Grade 7 math worksheets

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Model situations with equations and inequalities

Learning goals

Defines variables for unknown quantities in a context.
Writes equations or inequalities that represent the stated constraints.
Finds values satisfying the equation or inequality that models the situation.
Interprets the solution or solution set in context.
Checks whether contextual restrictions eliminate any mathematical solutions.

Difficulty: 1 – 5

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Model situations with equations and inequalities 1

A bakery has 24 muffins. Eight are set aside for the display case. The rest go equally into two trays. Let \(x\) be the number of muffins in each tray. Write an equation for the total. Do not solve it.

Hints

- How could you count the muffins in both trays without knowing \(x\)? - Which muffins are not included in the two trays?

Solution

1. The two trays hold \(x+x=2x\) muffins altogether. 2. Including the 8 display muffins gives \(8+2x\). 3. There are 24 muffins in all, so \(8+2x=24\).

Answer

\(8+2x=24\)

Model situations with equations and inequalities 2

A swimming pool charges \(\$4\) for entry and \(\$3\) per hour. Jules pays \(\$19\). Let \(h\) be the number of hours. Write an equation for the total charge. Do not solve it.

Hints

- If Jules stayed one hour, what would the total cost be? - Does your equation give that cost when \(h=1\)?

Solution

1. The hourly cost is \(3h\) dollars. 2. Add the entry fee to make the total: \(3h+4=19\).

Answer

\(3h+4=19\)

Model situations with equations and inequalities 3

A school auditorium has \(120\) seats. There are already \(42\) people seated. Each arriving group has \(6\) people. Let \(g\) be the number of groups that can still enter. Write an inequality that says the auditorium must not go over capacity. Do not solve it.

Hints

- How many people arrive with \(g\) groups? - Does the capacity permit more than \(120\), or up to and including \(120\)?

Solution

1. The new groups add \(6g\) people to the \(42\) already seated. 2. Capacity cannot exceed \(120\), so \(42+6g\le120\).

Answer

\(42+6g\le120\)

Model situations with equations and inequalities 4

A school concert will run only if at least \(20\) tickets have been sold. Let \(t\) be the number of tickets sold. Should the requirement be \(t>20\) or \(t\ge20\)? Explain whether selling exactly \(20\) tickets meets it.

Hints

- Imagine the concert sold the smallest number that should qualify. - Does the sign you choose include equality?

Solution

1. “At least \(20\)” includes \(20\) itself, as well as greater counts. 2. The correct condition is \(t\ge20\); selling exactly \(20\) tickets qualifies.

Answer

\(t\ge20\). Yes, exactly \(20\) tickets meets the requirement.

Model situations with equations and inequalities 5

Two rectangular garden beds have the dimensions shown in feet. The drawings are not to scale. Their combined area is \(24\) square feet. Write an equation using \(x\). Do not solve it.

Figure for problem 559669

Hints

- What does each product of two side lengths represent? - Does your equation account for both beds and the stated total area?

Solution

1. Both sections have height \(2\) feet. Their widths are \(x\) and \(4\) feet. 2. The total area is \(2x+2\cdot4=2x+8\) square feet. 3. The equation is \(2x+8=24\).

Answer

\(2x+8=24\)

Model situations with equations and inequalities 6

In the United States, temperature is commonly measured in degrees Fahrenheit. To convert a temperature from degrees Celsius to degrees Fahrenheit, multiply the Celsius value by \(1.8\) and add \(32\). a) Let \(c\) be the temperature in degrees Celsius and \(F\) be the temperature in degrees Fahrenheit. Write an equation that models the relationship. b) Use the equation to convert \(20\,^{\circ}\text{C}\), \(35\,^{\circ}\text{C}\), and \(-10\,^{\circ}\text{C}\) to degrees Fahrenheit.

Hints

- Translate the two operations in the order they are described. - Use one variable for the input temperature and one for the output temperature. - Be careful with the sign when substituting a negative Celsius temperature.

Solution

1. Following the stated operations gives \(F=1.8c+32\). 2. For \(20\,^{\circ}\text{C}\): \(F=1.8\cdot20+32=68\,^{\circ}\text{F}\). 3. For \(35\,^{\circ}\text{C}\): \(F=1.8\cdot35+32=95\,^{\circ}\text{F}\). 4. For \(-10\,^{\circ}\text{C}\): \(F=1.8\cdot(-10)+32=14\,^{\circ}\text{F}\).

Answer

a) \(F=1.8c+32\) b) \(68\,^{\circ}\text{F}\), \(95\,^{\circ}\text{F}\), and \(14\,^{\circ}\text{F}\)

Model situations with equations and inequalities 7

Write and solve an equation for each number riddle. a) What number must be increased by \(2.5\) to get \(1\frac{1}{4}\)? b) What number gives \(-1.5\) when \(\frac{2}{3}\) is subtracted from it? c) Three times what number is \(12.6\)?

Hints

- Translate each statement into an equation before solving. - Which operations are represented by “increased by” and “subtracted from”? - Use a variable for the unknown number. - Express fractions and decimals in compatible forms before calculating.

Solution

1. For part a, write \(x + 2.5 = 1.25\). Subtract \(2.5\): \(x = 1.25 - 2.5 = -1.25\). 2. For part b, write \(x - \frac{2}{3} = -1.5\). Since \(-1.5 = -\frac{3}{2}\), add \(\frac{2}{3}\): \(x = -\frac{3}{2} + \frac{2}{3} = -\frac{9}{6} + \frac{4}{6} = -\frac{5}{6}\). 3. For part c, write \(3x = 12.6\). Divide by \(3\): \(x = 4.2\).

Answer

a) \(x = -1.25\) b) \(x = -\frac{5}{6}\) c) \(x = 4.2\)

Model situations with equations and inequalities 8

Three piles of counters are on a table: a left pile \(L\), a middle pile \(M\), and a right pile \(R\). The left pile has at least \(6\) counters. The middle pile starts with \(z\) counters. Follow these directions: 1. Move \(6\) counters from the left pile to the middle pile. 2. Move \(4\) counters from the middle pile to the right pile. a) Write a simplified expression for the number of counters in the middle pile after both moves. b) The middle pile then has \(15\) counters. Write and solve an equation for its starting amount.

Hints

- Track only how each move changes the middle pile. - Combine the two changes to simplify the expression in part a. - In part b), set the expression for the final middle pile equal to the stated final amount. - Solve the equation for the starting value.

Solution

1. After the first move, the middle pile has \(z+6\) counters. 2. After the second move, it has \(z+6-4=z+2\) counters. 3. Set the final amount equal to \(15\): \(z+2=15\). 4. Subtract \(2\) from both sides to get \(z=13\).

Answer

a) \(z+2\) b) Equation: \(z+2=15\); \(z=13\) counters

Model situations with equations and inequalities 9

Write an equation and solve the number riddle: “Twice a number \(x\), increased by the difference between \(15.4\) and \(8.4\), equals \(25\).” What is \(x\)?

Hints

- How can you represent twice a number? - Which operation is indicated by “increased by”? - Evaluate the fixed difference before solving the equation.

Solution

1. Translate the statement into an equation: \(2x + (15.4 - 8.4) = 25\). 2. Evaluate the difference: \(15.4 - 8.4 = 7\). 3. Simplify: \(2x + 7 = 25\). 4. Subtract \(7\): \(2x = 18\). 5. Divide by \(2\): \(x = 9\).

Answer

The equation is \(2x + (15.4 - 8.4) = 25\), and \(x = 9\).

Model situations with equations and inequalities 10

Write and solve an equation for each number puzzle. a) Subtracting \(55\) from a number \(y\) gives \(-25\). b) Adding a number \(z\) to \(-140\) gives \(60\).

Hints

- Translate each sentence into an equation one phrase at a time. - Use inverse operations to isolate the variable. - Pay close attention to the signs when the value crosses zero.

Solution

1. In a), the equation is \(y-55=-25\). Add \(55\) to both sides to get \(y=30\). 2. In b), the equation is \(-140+z=60\). Add \(140\) to both sides to get \(z=200\).

Answer

a) \(y-55=-25\), so \(y=30\) b) \(-140+z=60\), so \(z=200\)

Model situations with equations and inequalities 11

A withdrawal of \(\$150\) changes an account balance. After the withdrawal, the balance is \(-\$35\). Write and solve an equation to find the balance before the withdrawal.

Hints

- Define a variable for the unknown starting balance. - How does a withdrawal change an account balance? - Which inverse operation will undo the withdrawal?

Solution

1. Let \(x\) represent the balance before the withdrawal. 2. The withdrawal gives the equation \(x-150=-35\). 3. Add \(150\) to both sides: \(x=-35+150=115\).

Answer

The balance before the withdrawal was \(\$115\).

Model situations with equations and inequalities 12

A gym charges a one-time enrollment fee of \(\$s\) and a monthly fee of \(\$m\). Let \(G\) be the total membership cost, in dollars, for \(x\) months. a) Write an equation that models \(G\). b) Find the total cost for one year when \(s=35.00\) and \(m=19.90\).

Hints

- Separate the one-time fee from the fee that repeats each month. - Use a variable for the total cost so the relationship can be written as an equation. - One year has \(12\) months.

Solution

1. The monthly fees total \(mx\), and the one-time fee is added once, so \(G=mx+s\). 2. For one year, \(x=12\). Substitute the values: \(G=19.90\cdot12+35.00=238.80+35.00=273.80\).

