35\) for each of \(f\) flyers. Write an expression for the total printing cost in dollars.

Hints

- Identify the charge that repeats for every flyer. - Keep the one-time charge separate from the repeated charge.

Solution

1. The variable part of the cost is \(0.35f\) dollars. 2. The one-time fee is added once. 3. The total cost is \(18+0.35f\) dollars.

Answer

\(18+0.35f\) dollars
5496046
Start with a number \(y\). Subtract \(6\), and then multiply the result by \(2\). Write an expression for the final result.

Hints

- Build the expression in the order the operations occur. - Use parentheses so the multiplication applies to the entire difference.

Solution

1. After subtracting \(6\), the result is \(y-6\). 2. Multiplying that entire result by \(2\) gives \(2(y-6)\).

Answer

\(2(y-6)\)
5496066
Write an expression for the total number of minutes in \(h\) hours plus an additional \(m\) minutes. Define what \(h\) and \(m\) represent.

Hints

- Name what each letter will represent before writing the expression. - Convert the larger time unit before adding the extra minutes.

Solution

1. Let \(h\) be the number of hours and \(m\) be the additional minutes. 2. The \(h\) hours contain \(60h\) minutes. 3. Adding the extra minutes gives a total of \(60h+m\) minutes.

Answer

Let \(h\) be the number of hours and \(m\) be the additional minutes. The expression is \(60h+m\) minutes.
5540746
Write a precise verbal phrase for the expression \(4(n+3)\). Then explain how its meaning differs from \(4n+3\).

Hints

- Identify the operation performed inside the parentheses. - Then identify the operation applied to the entire parenthesized expression. - Compare which part of each expression is multiplied by \(4\).

Solution

1. In \(4(n+3)\), the parentheses make \(n+3\) one grouped quantity. 2. The factor \(4\) multiplies that whole quantity, so one precise phrase is “four times the sum of \(n\) and \(3\).” 3. In \(4n+3\), only \(n\) is multiplied by \(4\), and then \(3\) is added. 4. The expressions therefore describe different operation structures.

Answer

One correct phrase is “four times the sum of \(n\) and \(3\).” In \(4(n+3)\), \(4\) multiplies the entire sum. In \(4n+3\), \(4\) multiplies only \(n\), and then \(3\) is added.
5124936
A quadrilateral has side lengths \(x\), \(x+2\), \(2x\), and \(5\) centimeters. a) Write and simplify an expression for its perimeter \(P\). b) Find the perimeter when \(x=3.5\).

Hints

- Perimeter is the sum of all side lengths. - Write the sum before combining any terms. - Combine like terms before substituting the value of \(x\).

Solution

1. Add the four side lengths: \(P=x+(x+2)+2x+5\). 2. Combine like terms: \(P=4x+7\). 3. Substitute \(x=3.5\): \(P=4\times3.5+7=21\,\text{cm}\).

Answer

a) \(P=4x+7\) b) \(21\,\text{cm}\)
5222096
At a movie theater, an order of popcorn costs \(\$4.50\) and a soft drink costs \(\$3.20\). 1) Find the total cost of \(2\) orders of popcorn and \(3\) soft drinks. 2) Write an expression for the total cost of \(p\) orders of popcorn and \(d\) soft drinks. 3) What purchase is represented by \(3\times 4.50+3.20\)? 4) A customer buys \(x\) orders of popcorn and pays with a \(\$20\) bill. Assume the cost is no more than \(\$20\). Write an expression for the change.

Hints

- Multiply each item's price by the number purchased. - Add the costs of the two types of items. - Variables can represent the number of items in the same way that specific numbers do. - Change equals the amount paid minus the purchase cost.

Solution

1. The total is \(2\times 4.50+3\times 3.20=9.00+9.60=18.60\). 2. The cost expression is \(4.50p+3.20d\). 3. The expression \(3\times 4.50+3.20\) represents \(3\) orders of popcorn and \(1\) soft drink. 4. The popcorn costs \(4.50x\), so the change is \(20-4.50x\).