Answer

a) \(G=mx+s\) b) \(\$273.80\)

Model situations with equations and inequalities 13

Write and solve an equation for each statement. 1) A number \(x\) is increased by \(8.4\), and the result is \(15\). 2) A number \(y\) is subtracted from \(42\), and the result is \(13\frac{1}{2}\). 3) Three times a number \(a\) is \(25.5\).

Hints

- Be careful about the order in “subtracted from.” Which amount comes first? - Which operation undoes each change to the unknown number? - Can you check each result in the original sentence?

Solution

1. For 1), \(x+8.4=15\), so subtracting \(8.4\) gives \(x=6.6\). 2. For 2), \(42-y=13\frac{1}{2}\), so \(y=42-13.5=28.5\). 3. For 3), \(3a=25.5\), so dividing by \(3\) gives \(a=8.5\).

Answer

1) \(x+8.4=15\); \(x=6.6\). 2) \(42-y=13\frac{1}{2}\); \(y=28.5\). 3) \(3a=25.5\); \(a=8.5\).

Model situations with equations and inequalities 14

Anna, Beth, and Clara collect a total of \(120\,\text{lb}\) of paper for a school recycling drive. Anna collects \(10\,\text{lb}\) more than Beth and \(10\,\text{lb}\) less than Clara. Let \(x\) be Anna’s amount. Write an equation for the three amounts, simplify it, solve it, and report how many pounds each person collects.

Hints

- Use Anna’s amount as the central variable because the other two amounts are symmetric around it. - Express Beth and Clara before adding the three amounts. - Notice what happens to the constant terms when the equation is simplified. - Check the total and both difference relationships.

Solution

1. Let \(x\) be the number of pounds Anna collects. 2. Beth collects \(x - 10\), and Clara collects \(x + 10\). 3. Write the total equation \(x + (x - 10) + (x + 10) = 120\). 4. Combine like terms: \(3x = 120\). 5. Divide by \(3\): \(x = 40\). 6. Beth collects \(40 - 10 = 30\,\text{lb}\), and Clara collects \(40 + 10 = 50\,\text{lb}\). 7. Check: \(40 + 30 + 50 = 120\).

Answer

Equation: \(x+(x-10)+(x+10)=120\), which simplifies to \(3x=120\). Anna: \(40\,\text{lb}\); Beth: \(30\,\text{lb}\); Clara: \(50\,\text{lb}\)

Model situations with equations and inequalities 15

Lucas, Mia, and Noah have \(200\) trading cards altogether. Lucas has \(20\) more cards than Mia, and Noah has twice as many cards as Mia. Let \(x\) be Mia’s number of cards. Write one equation for the total, solve it, and report each person’s number of cards.

Hints

- Use Mia’s count as the base variable for both relationships. - Translate “\(20\) more” and “twice as many” into separate expressions. - Add all three quantities before setting the total equal to \(200\). - Check both relationships after solving.

Solution

1. Let \(x\) be the number of cards Mia has. 2. Lucas has \(x + 20\), and Noah has \(2x\). 3. Write the total equation \(x + (x + 20) + 2x = 200\). 4. Combine like terms: \(4x + 20 = 200\). 5. Subtract \(20\): \(4x = 180\). Divide by \(4\): \(x = 45\). 6. Lucas has \(45 + 20 = 65\) cards, and Noah has \(2 \cdot 45 = 90\) cards.

Answer

Equation: \(x+(x+20)+2x=200\) Mia: \(45\) cards; Lucas: \(65\) cards; Noah: \(90\) cards

Model situations with equations and inequalities 16

Freyja has already earned \(26\) service points. Each community event adds \(9\) points. Freyja needs at least \(80\) points to qualify for an award. Write and solve an inequality for the number \(e\) of additional events needed. Interpret the solution.

Hints

- Translate “at least” into the appropriate comparison symbol. - Include both the points already earned and the points from future events. - Interpret the solution using the fact that events are counted in whole numbers.

Solution

1. “At least \(80\)” means the total must be greater than or equal to \(80\): \(26+9e\ge80\). 2. Subtract \(26\): \(9e\ge54\). 3. Divide by \(9\): \(e\ge6\). 4. Since \(e\) counts events, Freyja must attend at least \(6\) additional events.

Answer

\(26+9e\ge80\), so \(e\ge6\). Freyja needs at least \(6\) more events.

Model situations with equations and inequalities 17

A storage shelf can hold no more than \(38\,\text{lb}\). A fixed container weighs \(14\,\text{lb}\), and each identical supply box weighs \(4\,\text{lb}\). Write and solve an inequality for the number \(b\) of supply boxes that can be placed on the shelf. What is the greatest possible number of boxes?

Hints

- Translate “no more than” into an inequality symbol. - Include the fixed container weight before the box weight. - Use the whole-number meaning of \(b\) when stating the maximum.

Solution

1. “No more than \(38\,\text{lb}\)” gives \(14+4b\le38\). 2. Subtract \(14\): \(4b\le24\). 3. Divide by \(4\): \(b\le6\). 4. Since \(b\) is a count, the greatest possible number is \(6\).

Answer

\(14+4b\le38\), so \(b\le6\). The greatest possible number is \(6\) boxes.

Model situations with equations and inequalities 18

A delivery cart can carry at most \(53\) pounds. An empty crate weighs \(8\) pounds, and each sack adds \(6\) pounds. Write and solve an inequality to find the greatest whole number of sacks the cart can carry. Check your result.

Hints

- Include both the empty crate and the sacks in the total weight. - Translate “at most” into an inequality symbol. - If your solution is not a whole number, what does a sack count allow?

Solution

1. If \(s\) is the number of sacks, \(8+6s\le53\). 2. Subtract \(8\): \(6s\le45\). Divide by \(6\): \(s\le7.5\). 3. Because sacks are counted in whole numbers, the maximum is \(7\). 4. Seven sacks weigh \(8+6\cdot7=50\) pounds; eight weigh \(56\) pounds and exceed the limit.

Answer

\(8+6s\le53\), so \(s\le7.5\) and the maximum is \(7\) whole sacks. Check: \(7\) sacks weigh \(50\) pounds, while \(8\) weigh \(56\) pounds and exceed the limit.

Model situations with equations and inequalities 19

Hillside Pool charges \(\$3\) for each visit. A weekly pass costs \(\$14\). For how many visits in one week is the pass cheaper? Let \(v\) be the number of visits. Write and solve an inequality, then give the smallest whole-number answer.

Hints

- Write a cost expression for each way of paying. - Which total must be smaller for the weekly pass to be the better deal? - Can someone make a fraction of a visit?

Solution

1. A pass costs \(14\) dollars, and paying per visit costs \(3v\) dollars. 2. The pass is cheaper when \(14<3v\). Dividing by \(3\) gives \(v>\frac{14}{3}\). 3. Visits are whole numbers, so the pass is cheaper for \(5\) or more visits in the week.

Answer

\(14<3v\), so \(v>\frac{14}{3}\). The pass is cheaper for at least \(5\) visits in one week.

Model situations with equations and inequalities 20

Create one realistic story that can be modeled by \(5x+4=34\). Say what \(x\) means, then solve the equation and check your story.

Hints

- Think of a setting with a fixed starting amount and the same amount added each time. - What could the coefficient of \(x\) represent? - Make sure the result and units fit the story you create.

Solution

1. One possible story is: A skating rink charges \(\$4\) to rent skates and \(\$5\) per hour; the total is \(\$34\). Let \(x\) be the hours. 2. Subtract \(4\): \(5x=30\). Divide by \(5\): \(x=6\). 3. Six hours at \(\$5\) plus the rental fee costs \(5\cdot6+4=34\) dollars.

Answer

Answers may vary. Example: A skating rink charges \(\$4\) for skates and \(\$5\) per hour. Here \(x\) is the number of hours, \(x=6\), and the check is \(5\cdot6+4=34\) dollars.

Model situations with equations and inequalities 21

A cookbook gives this rule for roasting meat: “Allow \(45\) minutes per pound, plus \(20\) additional minutes for preheating and resting.” a) Write an equation for the total time \(T\), in minutes, for a roast weighing \(m\) pounds. b) Find the total time for a \(2.5\)-pound roast. c) A roast takes \(155\) minutes in all. Use your equation to find its weight.

Hints

- The phrase “per pound” indicates multiplication, and “plus” indicates a fixed addition. - Identify what the variable \(m\) represents. - To find the weight from the total time, reverse the operations in the equation.

Solution

1. The cooking time is \(45\) minutes for each pound plus \(20\) fixed minutes, so \(T=45m+20\). 2. For \(m=2.5\), \(T=45\cdot 2.5+20=112.5+20=132.5\) minutes. 3. Set \(T=155\): \(155=45m+20\). Subtract \(20\) to get \(135=45m\), then divide by \(45\): \(m=3\).

Answer

a) \(T=45m+20\) b) \(132.5\) minutes c) \(3\) pounds

Model situations with equations and inequalities 22

A student group has a budget of \(\$30.00\) for drinks. They first buy \(12\) bottles of water for \(\$0.75\) each. They will use the remaining money to buy as many bottles of apple juice as possible at \(\$1.40\) each. Write and solve an inequality for the number of juice bottles \(x\). How many bottles can the group buy at most?

Hints

- Identify the fixed amount already spent and the amount left for juice. - Translate “cannot exceed” into a relationship between the total cost and the budget. - After solving, check that the answer makes sense for a whole-number count of bottles.

Solution

1. The water costs \(12 \cdot \$0.75 = \$9.00\). 2. The total cost cannot exceed the budget, so \(9 + 1.40x \leq 30\). 3. Subtract \(9\): \(1.40x \leq 21\). 4. Divide by \(1.40\): \(x \leq 15\). 5. Since \(x\) counts bottles, the greatest possible value is \(15\).