Answer

1) \(\$18.60\) 2) \(4.50p+3.20d\) 3) The cost of \(3\) orders of popcorn and \(1\) soft drink 4) \(20-4.50x\) dollars
5222176
A model assumes that apples lose \(\frac{4}{5}\) of their original weight when dried. a) Write an expression for the weight of the dried apples when the fresh apples weigh \(x\) pounds. b) Use the expression to find the dried weight of \(15\) pounds of fresh apples.

Hints

- Subtract the fraction lost from the whole. - Express the remaining fraction as a fraction or decimal. - Multiply the remaining fraction by the original amount.

Solution

1. The weight lost is \(\frac{4}{5}x\). The remaining weight is \(x-\frac{4}{5}x=\frac{1}{5}x\), which is also \(0.2x\). 2. For \(x=15\), \(\frac{1}{5}\times 15=3\), so the dried apples weigh \(3\,\text{lb}\).

Answer

a) \(\frac{1}{5}x\), or \(0.2x\) b) \(3\,\text{lb}\)
5222476
A kayak rental company offers two plans. Plan A: a \(\$12\) base fee plus \(\$4\) per hour. Plan B: no base fee and \(\$7\) per hour. a) Write an expression for the total cost of each plan for \(t\) hours. b) Find the cost of each plan for \(2\) hours and for \(5\) hours. Which plan is less expensive each time? c) Explain what the number \(12\) and the variable \(t\) mean in Plan A.

Hints

- Separate each plan’s fixed cost from its per-hour cost. - Translate “per hour” as a multiplication involving the number of hours. - Substitute the requested values only after writing each expression. - Use the context to interpret both the constant and the variable.

Solution

1. Plan A is \(12+4t\), and Plan B is \(7t\). 2. For \(t=2\), Plan A costs \(12+4\times2=20\), and Plan B costs \(7\times2=14\). Plan B is less expensive. 3. For \(t=5\), Plan A costs \(12+4\times5=32\), and Plan B costs \(7\times5=35\). Plan A is less expensive. 4. In Plan A, \(12\) is the one-time base fee in dollars, and \(t\) is the rental time in hours.

Answer

a) Plan A: \(12+4t\); Plan B: \(7t\) b) At \(2\) hours: Plan A costs \(\$20\), Plan B costs \(\$14\); Plan B is less expensive. At \(5\) hours: Plan A costs \(\$32\), Plan B costs \(\$35\); Plan A is less expensive. c) The \(12\) is the one-time base fee, and \(t\) is the number of rental hours.
5222626
Investigate how parentheses affect an expression. a) Write an expression for: “Subtract the sum of \(b\) and \(c\) from \(a\).” b) Evaluate your expression for \(a=50\), \(b=15\), and \(c=5\). c) Evaluate \(a-b+c\) for the same values. Compare the results and explain the difference.

Hints

- Which entire quantity is being subtracted from \(a\)? - Substitute the values and follow the order of operations. - Compare whether \(c\) is subtracted or added in each expression.

Solution

1. The sum of \(b\) and \(c\) is \(b+c\), so subtracting that sum from \(a\) gives \(a-(b+c)\). 2. Substitution gives \(50-(15+5)=50-20=30\). 3. The comparison expression gives \(50-15+5=35+5=40\). 4. The expressions are not equivalent. In \(a-(b+c)\), both \(b\) and \(c\) are subtracted. In \(a-b+c\), \(c\) is added.

Answer

a) \(a-(b+c)\) b) \(30\) c) \(40\). The results differ because the first expression subtracts both \(b\) and \(c\), while the second adds \(c\).
5223226
A storage tank contains \(V\) gallons of oil. A machine uses \(12\) gallons each day. a) Write an expression for the amount of oil remaining after \(n\) days. b) Evaluate the expression for \(V=500\) and \(n=10\). c) What does it mean in this context if the expression has a value of \(0\)? d) How does the expression change if the machine uses an unknown amount of \(x\) gallons per day?

Hints

- Identify the starting quantity and the amount removed each day. - Express repeated daily use as a product before subtracting it. - Interpret a value of zero using the physical situation. - Replace the known daily rate with the new variable only after the original expression is clear.

Solution

1. Subtract the total amount used from the starting amount: \(V-12n\). 2. For \(V=500\) and \(n=10\), \(500-12\times10=500-120=380\) gallons. 3. A value of \(0\) means the oil supply has been completely used after \(n\) days. 4. Replacing the daily use of \(12\) gallons with \(x\) gallons gives \(V-xn\).

Answer

a) \(V-12n\) b) \(380\) gallons c) The oil supply is completely used up. d) \(V-xn\)
5223576
Write an algebraic expression for each description using \(x\), \(y\), and \(z\). 1) Five times \(x\) 2) The sum of twice \(y\) and \(z\) 3) The product of \(x\) and the difference of \(y\) and \(z\) 4) The quotient of three times \(z\) and the product of \(x\) and \(y\)

Hints

- Match sum, product, and quotient with their operations. - Use parentheses when a difference must be found before multiplication. - “Twice” and “three times” indicate multiplication by a number.

Solution

1. Five times \(x\) is \(5x\). 2. Twice \(y\) is \(2y\), so the sum is \(2y+z\). 3. The difference is \(y-z\), so the product is \(x(y-z)\). 4. Three times \(z\) is \(3z\), and the product of \(x\) and \(y\) is \(xy\). The quotient is \(\frac{3z}{xy}\).

Answer

1) \(5x\) 2) \(2y+z\) 3) \(x(y-z)\) 4) \(\frac{3z}{xy}\)
5223586
The order of operations matters when translating verbal descriptions. a) Write an expression for “three times the difference of \(a\) and \(b\).” b) Write an expression for “the difference between three times \(a\) and \(b\).” c) Explain the structural difference between the expressions. d) Evaluate both expressions when \(a=12\) and \(b=4\).

Hints

- Decide whether “three times” applies to one variable or to an entire difference. - Use parentheses to override the usual order of operations. - Substitute the values into each expression separately.

Solution

1. In part a), find the difference first, then multiply: \(3(a-b)\). 2. In part b), only \(a\) is multiplied by \(3\), then \(b\) is subtracted: \(3a-b\). 3. Parentheses in the first expression require subtraction before multiplication. In the second expression, multiplication is performed before subtraction. 4. Substitute \(a=12\) and \(b=4\): \(3\times(12-4)=24\), while \(3\times 12-4=32\).

Answer

a) \(3(a-b)\) b) \(3a-b\) c) In part a), the entire difference is tripled. In part b), only \(a\) is tripled before \(b\) is subtracted. d) Part a): \(24\); Part b): \(32\)
5224056
Translate each description into an algebraic expression. a) Six times the sum of \(x\) and \(y\) b) The quotient of the difference of \(a\) and \(b\) and \(5\) c) Add three times \(m\) to four times \(n\) d) The product of the sum of \(p\) and \(q\) and the difference of the same two numbers

Hints

- Use parentheses when an entire sum or difference is part of another operation. - Words such as “of” can help identify which quantities belong together. - Match sum, difference, product, and quotient with their operations. - In a quotient, identify the dividend and divisor in the stated order.

Solution

1. The sum is \(x+y\), so six times the sum is \(6(x+y)\). 2. The difference is \(a-b\), and dividing the entire difference by \(5\) gives \(\frac{a-b}{5}\). 3. Three times \(m\) is \(3m\), and four times \(n\) is \(4n\). Their sum is \(3m+4n\). 4. The sum is \(p+q\), and the difference is \(p-q\). Their product is \((p+q)(p-q)\).

Answer

a) \(6(x+y)\) b) \(\frac{a-b}{5}\) c) \(3m+4n\) d) \((p+q)(p-q)\)
5224066
Complete the following tasks with algebraic expressions. a) Subtract the sum of \(x\) and \(y\) from twice their product. b) Explain the difference between “the sum of twice \(a\) and \(b\)” and “twice the sum of \(a\) and \(b\).” Use expressions in your explanation. c) Evaluate the expression from part a) for \(x=5\) and \(y=2\).

Hints

- The phrase “subtract A from B” means \(B-A\). - Decide which operation must happen first and use parentheses when needed. - When evaluating, follow the order of operations.