Answer

The inequality is \(12 \cdot 0.75 + 1.40x \leq 30\), or \(9 + 1.40x \leq 30\). The group can buy at most \(15\) bottles of apple juice.

Model situations with equations and inequalities 23

Posters will be hung along a \(44\,\text{ft}\) wall. Each poster is \(4\,\text{ft}\) wide. All gaps—including the gaps from the end posters to the ends of the wall—must have the same width. a) What is the width of each gap when \(4\) posters are hung? b) Let \(x\) be the number of posters when every gap is exactly \(4\,\text{ft}\) wide. Write an equation that models the wall length, solve it, and state how many posters fit.

Hints

- Count the gaps at both ends as well as the gaps between posters. - For part a), subtract the poster widths from the wall length before sharing the remaining space among the gaps. - For part b), express both poster width and total gap width using the same variable. - The equation must account for the entire \(44\,\text{ft}\) wall.

Solution

1. Four posters use \(4 \cdot 4\,\text{ft} = 16\,\text{ft}\). 2. The remaining length is \(44\,\text{ft} - 16\,\text{ft} = 28\,\text{ft}\). 3. Four posters create \(5\) equal gaps: one at each end and three between posters. 4. Each gap is \(28\,\text{ft} \div 5 = 5.6\,\text{ft}\). 5. For b), let \(x\) be the number of posters. There are \(x + 1\) gaps, so \(4x + 4(x + 1) = 44\). 6. Solve: \(4x + 4x + 4 = 44\), so \(8x = 40\) and \(x = 5\).

Answer

a) \(5.6\,\text{ft}\) b) Equation: \(4x+4(x+1)=44\); \(5\) posters

Model situations with equations and inequalities 24

Square posters with side length \(2.5\,\text{ft}\) will be hung in one row along a \(43\,\text{ft}\) school hallway wall. Equal gaps of width \(x\) will be left between adjacent posters and at both ends of the row. a) Explain why the total-length model for \(n\) posters is \(L=2.5n+x(n+1)\). b) Exactly \(12\) posters will be used. Find \(x\) when the entire wall is filled. c) How would the model change if the posters touched each other but the row had a fixed \(4\,\text{ft}\) margin at each end?

Hints

- Visualize a short row with two or three posters and count the gaps. - Substitute the known values into the length model. - Subtract the total poster width before solving for the gap width.

Solution

1. a) The posters contribute \(2.5n\) feet. There is one more gap than posters, so the gaps contribute \(x(n+1)\) feet. 2. b) Substitute \(n=12\) and \(L=43\): \(2.5\cdot12+13x=43\). 3. Simplify: \(30+13x=43\), so \(13x=13\) and \(x=1\,\text{ft}\). 4. c) The posters contribute \(2.5n\), and the two fixed margins contribute \(8\). The model is \(L=2.5n+8\).

Answer

a) There are \(n\) posters and \(n+1\) equal gaps. b) \(x=1\,\text{ft}\) c) \(L=2.5n+8\)

Model situations with equations and inequalities 25

A water tank starts with \(1000\) gallons. Because of a leak, it loses water at a constant rate of \(15\) gallons per hour. a) Write an expression for the amount of water remaining after \(t\) hours. b) Use the expression to find how much water remains after \(12\) hours. c) Write and solve an equation for the time when the tank is empty. Round to the nearest tenth of an hour.

Hints

- Model the amount remaining as the starting amount minus repeated hourly loss. - Substitute a time only after writing the expression. - For an empty tank, set the remaining amount equal to the amount an empty tank contains. - Use the requested rounding only after solving the equation.

Solution

1. Start with \(1000\) and subtract \(15\) gallons for each hour: \(1000-15t\). 2. For \(t=12\), \(1000-15\cdot 12=1000-180=820\). 3. Set the remaining amount equal to \(0\): \(1000-15t=0\). Then \(15t=1000\), so \(t=\frac{1000}{15}=66.666\ldots\approx 66.7\).

Answer

a) \(1000-15t\) b) \(820\) gallons c) Equation: \(1000-15t=0\); \(t\approx66.7\) hours

Model situations with equations and inequalities 26

A bank statement shows a \(\$450.00\) rent payment and a \(\$1250.60\) paycheck deposit. After both transactions, the account balance is \(\$840.25\). Write and solve an equation to find the account balance immediately before the two transactions.

Hints

- Decide whether each transaction increases or decreases the account balance. - Write an equation that begins with the unknown starting balance and ends with the given final balance. - Combine the two known changes before isolating the variable.

Solution

1. Let \(x\) be the starting balance. Write the equation \(x - 450.00 + 1250.60 = 840.25\). 2. Combine the known changes: \(-450.00 + 1250.60 = 800.60\), so \(x + 800.60 = 840.25\). 3. Subtract \(800.60\) from both sides: \(x = 840.25 - 800.60\). 4. Calculate the starting balance: \(x = 39.65\).

Answer

The account balance was \(\$39.65\) before the two transactions.

Model situations with equations and inequalities 27

A hot-air balloon rises \(245\,\text{m}\) and then descends \(312\,\text{m}\). Its final altitude is \(128\,\text{m}\). Write and solve an equation to find the balloon’s altitude before these two changes.

Hints

- Decide which operation represents rising and which represents descending. - Find the net change in altitude after both movements. - Use an inverse operation to undo the net change and isolate the starting altitude.

Solution

1. Let \(h\) be the starting altitude. Write the equation \(h + 245 - 312 = 128\). 2. Combine the altitude changes: \(245 - 312 = -67\), so \(h - 67 = 128\). 3. Add \(67\) to both sides: \(h = 128 + 67\). 4. Calculate the starting altitude: \(h = 195\).

Answer

The balloon’s starting altitude was \(195\,\text{m}\).

Model situations with equations and inequalities 28

A row of adjacent squares is built with matchsticks. The first square uses \(4\) matchsticks. Each additional square shares one side with the square before it, so it requires only \(3\) more matchsticks. a) Write an equation for the number of matchsticks \(S\) needed for \(n\) squares. b) How many matchsticks are needed for \(15\) squares? c) A row uses \(76\) matchsticks. How many squares are in the row?

Hints

- Identify how many matchsticks each new square adds. - Account for the first square separately before simplifying. - Test your equation with \(2\) or \(3\) squares. - For part c), undo the added \(1\) before dividing by \(3\).

Solution

1. The first square uses \(4\) matchsticks, and the remaining \(n-1\) squares each add \(3\), so \(S=4+3(n-1)\). 2. Simplify: \(S=4+3n-3=3n+1\). 3. For \(n=15\), \(S=3\cdot15+1=46\). 4. For \(S=76\), solve \(76=3n+1\): \(75=3n\), so \(n=25\).

Answer

a) \(S=3n+1\) b) \(46\) matchsticks c) \(25\) squares

Model situations with equations and inequalities 29

A wireframe model of a rectangular prism is made from \(20\,\text{ft}\) of wire. One base edge is twice the length of the other base edge, and the height equals the longer base edge. Let \(x\) be the shorter base edge. Write a one-variable equation for the total length of all \(12\) edges, solve it, and report the three edge lengths.

Hints

- A rectangular prism has four edges of each of its three dimensions. - Express both longer dimensions in terms of the shorter edge. - Build one equation whose left side represents all twelve edge lengths. - Solve only after the wire-length model is complete.

Solution

1. Let \(x\) feet be the shorter base edge. Then the longer base edge and the height are each \(2x\) feet. 2. A rectangular prism has four edges of each of the three lengths. Therefore, the total wire length is \(4x + 4(2x) + 4(2x)\). 3. Write and simplify the equation: \(4x + 8x + 8x = 20\), so \(20x = 20\). 4. Divide by \(20\): \(x = 1\). 5. The shorter base edge is \(1\,\text{ft}\), and the longer base edge and height are each \(2\,\text{ft}\).

Answer

Equation: \(4x+4(2x)+4(2x)=20\) The three dimensions are \(1\,\text{ft}\), \(2\,\text{ft}\), and \(2\,\text{ft}\).

Model situations with equations and inequalities 30

A wireframe square pyramid is made from a \(160\,\text{in.}\) aluminum rod. Its four base edges have equal length. Each of the four edges from the base to the apex is exactly \(5\,\text{in.}\) longer than a base edge. Let \(a\) be a base-edge length. Write an equation for the total rod length, solve it, and report both edge lengths.

Hints

- Count how many edges have each of the two lengths. - Express an apex edge using the base-edge variable. - The sum of all eight edge lengths must equal the rod length. - Solve the model before calculating the longer edge.

Solution

1. Let \(a\) inches be the length of a base edge. Each edge to the apex has length \(a + 5\). 2. The pyramid has \(4\) base edges and \(4\) edges to the apex, so \(4a + 4(a + 5) = 160\). 3. Distribute and combine like terms: \(4a + 4a + 20 = 160\), so \(8a = 140\). 4. Divide by \(8\): \(a = 17.5\). 5. Each edge to the apex is \(17.5\,\text{in.} + 5\,\text{in.} = 22.5\,\text{in.}\).

Answer

Equation: \(4a+4(a+5)=160\) Base edge: \(17.5\,\text{in.}\) Edge to the apex: \(22.5\,\text{in.}\)

Model situations with equations and inequalities 31

Paul is selling \(40\) used video games at a flea market. His total cost for the booth and the games was \(\$120.00\). He has already sold \(15\) games for \(\$2.50\) each. At what price must he sell each of the remaining games to recover exactly his total cost? Write and solve an equation.

Hints

- Find the revenue from the games already sold. - Determine how many games remain. - Set total revenue equal to the total cost. - Divide the remaining amount by the number of remaining games.

Solution

1. There are \(40 - 15 = 25\) games remaining. Let \(x\) dollars be the price of each remaining game. 2. The total revenue must equal \(\$120.00\): \(15 \cdot 2.50 + 25x = 120\). 3. Simplify: \(37.50 + 25x = 120\). 4. Subtract \(37.50\): \(25x = 82.50\). 5. Divide by \(25\): \(x = 3.30\).

Answer

Paul must sell each remaining game for \(\$3.30\).

Model situations with equations and inequalities 32

A student newspaper club prints \(200\) copies of a new issue. Printing costs \(\$180.00\), and the club wants to earn an additional \(\$60.00\) for the class fund. The first \(120\) copies were sold for \(\$1.00\) each. At what price must the remaining copies be sold? Write and solve an equation.

Hints

- Add the printing cost and desired profit to find the target revenue. - Find the revenue already earned. - Determine how many copies remain. - Use an equation to distribute the remaining revenue across those copies.

Solution

1. The target revenue is \(180 + 60 = 240\) dollars. There are \(200 - 120 = 80\) copies remaining. 2. Let \(x\) dollars be the price of each remaining copy. Write \(120 \cdot 1.00 + 80x = 240\). 3. Simplify: \(120 + 80x = 240\). 4. Subtract \(120\): \(80x = 120\). 5. Divide by \(80\): \(x = 1.50\).

Answer

The remaining copies must be sold for \(\$1.50\) each.

Model situations with equations and inequalities 33

A raffle has \(500\) tickets. The original plan was to sell every ticket for \(\$1.50\). During a promotion, \(300\) tickets were sold for an average of \(\$1.10\) each. What price must be charged for each of the remaining \(200\) tickets to reach exactly the originally planned total revenue? Write and solve an equation.

Hints

- Find the originally planned total revenue. - Calculate the revenue already earned during the promotion. - Subtract to find the revenue still needed. - Divide that amount among the remaining tickets.

Solution

1. The originally planned revenue was \(500 \cdot 1.50 = 750\) dollars. 2. Let \(x\) dollars be the price of each remaining ticket. Write \(300 \cdot 1.10 + 200x = 750\). 3. Simplify: \(330 + 200x = 750\). 4. Subtract \(330\): \(200x = 420\). 5. Divide by \(200\): \(x = 2.10\).

Answer

Each remaining ticket must cost \(\$2.10\).

Model situations with equations and inequalities 34

Lena is saving for a mountain bike that costs \(\$540\). She already has \(\$155\) and can save \(\$25\) each month. After how many full months will she have enough money? Solve using an inequality.

Hints

- Write an expression for the total amount saved after \(n\) months. - “Enough money” means the amount must be at least the price. - Round up to the next whole month after solving.

Solution

1. Let \(n\) be the number of months. 2. After \(n\) months, Lena will have \(155 + 25n\) dollars. 3. She needs at least \(\$540\), so \(155 + 25n \geq 540\). 4. Subtract \(155\): \(25n \geq 385\). 5. Divide by \(25\): \(n \geq 15.4\). 6. Since only full months count, the least possible value is \(16\).

Answer

Lena will have enough money after \(16\) months.

Model situations with equations and inequalities 35

A rectangle has a perimeter of at most \(48\,\text{in.}\). One side is exactly three times as long as the other side. Let \(x\) be the shorter side. Write and solve an inequality for the perimeter, then give the greatest possible length of each side.

Hints

- Express the longer side in terms of the shorter side. - Use the rectangle perimeter formula to build the left side of the inequality. - Translate “at most” as an upper-bound inequality. - After solving, use the boundary value to find the longest possible sides.

Solution

1. Let \(x\) inches be the shorter side. The longer side is \(3x\) inches. 2. The perimeter is \(2(x + 3x) = 8x\). 3. Because the perimeter is at most \(48\,\text{in.}\), write \(8x \leq 48\). 4. Divide by \(8\): \(x \leq 6\). 5. The greatest possible shorter side is \(6\,\text{in.}\), and the greatest possible longer side is \(3 \cdot 6\,\text{in.} = 18\,\text{in.}\).

Answer

Inequality: \(2(x+3x)\le48\), or \(8x\le48\) Greatest possible side lengths: \(6\,\text{in.}\) and \(18\,\text{in.}\)

Model situations with equations and inequalities 36

A rectangular flower bed can use at most \(40\,\text{ft}\) of edging. Its length will be exactly \(4\,\text{ft}\) greater than its width. Let \(w\) be the width. Write and solve an inequality for the edging constraint, then state the greatest possible width.

Hints

- Express the length using the width variable. - Build a perimeter expression before applying the edging limit. - Translate “at most” into the correct inequality direction. - Interpret the inequality boundary in the physical context.

Solution

1. Let \(w\) feet be the width. The length is \(w + 4\) feet. 2. The perimeter is \(2(w + w + 4) = 4w + 8\). 3. Because no more than \(40\,\text{ft}\) of edging is available, write \(4w + 8 \leq 40\). 4. Subtract \(8\): \(4w \leq 32\). 5. Divide by \(4\): \(w \leq 8\).

Answer

Inequality: \(2(w+w+4)\le40\), or \(4w+8\le40\) Greatest possible width: \(8\,\text{ft}\)

Model situations with equations and inequalities 37

A small birdbath starts with \(15\,\text{L}\) of water. During a rain shower, \(0.5\,\text{L}\) flows in each minute. At the same time, a leak drains a constant amount \(y\), in liters per minute. After \(20\) minutes, the birdbath contains exactly \(21\,\text{L}\). Niran proposes the equation \(15 + 0.5 \cdot 20 - y \cdot 20 = 21\). a) Does the equation correctly model the situation? Briefly explain. b) Find the water loss \(y\) in liters per minute.

Hints

- Identify what the \(15\) and \(21\) represent. - Match each term in the equation to a change during the \(20\) minutes. - Simplify the numerical terms before isolating \(y\).

Solution

1. The equation is correct: \(15\) is the starting amount, \(0.5 \cdot 20\) is the rain added in \(20\) minutes, \(y \cdot 20\) is the amount lost, and \(21\) is the final amount. 2. Simplify: \(15 + 10 - 20y = 21\). 3. Combine constants: \(25 - 20y = 21\). 4. Subtract \(21\) from \(25\): \(4 = 20y\). 5. Divide by \(20\): \(y = 0.2\).

Answer

a) Yes. The equation combines the starting amount, the rain added, and the leak loss to produce the final amount. b) The birdbath loses \(0.2\,\text{L}\) per minute.

Model situations with equations and inequalities 38

A school library receives a shipment of new books. One-half are novels, one-fifth are nonfiction books, and one-eighth are graphic novels. The remaining \(63\) books are children’s books. Write and solve an equation to find the total number of new books \(x\). Briefly explain what \(1 - \left(\frac{1}{2} + \frac{1}{5} + \frac{1}{8}\right)\) represents in this situation.

Hints

- Subtract the known category fractions from one whole to find the children’s-book fraction. - Express the fractions with a common denominator. - Set the remaining fraction of the total equal to \(63\).

Solution

1. Write the equation \(x\left(1 - \left(\frac{1}{2} + \frac{1}{5} + \frac{1}{8}\right)\right) = 63\). 2. Use denominator \(40\): \(\frac{1}{2} + \frac{1}{5} + \frac{1}{8} = \frac{20}{40} + \frac{8}{40} + \frac{5}{40} = \frac{33}{40}\). 3. Find the remaining fraction: \(1 - \frac{33}{40} = \frac{7}{40}\). 4. Solve \(\frac{7}{40}x = 63\): \(x = 63 \cdot \frac{40}{7}\). 5. Calculate: \(x = 360\). 6. The expression represents the fraction of all new books that are children’s books.

Answer

The equation can be written as \(\frac{7}{40}x = 63\), and the total is \(x = 360\) books. The expression \(1 - \left(\frac{1}{2} + \frac{1}{5} + \frac{1}{8}\right)\) is the fraction of the shipment made up of children’s books.

Model situations with equations and inequalities 39

Max and Julia start \(18\) miles apart and walk toward each other. Max walks at \(4\,\text{mph}\). Julia walks \(50\%\) faster than Max. They start at the same time. Use a linear equation to determine who travels farther before they meet and by how many miles.

Hints

- Find Julia's speed first. - Write each distance as rate times the same meeting time. - The two distances add to the original separation. - Subtract the smaller distance from the larger distance.

Solution

1. Julia's speed is \(4 \cdot 1.5 = 6\,\text{mph}\). 2. Let \(t\) be the meeting time in hours. Their distances add to \(18\) miles, so \(4t + 6t = 18\). 3. Combine like terms: \(10t = 18\), so \(t = 1.8\) hours. 4. Max travels \(4 \cdot 1.8 = 7.2\) miles, and Julia travels \(6 \cdot 1.8 = 10.8\) miles. 5. Julia travels farther by \(10.8 - 7.2 = 3.6\) miles.

Answer

Julia travels farther, by \(3.6\) miles.

Model situations with equations and inequalities 40

A class has a budget of \(\$280.00\) for a field trip. The bus costs a flat \(\$135.00\), and zoo admission costs \(\$5.20\) per person. a) Let \(x\) be the number of people. Write a two-step inequality that models the complete budget constraint. b) Solve the inequality. c) Use the solution set to state the greatest whole-number value of \(x\) that is feasible.

Hints

- Account for both the fixed cost and the amount that changes with the number of people. - Translate the budget ceiling into an inequality before solving. - After solving algebraically, apply the fact that \(x\) counts whole people.