Solution

1. The product of \(x\) and \(y\) is \(xy\), so twice the product is \(2xy\). Subtracting the sum \(x+y\) gives \(2xy-(x+y)\). 2. “The sum of twice \(a\) and \(b\)” is \(2a+b\). “Twice the sum of \(a\) and \(b\)” is \(2(a+b)\). In the first expression, only \(a\) is doubled; in the second, the entire sum is doubled. 3. Substitute the values: \(2\times 5\times 2-(5+2)=20-7=13\).

Answer

a) \(2xy-(x+y)\) b) \(2a+b\) doubles only \(a\), while \(2(a+b)\) doubles the entire sum. c) \(13\)
5224186
Translate each verbal description into an algebraic expression. 1) The square of half the sum of \(a\) and \(b\) 2) The difference between the sum of the squares of \(x\) and \(y\) and the square of their sum 3) Three times the difference of the cubes of \(m\) and \(n\)

Hints

- Read each description from the inside operation outward. - Distinguish “sum of the squares” from “square of the sum.” - A cube has exponent \(3\). - Use parentheses when the entire difference is multiplied.

Solution

1. The sum is \(a+b\), half the sum is \(\frac{a+b}{2}\), and its square is \(\left(\frac{a+b}{2}\right)^2\). 2. The sum of the squares is \(x^2+y^2\), and the square of the sum is \((x+y)^2\). Their difference is \((x^2+y^2)-(x+y)^2\). 3. The cubes are \(m^3\) and \(n^3\). Their difference is \(m^3-n^3\), and three times that difference is \(3(m^3-n^3)\).

Answer

1) \(\left(\frac{a+b}{2}\right)^2\) 2) \((x^2+y^2)-(x+y)^2\) 3) \(3(m^3-n^3)\)
5224196
Write an algebraic expression for each description. 1) The sum of three times a number \(x\) and \(7\) 2) Three times the sum of a number \(x\) and \(7\) 3) The square of the difference of \(a\) and \(b\) 4) The difference of the squares of \(a\) and \(b\)

Hints

- Identify which operation is applied to the entire expression. - Use parentheses when the usual order of operations must be changed. - “The square of” applies to the complete quantity named. - “The difference of the squares” means square first and subtract afterward.

Solution

1. Three times \(x\) is \(3x\), so the sum is \(3x+7\). 2. The sum is \(x+7\), so three times the entire sum is \(3(x+7)\). 3. The difference is \(a-b\), and squaring the entire difference gives \((a-b)^2\). 4. Square each number first, then subtract: \(a^2-b^2\).

Answer

1) \(3x+7\) 2) \(3(x+7)\) 3) \((a-b)^2\) 4) \(a^2-b^2\)
5224206
Translate each verbal description into an algebraic expression. 1) The product of the sum of \(x\) and \(4\) and the difference of \(x\) and \(4\) 2) Half the sum of a number \(x\) and the square of a number \(y\) 3) The square of half the sum of \(x\) and \(y\)

Hints

- Identify the entire expression that is multiplied, halved, or squared. - Put sums and differences in parentheses. - Decide whether the square applies to one term or to a complete expression.

Solution

1. The sum is \(x+4\), and the difference is \(x-4\). Their product is \((x+4)(x-4)\). 2. The sum is \(x+y^2\), so half the sum is \(\frac{x+y^2}{2}\). 3. Half the sum is \(\frac{x+y}{2}\), and its square is \(\left(\frac{x+y}{2}\right)^2\).

Answer

1) \((x+4)(x-4)\) 2) \(\frac{x+y^2}{2}\) 3) \(\left(\frac{x+y}{2}\right)^2\)
5225356
At the start of a dry period, a rainwater reservoir contains \(V\) cubic meters of water. During the first week, \(v\) cubic meters are used for irrigation. The remaining water must last \(d\) more days. Write an expression for the average number of liters that may be used per day during the remaining time. Use \(1\,\text{m}^3=1000\,\text{L}\).

Hints

- First find the amount of water remaining. - Convert cubic meters to liters before finding the daily amount. - Divide the total remaining amount equally among the \(d\) days.

Solution

1. The remaining volume is \(V-v\) cubic meters. 2. Convert the remaining amount to liters: \(1000(V-v)\). 3. Divide by the number of remaining days: \(\frac{1000(V-v)}{d}\) liters per day.