Solution

1. The complete cost is \(135+5.20x\), and it may not exceed \(280\), so \(135+5.20x\le280\). 2. Subtract \(135\): \(5.20x\le145\). 3. Divide by \(5.20\): \(x\le\frac{725}{26}\approx27.8846\). 4. Since \(x\) counts people, the greatest feasible whole number is \(27\).

Answer

a) \(135+5.20x\le280\) b) \(x\le\frac{725}{26}\approx27.8846\) c) At most \(27\) people can attend.

Model situations with equations and inequalities 41

Ms. Weber’s bank account balance is \(-\$120\). After her garage rent is withdrawn, the new balance is \(-\$185\). Write an equation using \(x\) for the garage rent, and solve for \(x\).

Hints

- The rent decreases the account balance. - Write the relationship as previous balance minus rent equals new balance. - Use inverse operations to isolate \(x\). - Check that subtracting your answer produces the stated new balance.

Solution

1. The previous balance minus the rent equals the new balance, so \(-120-x=-185\). 2. Add \(120\) to both sides: \(-x=-65\). 3. Multiply both sides by \(-1\): \(x=65\). 4. The garage rent is \(\$65\).

Answer

Equation: \(-120-x=-185\) Solution: \(x=65\) The garage rent is \(\$65\).

Model situations with equations and inequalities 42

The total cost of manufacturing a part includes material cost, labor cost, and a fixed \(\$100\) machine fee. Labor costs \(25\%\) more than materials. The total cost is \(\$1900\). Let \(x\) be the material cost. Write a one-variable equation that models all three cost components, solve it, and report the material cost.

Hints

- Express the labor cost as a multiple of the material cost. - Include both variable cost components and the fixed machine fee in one total-cost equation. - Combine like terms only after the model is written. - Check that the three cost components sum to the stated total.

Solution

1. Let \(x\) dollars be the material cost. 2. Labor costs \(25\%\) more, so the labor cost is \(1.25x\). 3. Add all cost components: \(x + 1.25x + 100 = 1900\). 4. Combine like terms: \(2.25x + 100 = 1900\). 5. Subtract \(100\): \(2.25x = 1800\). 6. Divide by \(2.25\): \(x = 800\).

Answer

Equation: \(x+1.25x+100=1900\) Material cost: \(\$800\)

Model situations with equations and inequalities 43

A bus for a class field trip costs a flat \(\$450\). Zoo admission costs \(\$12\) per person. The total trip cost is \(\$774\). Let \(x\) be the number of people. Write and solve an equation for the total trip cost, then report how many people are attending.

Hints

- Separate the fixed bus cost from the per-person admission cost. - Use one variable for the number of people and build the complete total-cost equation. - Remove the fixed cost before isolating the variable. - Check the total cost with the resulting attendance.

Solution

1. Let \(x\) be the number of people. 2. Add the fixed bus cost and the admission cost: \(450 + 12x = 774\). 3. Subtract \(450\): \(12x = 324\). 4. Divide by \(12\): \(x = 27\).

Answer

Equation: \(450+12x=774\) \(27\) people

Model situations with equations and inequalities 44

Two adjacent angles form a straight angle, so their measures add to \(180^\circ\). One angle is \(44^\circ\) greater than the other. Find both angle measures by writing and solving an equation.

Hints

- Recall the sum of two angles that form a straight angle. - Represent the smaller angle with a variable. - Express the larger angle using the given difference. - Add the two expressions and solve.

Solution

1. Let \(x\) degrees be the smaller angle. The larger angle is \(x + 44\) degrees. 2. Their sum is \(180^\circ\), so \(x + (x + 44) = 180\). 3. Combine like terms: \(2x + 44 = 180\). 4. Subtract \(44\): \(2x = 136\). 5. Divide by \(2\): \(x = 68\). 6. The larger angle is \(68^\circ + 44^\circ = 112^\circ\).

Answer

The angle measures are \(68^\circ\) and \(112^\circ\).

Model situations with equations and inequalities 45

Three adjacent angles lie along a straight line. The second angle is three times the first angle. The third angle equals the sum of the first two angles. Find all three angle measures by writing and solving an equation.

Hints

- A straight angle measures \(180^\circ\). - Express all three angles using the first angle. - Use the relationship between the third angle and the first two. - Check that the three results add to \(180^\circ\).

Solution

1. Let \(x\) degrees be the first angle. 2. The second angle is \(3x\), and the third angle is \(x + 3x = 4x\). 3. The three angles form a straight angle, so \(x + 3x + 4x = 180\). 4. Combine like terms: \(8x = 180\). 5. Divide by \(8\): \(x = 22.5\). 6. The angles are \(22.5^\circ\), \(3 \cdot 22.5^\circ = 67.5^\circ\), and \(4 \cdot 22.5^\circ = 90^\circ\).

Answer

The three angle measures are \(22.5^\circ\), \(67.5^\circ\), and \(90^\circ\).

Model situations with equations and inequalities 46

The sum of four consecutive integers is \(66\). Use an equation to find the four integers.

Hints

- Express each integer after the first in terms of \(x\). - Add the four expressions and set their sum equal to \(66\). - Combine like terms before solving.

Solution

1. Let \(x\) be the first integer. 2. The next three integers are \(x + 1\), \(x + 2\), and \(x + 3\). 3. Write the equation \(x + (x + 1) + (x + 2) + (x + 3) = 66\). 4. Combine like terms: \(4x + 6 = 66\). 5. Subtract \(6\): \(4x = 60\). 6. Divide by \(4\): \(x = 15\). 7. The integers are \(15\), \(16\), \(17\), and \(18\).

Answer

The four integers are \(15\), \(16\), \(17\), and \(18\).

Model situations with equations and inequalities 47

A \(100\,\text{in.}\) rope is cut into three pieces. The first piece is twice as long as the second piece. The third piece is exactly \(5\,\text{in.}\) shorter than the first piece. Let \(x\) be the second piece’s length. Write one equation that models all three pieces and the total length, solve it, and report every piece length.

Hints

- Choose the second piece as the base quantity because the other two are described from it. - Express all three lengths with the same variable before adding them. - Set their sum equal to the original rope length. - Use the solved variable to recover all three piece lengths.

Solution

1. Let \(x\) be the length of the second piece in inches. 2. The first piece is \(2x\), and the third piece is \(2x - 5\). 3. Write the total-length equation \(2x + x + (2x - 5) = 100\). 4. Combine like terms: \(5x - 5 = 100\). 5. Add \(5\): \(5x = 105\). Divide by \(5\): \(x = 21\). 6. The first piece is \(2 \cdot 21 = 42\,\text{in.}\), the second is \(21\,\text{in.}\), and the third is \(42 - 5 = 37\,\text{in.}\).

Answer

Equation: \(2x+x+(2x-5)=100\) Piece lengths: \(42\,\text{in.}\), \(21\,\text{in.}\), and \(37\,\text{in.}\)

Model situations with equations and inequalities 48

Three seventh-grade classes raise a total of \(\$510\) for an environmental project. Class B raises exactly two-thirds as much as Class A. Class C raises \(\$15\) less than Class B. Let \(x\) be Class A’s amount. Write one equation that models the three class amounts, solve it, and report how much each class raised.

Hints

- Use Class A as the base quantity for the fractional relationship. - Express Class C from the Class B expression, not directly from Class A. - Add all three expressions and set the sum equal to the fundraiser total. - Keep the fractional coefficient exact while solving.

Solution

1. Let \(x\) be the amount raised by Class A. 2. Class B raises \(\frac{2}{3}x\), and Class C raises \(\frac{2}{3}x - 15\). 3. Write the total equation \(x + \frac{2}{3}x + \left(\frac{2}{3}x - 15\right) = 510\). 4. Combine like terms: \(\frac{7}{3}x - 15 = 510\). 5. Add \(15\): \(\frac{7}{3}x = 525\). 6. Multiply by \(\frac{3}{7}\): \(x = 225\). 7. Class B raises \(\$150\), and Class C raises \(\$135\).

Answer

Equation: \(x+\frac{2}{3}x+\left(\frac{2}{3}x-15\right)=510\) Class A: \(\$225\); Class B: \(\$150\); Class C: \(\$135\)

Model situations with equations and inequalities 49

A school cafeteria sells sandwiches for \(\$2.00\) and bottles of apple juice for \(\$1.50\). Total sales were \(\$110.00\), and exactly twice as many sandwiches as juice bottles were sold. Let \(x\) be the number of juice bottles. Write a one-variable revenue equation, solve it, and report both quantities.

Hints

- Represent the smaller item count with a variable. - Use the “twice as many” condition to express the other count. - Multiply each item count by its price before adding the revenues. - Check both the count relationship and the total sales.

Solution

1. Let \(x\) be the number of juice bottles sold. Then \(2x\) sandwiches were sold. 2. Write the revenue equation: \(2.00(2x) + 1.50x = 110\). 3. Combine like terms: \(4x + 1.5x = 110\), so \(5.5x = 110\). 4. Divide by \(5.5\): \(x = 20\). 5. The number of sandwiches is \(2 \cdot 20 = 40\).

Answer

Equation: \(2.00(2x)+1.50x=110\) \(40\) sandwiches and \(20\) bottles of apple juice

Model situations with equations and inequalities 50

A hiking group travels a total of \(46\,\text{miles}\) over three days. On the second day, the group hikes \(1.5\) times the distance from the first day. On the third day, the group hikes \(2\,\text{miles}\) less than on the second day. Let \(x\) be the first-day distance. Write one equation for the three-day total, solve it, and report the distance hiked each day.