Answer

\(\frac{1000(V-v)}{d}\) liters per day
5225366
A youth group buys \(k\) kilograms of pasta for a camp. During the first two days, the group uses \(m\) kilograms. The remaining pasta will be divided equally among the final \(n\) days. a) Write an expression for the daily amount of pasta, in grams, during the remaining days. b) If the remaining amount stays the same while \(n\) increases, how does the value of the expression change? Explain.

Hints

- Consider what happens to a quotient when its positive divisor increases. - Imagine sharing the same amount among more groups. - Use parentheses so the remaining amount is found before conversion and division.

Solution

1. The remaining amount is \(k-m\) kilograms. 2. Convert to grams: \(1000(k-m)\). 3. Divide equally among \(n\) days: \(\frac{1000(k-m)}{n}\) grams per day. 4. With a fixed numerator, increasing the positive denominator makes the quotient smaller. The same amount of pasta must last more days, so less is available each day.

Answer

a) \(\frac{1000(k-m)}{n}\) grams per day b) The value decreases because the same remaining amount is divided among more days.
5225436
Lucas and Maya live in towns that are \(s\) miles apart. They start biking toward each other at the same time. Lucas rides at \(v_L\) miles per hour, and Maya rides at \(v_M\) miles per hour. a) Write an expression for the distance \(e\) between them after \(t\) hours, assuming they have not met yet. b) Find the remaining distance when \(s=38\), \(v_L=16\), \(v_M=14\), and \(t=0.5\).

Hints

- Find their combined closing rate. - Determine how far each rider travels in \(t\) hours. - Subtract the total distance traveled from the original separation. - Keep the time and rate units consistent.

Solution

1. In \(t\) hours, Lucas travels \(v_Lt\) miles, and Maya travels \(v_Mt\) miles. 2. Together, they reduce the distance by \((v_L+v_M)t\). 3. Therefore, \(e=s-(v_L+v_M)t\). 4. Substitute the values: \(e=38-(16+14)\times 0.5=38-15=23\) miles.

Answer

a) \(e=s-(v_L+v_M)t\) b) \(23\) miles
5225636
Lucas and Maya walk along a straight path in the same direction. Maya has a head start of \(d\) meters. Lucas walks at \(v_1\) meters per minute, and Maya walks at \(v_2\) meters per minute, where \(v_1>v_2\). a) Write an expression for the time \(t\), in minutes, that Lucas needs to catch Maya. b) Find \(t\) when \(d=450\), \(v_1=80\), and \(v_2=50\).

Hints

- Find how many meters Lucas gains each minute. - Divide the entire head start by the rate at which the gap closes. - Write the head start and the difference in speeds as a quotient.

Solution

1. Lucas closes the gap at \(v_1-v_2\) meters per minute. 2. Time equals distance divided by rate, so \(t=\frac{d}{v_1-v_2}\). 3. Substitute the values: \(t=\frac{450}{80-50}=\frac{450}{30}=15\) minutes.

Answer

a) \(t=\frac{d}{v_1-v_2}\) b) \(15\) minutes
5226026
A streaming service charges a monthly base fee of \(\$8.50\) plus \(\$2.50\) for each rented movie. a) Write an expression for the monthly cost when \(n\) movies are rented. b) Find the cost for a month with \(6\) rentals. c) A competing service has no base fee but charges \(\$4.00\) per rental. Which service costs less for \(5\) rentals? Show the calculation.

Hints

- Separate the fixed monthly charge from the per-rental charge. - Use the number of rentals as the variable in the expression. - Substitute a rental count only after the expression has been written. - Calculate both plans independently before comparing them.

Solution

1. The monthly cost is \(8.50+2.50n\). 2. For \(n=6\), \(8.50+2.50\times6=8.50+15.00=23.50\). 3. For \(5\) rentals, the first service costs \(8.50+2.50\times5=21.00\). The competing service costs \(4.00\times5=20.00\). The competing service costs less.

Answer

a) \(8.50+2.50n\) dollars b) \(\$23.50\) c) The competing service is less expensive: \(\$20.00\) instead of \(\$21.00\).
5238076
Ava and Lucas walk a distance of \(s\) feet. Ava's step length is \(L\) inches, where \(L>2\). Lucas's step length is \(2\) inches shorter than Ava's. a) Write an expression for the number of steps each person takes. Remember that the distance and step length must use the same unit. b) Write an expression for how many more steps Lucas takes than Ava. c) Find each number of steps and the difference when \(s=4200\) feet and \(L=30\) inches.