Hints

- Use the first-day distance as the base variable. - Build the second- and third-day expressions from that same variable. - Add the three daily expressions and equate them to the total distance. - Check that the solved distances satisfy both stated relationships.

Solution

1. Let \(x\) be the distance hiked on the first day. 2. The second-day distance is \(1.5x\), and the third-day distance is \(1.5x - 2\). 3. Write the total equation \(x + 1.5x + (1.5x - 2) = 46\). 4. Combine like terms: \(4x - 2 = 46\). 5. Add \(2\): \(4x = 48\). Divide by \(4\): \(x = 12\). 6. The second-day distance is \(1.5 \cdot 12 = 18\,\text{miles}\), and the third-day distance is \(18 - 2 = 16\,\text{miles}\).

Answer

Equation: \(x+1.5x+(1.5x-2)=46\) Day 1: \(12\,\text{miles}\); Day 2: \(18\,\text{miles}\); Day 3: \(16\,\text{miles}\)

Model situations with equations and inequalities 51

Three classes raise a total of \(\$720\) for a community project. Class B raises \(\$40\) more than Class A. Class C raises the same amount as Classes A and B combined. 1) Let \(x\) be Class A’s amount. Write and solve one equation that models all three contributions, then report each amount. 2) What fraction or percent of the total is raised by Class C?

Hints

- Express Class B from Class A before building Class C’s expression. - Class C equals the sum of the first two class expressions. - Add all three contributions to create the total-fundraising equation. - For part 2, compare Class C’s solved amount with the full total.

Solution

1. Let \(x\) be the amount raised by Class A. 2. Class B raises \(x + 40\), and Class C raises \(x + (x + 40) = 2x + 40\). 3. Write \(x + (x + 40) + (2x + 40) = 720\). 4. Combine like terms: \(4x + 80 = 720\). 5. Subtract \(80\): \(4x = 640\). Divide by \(4\): \(x = 160\). 6. Class B raises \(\$200\), and Class C raises \(\$360\). 7. Class C’s share is \(\frac{360}{720} = \frac{1}{2} = 50\%\).

Answer

1) Equation: \(x+(x+40)+(2x+40)=720\). Class A: \(\$160\); Class B: \(\$200\); Class C: \(\$360\). 2) \(\frac{1}{2}\), or \(50\%\)

Model situations with equations and inequalities 52

A triangle has a perimeter of \(27\,\text{in.}\). The second side is \(3\,\text{in.}\) longer than the first side, and the third side is twice as long as the first side. a) Write an equation that can be used to find the first side length \(x\). b) Find all three side lengths. c) How would the equation in part a) change if the third side were \(3\,\text{in.}\) shorter than the first side instead of twice as long?

Hints

- Add the three side lengths to represent the perimeter. - Translate “twice as long” and “shorter than” into algebraic expressions. - Express every side in terms of \(x\). - For part c), replace only the expression for the third side.

Solution

1. Add the three side expressions: \(x + (x + 3) + 2x = 27\). 2. Combine like terms: \(4x + 3 = 27\). 3. Subtract \(3\): \(4x = 24\). Divide by \(4\): \(x = 6\). 4. The three side lengths are \(6\,\text{in.}\), \(9\,\text{in.}\), and \(12\,\text{in.}\). 5. Under the changed condition, the third side would be \(x - 3\), so the equation would be \(x + (x + 3) + (x - 3) = 27\).

Answer

a) \(x + (x + 3) + 2x = 27\), or \(4x + 3 = 27\) b) The side lengths are \(6\,\text{in.}\), \(9\,\text{in.}\), and \(12\,\text{in.}\). c) The equation would be \(x + (x + 3) + (x - 3) = 27\).

Model situations with equations and inequalities 53

Consider the equation \(x+(x-2)+3x=43\). a) Create a number puzzle or real-world situation represented by this equation. Briefly explain what \(x\), \(x-2\), and \(3x\) mean in your situation. b) Solve the equation. Give the value of \(x\) and the values of the other two expressions.

Hints

- Think of three related quantities whose total is \(43\). - In a context, what could “\(2\) fewer” and “three times as many” describe? - Combine all the \(x\)-terms before solving. - Check that the three values add to \(43\).

Solution

1. One possible situation is: Three boxes contain \(43\) apples altogether. The first box contains \(x\) apples, the second contains \(2\) fewer than the first, and the third contains three times as many as the first. 2. Combine like terms: \(x+x-2+3x=43\), so \(5x-2=43\). 3. Add \(2\) to both sides to get \(5x=45\), then divide by \(5\) to get \(x=9\). 4. The three values are \(x=9\), \(x-2=7\), and \(3x=27\). The check is \(9+7+27=43\).

Answer

a) Answers will vary. One example is three boxes containing \(x\), \(x-2\), and \(3x\) apples, for a total of \(43\). b) \(x=9\). The three values are \(9\), \(7\), and \(27\).

Model situations with equations and inequalities 54

Consider the formula \(x=\frac{S-b}{k}\). a) Create a real-world problem that can be modeled by this formula. Explain what \(S\), \(b\), and \(k\) represent in your problem. b) Find \(x\) when \(S=85\), \(b=13\), and \(k=12\).

Hints

- Think of a situation in which part of a total is removed first. - What quantity could be divided into \(k\) equal groups? - The fraction bar groups the entire numerator, so subtract before dividing.

Solution

1. One possible context is dividing a remaining balance into equal payments. Let \(S\) be the total price, \(b\) the amount paid at the start, \(k\) the number of equal payments, and \(x\) the amount of each payment. 2. Substitute the given values: \(x=\frac{85-13}{12}\). 3. Simplify the numerator: \(85-13=72\). 4. Divide: \(x=\frac{72}{12}=6\).

Answer

a) Answers will vary. Example: A used bicycle costs \(\$85\). A buyer pays \(\$13\) at the start and divides the remaining balance into \(12\) equal monthly payments. Here, \(S\) is the total price, \(b\) is the initial payment, \(k\) is the number of monthly payments, and \(x\) is one payment. b) \(x=6\), so each payment is \(\$6\).

Model situations with equations and inequalities 55

A cycling group plans a three-day, \(180\,\text{mile}\) trip. The second-day distance will be twice the first-day distance. The third-day distance will be \(15\,\text{miles}\) shorter than the second-day distance. a) Write an equation to find the first-day distance \(x\). b) Find the distance for each day. c) Without solving again, explain how the equation from part a would change if the third-day distance were instead \(10\,\text{miles}\) longer than the first-day distance.

Hints

- The third-day distance in parts a and b is compared with the second-day distance. - Add the three daily expressions to equal the total distance. - In part c, change only the expression for the third day.

Solution

1. For part a, the three daily distances are \(x\), \(2x\), and \(2x - 15\). Write \(x + 2x + (2x - 15) = 180\). 2. Combine like terms: \(5x - 15 = 180\). 3. Add \(15\): \(5x = 195\). Divide by \(5\): \(x = 39\). 4. The second-day distance is \(2 \cdot 39 = 78\,\text{miles}\), and the third-day distance is \(78 - 15 = 63\,\text{miles}\). 5. For part c, replace the third-day expression with \(x + 10\). The new equation is \(x + 2x + (x + 10) = 180\), or \(4x + 10 = 180\).

Answer

a) \(x + 2x + (2x - 15) = 180\) b) The first-day distance is \(39\,\text{miles}\), the second-day distance is \(78\,\text{miles}\), and the third-day distance is \(63\,\text{miles}\). c) The equation becomes \(x + 2x + (x + 10) = 180\), or \(4x + 10 = 180\).

Model situations with equations and inequalities 56

Ms. Byrne buys \(6\) packs of highlighters and \(12\) pencils for \(\$54.00\). One pack of highlighters costs four times as much as one pencil. a) Use an equation to find the price of one pencil and one pack of highlighters. b) Kavindu claims, “If a pack of highlighters cost only three times as much as a pencil, the savings would buy exactly \(6\) more pencils.” Determine whether the claim is correct.

Hints

- Use the smaller price as the variable. - Express the larger price as a multiple of that variable. - Write an equation for the original total cost. - For part b), compare the total savings with the cost of six pencils.

Solution

1. Let \(x\) dollars be the price of one pencil. A pack of highlighters costs \(4x\) dollars. 2. Write \(12x + 6(4x) = 54\). 3. Combine like terms: \(12x + 24x = 54\), so \(36x = 54\) and \(x = 1.50\). 4. A pencil costs \(\$1.50\), and a pack of highlighters costs \(4 \cdot 1.50 = 6.00\) dollars. 5. Under the changed ratio, a pack would cost \(3 \cdot 1.50 = 4.50\) dollars. The savings per pack would be \(6.00 - 4.50 = 1.50\) dollars. 6. Across \(6\) packs, the savings would be \(6 \cdot 1.50 = 9.00\) dollars. 7. Six pencils cost \(6 \cdot 1.50 = 9.00\) dollars, so Kavindu’s claim is correct.

Answer

a) One pencil costs \(\$1.50\), and one pack of highlighters costs \(\$6.00\). b) The claim is correct because the \(\$9.00\) savings equals the cost of \(6\) pencils.

Model situations with equations and inequalities 57

A workshop uses \(8\) identical LED floodlights and \(2\) identical heaters. Together, they draw \(3600\,\text{W}\). Each heater draws five times as much power as one floodlight. a) Let \(s\) be the power draw of one floodlight. Write and solve an equation for the original system, then report the power draw of one floodlight and one heater. b) The heaters are upgraded so each uses \(20\%\) less power. What is the new total power draw of all \(8\) floodlights and \(2\) heaters?

Hints

- Express a heater’s power draw using the floodlight variable. - Model the original total before solving for either device’s power. - A \(20\%\) reduction means the upgraded heater uses \(80\%\) of its original power. - Recalculate the system total with the solved device powers.

Solution

1. Let \(s\) be the power draw of one floodlight. Each heater draws \(5s\). 2. Write \(8s + 2(5s) = 3600\). 3. Combine like terms: \(18s = 3600\), so \(s = 200\). 4. One heater draws \(5 \cdot 200 = 1000\,\text{W}\). 5. A \(20\%\) reduction is \(0.20 \cdot 1000 = 200\,\text{W}\), so each upgraded heater draws \(800\,\text{W}\). 6. The new total is \(8 \cdot 200 + 2 \cdot 800 = 1600 + 1600 = 3200\,\text{W}\).

Answer

a) Equation: \(8s+2(5s)=3600\). One floodlight draws \(200\,\text{W}\), and one heater draws \(1000\,\text{W}\). b) \(3200\,\text{W}\)

Model situations with equations and inequalities 58

A smartphone battery is at \(80\%\) at the start of an observation. During intensive use, its charge decreases by \(12\) percentage points per hour. Write an inequality and find the times \(t\) when the charge is below \(20\%\). Consider only times up to complete discharge.

Hints

- Write a linear expression for the battery charge after \(t\) hours. - Translate “below” with a strict inequality. - Reverse the inequality sign when dividing by a negative number. - Find when the battery reaches \(0\%\) to set the physical endpoint.

Solution

1. The charge after \(t\) hours is \(80 - 12t\). 2. Write the inequality \(80 - 12t < 20\). 3. Subtract \(80\): \(-12t < -60\). 4. Divide by \(-12\) and reverse the inequality sign: \(t > 5\). 5. Complete discharge occurs when \(80 - 12t = 0\), so \(t = \frac{20}{3} \approx 6.67\). 6. Therefore, within the physical time interval, the charge is below \(20\%\) when \(5 < t \le \frac{20}{3}\).

Answer

The inequality is \(80 - 12t < 20\), and the relevant solution is \(5 < t \le \frac{20}{3}\) hours.

Model situations with equations and inequalities 59

Maya is reading a new book. On the first day, she reads exactly one-third of the pages. On the second day, she reads one-fourth of the total number of pages. She then has exactly \(125\) pages left. Write an equation and find the total number of pages in the book.

Hints

- Use a variable for the total number of pages. - Find the fraction of the book that remains unread. - Relate that remaining fraction directly to \(125\) pages.

Solution

1. Let \(x\) be the total number of pages. 2. The fraction left is \(1-\frac{1}{3}-\frac{1}{4}=\frac{5}{12}\). 3. Model the remaining pages directly: \(\frac{5}{12}x=125\). 4. Multiply both sides by \(\frac{12}{5}\): \(x=300\).

Answer

A suitable equation is \(\frac{5}{12}x=125\). The book has \(300\) pages.

Model situations with equations and inequalities 60

Two trains leave the same station at the same time and travel in opposite directions. After \(1.5\) hours, the trains are \(150\) miles apart. The slower train travels at exactly \(\frac{2}{3}\) the speed of the faster train. Let \(v\) be the faster train’s speed. Write one equation using distance equals rate times time, solve it, and report both speeds.

Hints

- Express the slower speed as a fraction of the faster speed. - Write each train’s distance using the common travel time. - Because the trains move in opposite directions, their distances add to the separation. - Check that the two speeds have the required ratio.

Solution

1. Let \(v\,\text{mph}\) be the faster train's speed. The slower train's speed is \(\frac{2}{3}v\). 2. In \(1.5\) hours, their distances add to \(150\) miles: \(1.5v + 1.5(\frac{2}{3}v) = 150\). 3. Simplify: \(1.5v + v = 150\), so \(2.5v = 150\). 4. Divide by \(2.5\): \(v = 60\). 5. The slower train's speed is \(\frac{2}{3} \cdot 60 = 40\,\text{mph}\).

Answer

Equation: \(1.5v+1.5\left(\frac{2}{3}v\right)=150\) Faster train: \(60\,\text{mph}\); slower train: \(40\,\text{mph}\)

Model situations with equations and inequalities 61

A recreation center charges a \(\$25\) room fee plus \(\$7\) per participant. Every group has at least one participant. Situation A: A group is charged exactly \(\$74\). Situation B: Another group has a budget of no more than \(\$74\). For each situation, choose an equation or inequality, write it, solve it, and explain what the answer means.

Hints

- Compare “exactly” with “no more than.” - Use one cost expression, but a different comparison symbol. - Remember the count must be a whole number and the group must have someone in it.

Solution

1. Situation A requires \(25+7p=74\), so \(7p=49\) and \(p=7\) participants. 2. Situation B requires \(25+7p\le74\), so \(p\le7\). 3. Since each group has at least one participant and counts must be whole, Situation B allows \(1,2,3,4,5,6,7\) participants. 4. The equation identifies one count; the inequality allows a range of counts.

Answer

Situation A: \(25+7p=74\), so \(p=7\). Situation B: \(25+7p\le74\), so any whole-number count from \(1\) through \(7\).

Model situations with equations and inequalities 62

The Pinecrest Middle School fundraiser already has \(\$32\) and earns \(\$6\) for each batch of cards sold. The goal is to raise at least \(\$80\). Mariam writes \(32+6b\le80\). a) Explain the modeling error. b) Write the correct inequality. c) Solve it and interpret the answer.

Hints

- Translate “at least” before examining Mariam's symbol. - Ask whether a total greater than \(80\) should satisfy the model. - Solve the corrected inequality and interpret \(b\) as a count.

Solution

1. The symbol \(\le\) represents totals no greater than \(80\), but the fundraiser needs a total of \(80\) or more. 2. The correct model is \(32+6b\ge80\). 3. Subtract \(32\): \(6b\ge48\). 4. Divide by \(6\): \(b\ge8\). 5. The fundraiser must sell at least \(8\) batches.

Answer

a) Mariam's inequality points in the wrong direction for “at least.” b) \(32+6b\ge80\) c) \(b\ge8\), so at least \(8\) batches must be sold.

Model situations with equations and inequalities 63

A school club has a total budget of \(\$28\) for an order that includes an \(\$8\) shipping charge. Each notebook costs \(\$4\). The graph shows the continuous solution set of the resulting budget inequality within the displayed window; the highlighted segment continues to the left. a) Write the budget inequality. b) Explain why not every value shown by the continuous graph can represent a possible number of notebooks. c) List all feasible notebook counts.

Figure for problem 554704

Hints

- Start with cost per notebook times the number of notebooks. - Add the fixed shipping charge. - Use “at most” to decide the inequality symbol. - The graph is continuous, but the context is discrete. - Count only whole-number values that satisfy the inequality.

Solution

1. Let \(n\) be the number of notebooks. 2. The total cost is \(4n + 8\), and it must be at most \(28\), so the inequality is \(4n + 8 \le 28\). 3. Subtract \(8\): \(4n \le 20\). 4. Divide by \(4\): \(n \le 5\). 5. The continuous graph shows all real values \(n \le 5\), but a notebook count must be a whole number and cannot be negative. 6. Therefore, the feasible notebook counts are \(0, 1, 2, 3, 4,\) and \(5\).

Answer

a) \(4n + 8 \le 28\) b) A notebook count must be a nonnegative whole number, so fractional and negative values are not possible. c) \(0, 1, 2, 3, 4, 5\)

Model situations with equations and inequalities 64

A food pantry puts the same number of oranges in each of \(4\) bags. It then takes \(7\) oranges from each bag for lunches. There are \(52\) oranges left altogether. How many oranges were in each bag at first? Write and solve an equation, then check your answer.

Hints

- What should \(x\) count in one bag before any oranges are taken out? - How could you describe the oranges left in all four bags with one expression? - Does your starting number give the stated amount left after the lunches?

Solution

1. Let \(x\) be the number of oranges in each bag at first. After removing \(7\), each bag has \(x-7\) oranges. 2. The four bags hold \(52\) oranges in total, so \(4(x-7)=52\). 3. Divide by \(4\): \(x-7=13\). Add \(7\): \(x=20\). 4. Check: \(4(20-7)=52\) oranges remain, as stated.

Answer

\(4(x-7)=52\); \(x=20\) oranges per bag at first. Check: \(4(20-7)=52\).

Model situations with equations and inequalities 65

Hikers use this rule to estimate travel time: “Allow one hour for every \(3\) miles of horizontal distance. Add one more hour for every \(2000\) feet of elevation gain.” a) Write an equation for the total time \(T\), in hours. Let \(s\) be the horizontal distance in miles and \(h\) be the elevation gain in feet. b) Estimate the time for a hike of \(7.5\) miles with \(3000\) feet of elevation gain. c) A group claims, “If we double the distance but keep the elevation gain the same, the estimated time also doubles.” Use your equation to evaluate the claim.

Hints

- Express “one hour for every \(3\) miles” as a quotient. - Keep track of the units: \(s\) is in miles and \(h\) is in feet. - Compare the new expression with twice the original expression.

Solution

1. The horizontal-distance time is \(\frac{s}{3}\), and the elevation-gain time is \(\frac{h}{2000}\). Therefore, \(T=\frac{s}{3}+\frac{h}{2000}\). 2. Substitute \(s=7.5\) and \(h=3000\): \(T=\frac{7.5}{3}+\frac{3000}{2000}=2.5+1.5=4\) hours. 3. The original time is \(T_1=\frac{s}{3}+\frac{h}{2000}\). Doubling only the distance gives \(T_2=\frac{2s}{3}+\frac{h}{2000}\). But twice the original time is \(2T_1=\frac{2s}{3}+\frac{h}{1000}\). These are not equal unless \(h=0\), so the claim is generally false.