Hints

- Convert feet to inches before dividing by a step length. - Divide the total distance by the length of one step. - Decide which person takes more steps before writing the difference.

Solution

1. Convert the walking distance to inches: \(12s\) inches. 2. Ava takes \(\frac{12s}{L}\) steps. 3. Lucas's step length is \(L-2\) inches, so he takes \(\frac{12s}{L-2}\) steps. 4. Lucas takes \(\frac{12s}{L-2}-\frac{12s}{L}\) more steps than Ava. 5. For \(s=4200\) and \(L=30\), Ava takes \(\frac{12\times 4200}{30}=1680\) steps. Lucas takes \(\frac{12\times 4200}{28}=1800\) steps. The difference is \(1800-1680=120\) steps.

Answer

a) Ava: \(\frac{12s}{L}\) steps; Lucas: \(\frac{12s}{L-2}\) steps b) \(\frac{12s}{L-2}-\frac{12s}{L}\) steps c) Ava takes \(1680\) steps, Lucas takes \(1800\) steps, and Lucas takes \(120\) more steps.
5238226
A copier prints \(n\) pages in \(m\) minutes. a) Write an expression for its printing rate \(v\), in pages per minute. b) Write an expression for the time \(T\) needed to print \(x\) pages. c) Find \(v\) when the copier prints \(180\) pages in \(4.5\) minutes. d) At the rate from part c), how long will it take to print \(540\) pages?

Hints

- First determine how many pages the copier prints in one minute. - Divide the number of pages by the printing rate to find the time. - Substitute the expression from part a) into the formula for time.

Solution

1. The printing rate is pages divided by minutes: \(v=\frac{n}{m}\). 2. The time for \(x\) pages is \(T=\frac{x}{v}=\frac{xm}{n}\). 3. For \(n=180\) and \(m=4.5\), \(v=180\div 4.5=40\) pages per minute. 4. For \(x=540\), \(T=540\div 40=13.5\) minutes.

Answer

a) \(v=\frac{n}{m}\) pages per minute b) \(T=\frac{x}{v}=\frac{xm}{n}\) minutes c) \(40\) pages per minute d) \(13.5\) minutes
5244606
Let \(k\) be any positive integer. Write an expression for each description. a) A positive multiple of \(8\) b) A number that is \(5\) more than a multiple of \(8\) c) A positive odd number d) A number that is \(2\) less than a positive multiple of \(3\)

Hints

- A multiple of a number can be written as that number times an integer. - Translate “more than” as addition and “less than” as subtraction. - Begin with an even-number expression before making it odd. - Check your expressions with \(k=1\) and \(k=2\).

Solution

1. A positive multiple of \(8\) has the form \(8k\). 2. A number that is \(5\) more than a multiple of \(8\) has the form \(8k+5\). 3. An even number has the form \(2k\). Adding \(1\) gives the odd number \(2k+1\). 4. A positive multiple of \(3\) has the form \(3k\), so a number \(2\) less has the form \(3k-2\).

Answer

a) \(8k\) b) \(8k+5\) c) \(2k+1\) d) \(3k-2\)
5279286
Three whole numbers have a sum of \(150\). Two of the numbers are \(a\) and \(b\). a) Write an expression for the third number. b) If \(a\) increases by \(15\), how must the third number change so that the total remains \(150\)? Explain.

Hints

- Subtract the two known parts from the total. - If one addend increases while the sum stays fixed, another addend must compensate. - Test the change with a simple numerical example.

Solution

1. Let the third number be \(z\). Then \(a + b + z = 150\), so \(z = 150 - a - b\). 2. If \(a\) increases by \(15\), the sum would increase by \(15\) unless another addend changes. 3. To keep the total at \(150\), the third number must decrease by \(15\).

Answer

a) \(150 - a - b\) b) The third number must decrease by \(15\).
5279296
Two numbers have a sum of \(s\). One addend is \(15\). 1. Write an expression for the other addend. 2. Find the other addend when \(s = 42\). 3. State a rule for finding an unknown addend when the sum and the other addend are known.

Hints

- Write an equation for the two addends and their sum. - Which operation undoes addition? - Substitute \(s = 42\) into your expression.

Solution

1. If the unknown addend is \(x\), then \(15 + x = s\), so \(x = s - 15\). 2. When \(s = 42\), \(x = 42 - 15 = 27\). 3. Subtract the known addend from the sum.

Answer

1. \(s - 15\) 2. \(27\) 3. Subtract the known addend from the sum.
5279306
A division has a quotient of \(7\) and a divisor of \(d\). 1. Write an expression for the dividend \(x\). 2. Find \(x\) when \(d = 13\). 3. State a rule for finding the dividend from the quotient and divisor.

Hints

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

Solution

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

Answer

1. \(x = 7d\) 2. \(x = 91\) 3. \(\text{dividend} = \text{quotient} \times \text{divisor}\)
5495956
A board is \(b\) feet long. It is cut into \(6\) equal pieces, and then each piece is shortened by \(\frac{1}{4}\) foot. Write an expression for the final length of one piece.

Hints

- Follow the events in the order they occur. - Be clear whether the shortening happens before or after the equal split.

Solution

1. Cutting the board into \(6\) equal pieces gives \(\frac{b}{6}\) feet per piece. 2. Shortening one piece by \(\frac{1}{4}\) foot gives \(\frac{b}{6}-\frac{1}{4}\) feet. 3. The final length is \(\left(\frac{b}{6}-\frac{1}{4}\right)\,\text{ft}\).

Answer

\(\left(\frac{b}{6}-\frac{1}{4}\right)\,\text{ft}\)
5495976
The temperature is \(t\) degrees Fahrenheit at noon. By evening it is \(6\) degrees lower, and overnight it drops another \(h\) degrees. Write an expression for the overnight temperature.

Hints

- Track each change from the starting temperature. - Both changes lower the temperature, so check the signs in your expression.

Solution

1. The evening temperature is \(t-6\) degrees Fahrenheit. 2. A further drop of \(h\) degrees gives \((t-6)-h\) degrees Fahrenheit. 3. An equivalent simplified expression for the overnight temperature is \(t-6-h\) degrees Fahrenheit.

Answer

\(t-6-h\) degrees Fahrenheit
5495986
A snack mix uses \(c\) cups of cereal. It uses twice as many cups of pretzels as cereal and \(\frac{1}{2}\) cup of seeds. Write an expression for the total number of cups in the mix.

Hints

- Write a separate expression for each ingredient first. - The question asks for a total, so combine all ingredient amounts.

Solution

1. The cereal amount is \(c\) cups. 2. The pretzel amount is \(2c\) cups. 3. Adding cereal, pretzels, and seeds gives \(c+2c+\frac{1}{2}\) cups.

Answer

\(c+2c+\frac{1}{2}\) cups
5495996
A rectangular banner has width \(w\) feet. Its length is \(3\) feet more than twice its width. Write an expression for the area of the banner.

Hints

- Express the length in terms of the width before finding area. - Keep the entire length expression together when multiplying.

Solution

1. The width is \(w\) feet. 2. The length is \(2w+3\) feet. 3. Area is width times length, so the area is \(w(2w+3)\,\text{ft}^2\).

Answer

\(w(2w+3)\,\text{ft}^2\)
5496006
Mila says that “five times the sum of \(q\) and \(2\)” is \(5q+2\). Write the correct expression and state what Mila’s expression actually means.

Hints

- Locate the quantity named immediately after “five times.” - Compare what the parentheses would cause to be multiplied.

Solution

1. The sum of \(q\) and \(2\) is \(q+2\). 2. Five times that entire sum is \(5(q+2)\). 3. Mila’s expression \(5q+2\) means “the sum of five times \(q\) and \(2\).”

Answer

Correct expression: \(5(q+2)\) Mila’s expression means: the sum of five times \(q\) and \(2\).
5496016
Two expressions are proposed for “divide the difference of \(a\) and \(9\) by \(4\).” Expression A: \(a-\frac{9}{4}\) Expression B: \(\frac{a-9}{4}\) Which expression matches the phrase? Explain the role of the grouping.

Hints

- Identify the named difference before applying division. - Ask which expression divides both parts of the difference by the same number.

Solution

1. The difference of \(a\) and \(9\) is \(a-9\). 2. The instruction divides that entire difference by \(4\). 3. Expression B, \(\frac{a-9}{4}\), matches the phrase.

Answer

Expression B: \(\frac{a-9}{4}\). The grouping places the entire difference in the numerator.
5496026
Write an expression for the result of this process: 1. Start with \(x\). 2. Add \(8\). 3. Divide the result by \(3\). 4. Square the quotient.

Hints

- Build the expression one stage at a time. - Use grouping so each later operation acts on the entire previous result.

Solution

1. After adding \(8\), the quantity is \(x+8\). 2. Dividing by \(3\) gives \(\frac{x+8}{3}\). 3. Squaring the quotient gives \(\left(\frac{x+8}{3}\right)^2\).

Answer

\(\left(\frac{x+8}{3}\right)^2\)
5496036
A library has \(b\) books on one cart. A second cart has \(12\) fewer books than the first. A third cart has half as many books as the first. Assume \(b\) is an even whole number and \(b\ge 12\). Write an expression for the total number of books on the three carts.

Hints

- Express the amount on each cart separately. - Pay attention to which cart is described as fewer than or half of the first cart. - Add the three cart expressions only after all relationships are represented.

Solution

1. The first cart has \(b\) books. 2. The second cart has \(b-12\) books. 3. The third cart has \(\frac{b}{2}\) books. 4. The total is \(b+(b-12)+\frac{b}{2}\) books.

Answer

\(b+(b-12)+\frac{b}{2}\) books
5496056
A music app stores \(s\) songs. After \(25\) songs are deleted, the remaining songs are arranged equally into \(5\) playlists. Assume \(s\) is a whole number at least \(25\) and \(s-25\) is divisible by \(5\). Write an expression for the number of songs in each playlist.

Hints

- Follow the time order in the situation. - First represent how many songs remain after the deletion. - The division applies to the entire remaining amount.

Solution

1. After deletion, \(s-25\) songs remain. 2. Equal arrangement into \(5\) playlists means divide the remaining amount by \(5\). 3. The number of songs in each playlist is \(\frac{s-25}{5}\).

Answer

\(\frac{s-25}{5}\) songs per playlist
5540756
The expression is \(\frac{3m-5}{2}\). Write a precise verbal description of the expression. Then explain why “three times \(m\) minus five halves” does not describe the same structure.

Hints

- Treat the numerator as one grouped quantity. - Decide what the denominator \(2\) divides in the displayed fraction. - Compare that structure with a phrase in which only \(5\) is divided by \(2\).

Solution

1. The numerator is the entire difference \(3m-5\). 2. That whole difference is divided by \(2\). 3. A precise description is “the difference between three times \(m\) and \(5\), divided by \(2\).” 4. The phrase “three times \(m\) minus five halves” describes \(3m-\frac{5}{2}\), because only \(5\) is divided by \(2\).

Answer

One correct description is “the difference between three times \(m\) and \(5\), divided by \(2\).” The phrase “three times \(m\) minus five halves” instead describes \(3m-\frac{5}{2}\).
5223566
Algebraic expressions can describe whole groups of numbers. a) Explain why \(2n+1\) is odd for every whole number \(n\). b) Write an expression that represents all positive integers divisible by both \(3\) and \(4\). State the allowed values of your variable.

Hints

- What kind of number results when a whole number is multiplied by \(2\)? - What is the least positive integer divisible by both \(3\) and \(4\)? - How can a variable represent all multiples of one number?

Solution

1. For every whole number \(n\), \(2n\) is even. Adding \(1\) produces an odd number, so \(2n+1\) is always odd. 2. A number divisible by both \(3\) and \(4\) must be a multiple of their least common multiple. The least common multiple of \(3\) and \(4\) is \(12\). 3. Therefore, \(12k\), where \(k\) is any positive integer, represents all positive integers divisible by both \(3\) and \(4\).

Answer

a) \(2n\) is even, so \(2n+1\) is odd. b) \(12k\), where \(k\) is a positive integer.

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