Answer

a) \(T=\frac{s}{3}+\frac{h}{2000}\) b) \(4\) hours c) The claim is false in general. Only the distance portion doubles; the elevation portion stays the same.

Model situations with equations and inequalities 66

A temporary construction fence uses \(12\,\text{ft}\) panels and connecting posts that are \(4\,\text{in.}\) wide. A fence section begins and ends with a post, with one panel between each pair of posts. a) Write a formula for the total length \(L\), in feet, of a fence with \(n\) panels. b) How many panels are needed to cover at least \(160\,\text{ft}\)? c) How many posts are needed for that number of panels? Explain the relationship.

Hints

- Convert the post width to feet. - Count the posts for a small number of panels to identify the pattern. - “At least” requires an inequality and a whole-number result.

Solution

1. Convert the post width: \(4\,\text{in.}=\frac{1}{3}\,\text{ft}\). A fence with \(n\) panels has \(n+1\) posts, so \(L=12n+\frac{1}{3}(n+1)\). 2. To cover at least \(160\,\text{ft}\), solve \(12n+\frac{1}{3}(n+1)\ge160\). 3. Multiply by \(3\): \(36n+n+1\ge480\), so \(37n\ge479\) and \(n\ge\frac{479}{37}\approx12.95\). 4. Since \(n\) must be a whole number, at least \(13\) panels are needed. 5. The number of posts is \(n+1=14\).

Answer

a) \(L=12n+\frac{1}{3}(n+1)\) b) \(13\) panels c) \(14\) posts; there is one more post than panel.

Model situations with equations and inequalities 67

Noah and Maya each have exactly \(48\,\text{ft}\) of wire for a wireframe model. Noah builds a cube. Maya builds a rectangular prism with a square base whose height is twice a base edge. a) Let \(s\) be Noah's cube edge length. Write and solve an equation for \(s\). b) Let \(a\) be Maya's base-edge length. Write and solve an equation for \(a\) using the wire lengths of all \(12\) prism edges. c) Use your two equations to find the difference between the model heights.

Hints

- Count all edges that use wire before writing each equation. - For Maya's model, express the vertical edges in terms of the same variable used for a base edge. - Your response must include an equation for each model, not only arithmetic calculations. - Interpret the solved variable before comparing the heights.

Solution

1. A cube has \(12\) equal edges, so Noah's equation is \(12s=48\). Thus \(s=4\), so Noah's model is \(4\,\text{ft}\) high. 2. Maya's prism has \(8\) horizontal base/top edges of length \(a\) and \(4\) vertical edges of length \(2a\). Her equation is \( 8a+4(2a)=48. \) 3. Simplify: \(16a=48\), so \(a=3\). Maya's height is \(2a=6\,\text{ft}\). 4. The height difference is \(6-4=2\,\text{ft}\).

Answer

a) \(12s=48\), so \(s=4\) b) \(8a+4(2a)=48\), so \(a=3\) c) Maya's model is \(2\,\text{ft}\) taller.

Model situations with equations and inequalities 68

A class sells waffles at a school carnival. Fixed costs for equipment, batter, and supplies are \(\$42.00\). Each waffle also costs \(\$0.35\) in toppings and is sold for \(\$1.50\). a) First find the net amount from each waffle that is available to cover fixed costs and profit. Use that net amount to write and solve an inequality for avoiding a loss. Then give the least whole number of waffles the class must sell. b) Write and solve an inequality for earning a profit of at least \(\$120.00\). Then give the least whole number of waffles the class must sell.

Hints

- Find how much of each sale remains after paying the per-waffle topping cost. - Represent the number of waffles with one variable and use the net amount as its coefficient. - Translate “at least” with a lower-bound inequality. - Interpret a decimal boundary using the fact that the number of waffles must be whole.

Solution

1. Let \(w\) be the number of waffles sold. The net amount from each waffle after toppings is \(1.50-0.35=1.15\) dollars. 2. To avoid a loss, the net waffle revenue must cover the fixed costs: \(1.15w\ge42\). 3. Solve: \(w\ge36.521\ldots\). Because \(w\) is a whole number, at least \(37\) waffles are needed. 4. A profit of at least \(\$120.00\) means net waffle revenue minus fixed costs is at least \(120\): \(1.15w-42\ge120\). 5. Add \(42\): \(1.15w\ge162\), so \(w\ge140.869\ldots\). 6. Therefore, at least \(141\) waffles are needed.

Answer

a) Net amount per waffle: \(\$1.15\). Inequality: \(1.15w\ge42\); at least \(37\) waffles. b) Inequality: \(1.15w-42\ge120\); at least \(141\) waffles.

Model situations with equations and inequalities 69

Isabelle has \(40\) counters divided among three piles: left \(L\), middle \(M\), and right \(R\). She does not know the amount in each pile. The left pile has at least \(7\) counters, and the right pile has at least \(5\). She moves \(7\) counters from \(L\) to \(M\), then moves \(5\) counters from \(R\) to \(M\). After the moves, Leandro reports that the left and right piles contain \(15\) counters altogether. Isabelle immediately says, “Then the middle pile started with exactly \(13\) counters.” Use equations with variables to show that Isabelle is correct.

Hints

- Write an equation for the total number of counters before any moves. - Express the reported outside-pile total using the original amounts \(L\) and \(R\). - How are \(L+R\), \(M\), and the total of \(40\) related?

Solution

1. At the start, \(L+M+R=40\). 2. After the moves, the two outside piles contain \((L-7)+(R-5)\) counters. Leandro reports that this total is \(15\), so \((L-7)+(R-5)=15\). 3. Simplifying gives \(L+R-12=15\), so \(L+R=27\). 4. Substitute this into the total-count equation: \(27+M=40\). 5. Solving gives \(M=13\). Therefore, Isabelle is correct.

Answer

From \((L-7)+(R-5)=15\), we get \(L+R=27\). Since \(L+M+R=40\), it follows that \(27+M=40\), so \(M=13\).

Model situations with equations and inequalities 70

In a class of \(40\) students, \(60\%\) regularly play on organized sports teams. Of those team athletes, \(25\%\) play on a soccer team. a) How many students play on a soccer team? b) Some students who do not play organized sports hike regularly. Let \(h\) be the number of these hikers. The team athletes and hikers together must make up more than \(80\%\) of the class. Write and solve an inequality for \(h\), then give the least whole number of hikers that satisfies the condition. c) If the number of hikers is the least value from part b), how many students are neither team athletes nor hikers?

Hints

- Find the number of team athletes before finding the soccer subset. - Convert \(80\%\) of the class into a student-count threshold. - Add the unknown hiker count to the known number of team athletes, then translate “more than” with a strict inequality. - Interpret the inequality using the fact that a student count must be a whole number.

Solution

1. There are \(0.60\cdot40=24\) team athletes. Of these, \(0.25\cdot24=6\) play soccer. 2. More than \(80\%\) of \(40\) students means more than \(0.80\cdot40=32\) students altogether. 3. With \(h\) non-athlete hikers, write \(24+h>32\). 4. Subtract \(24\): \(h>8\). Since \(h\) is a whole number, the least possible value is \(9\). 5. With \(9\) hikers, \(40-(24+9)=7\) students are neither team athletes nor hikers.

Answer

a) \(6\) students b) Inequality: \(24+h>32\), so \(h>8\); least value: \(9\) hikers c) \(7\) students

Model situations with equations and inequalities 71

A new candle is \(10\,\text{in.}\) tall. Once lit, it burns down at a constant rate of \(0.4\,\text{in.}\) per hour. For what burning times \(h\) is the candle shorter than \(4\,\text{in.}\)? Consider only times up to the moment when the candle burns out completely.

Hints

- Write a linear expression for the candle's height after \(h\) hours. - Use a strict inequality because the candle must be shorter than \(4\,\text{in.}\). - Reverse the inequality sign when dividing by a negative number. - Find the time when the candle reaches a height of \(0\).

Solution

1. The candle's height after \(h\) hours is \(10 - 0.4h\). 2. Write the inequality for a height below \(4\,\text{in.}\): \(10 - 0.4h < 4\). 3. Subtract \(10\): \(-0.4h < -6\). 4. Divide by \(-0.4\) and reverse the inequality sign: \(h > 15\). 5. The candle burns out when \(10 - 0.4h = 0\), which gives \(h = 25\). 6. Therefore, within the physical time interval, the candle is shorter than \(4\,\text{in.}\) when \(15 < h \le 25\).

Answer

\(15 < h \le 25\) hours

Model situations with equations and inequalities 72

A rancher is asked how many sheep are in the flock and replies, “If I tripled my flock, sold one-fourth of the original number of sheep, and then received \(2\) more sheep, I would have exactly \(68\) sheep.” a) Write and solve an equation to find the original number of sheep. b) An apprentice claims the statement must be impossible because calculations involving fourths can produce fractional sheep. Determine whether that concern applies here.

Hints

- Translate “triple” and “one-fourth of the original number” into algebraic terms. - Every fractional part refers to the original flock size. - A solution involving animals must be checked for whole-number reasonableness. - Verify that one-fourth of your result is a whole number.

Solution

1. Let \(x\) be the original number of sheep. 2. Write the equation \(3x - \frac{1}{4}x + 2 = 68\). 3. Combine like terms: \(\frac{11}{4}x + 2 = 68\). 4. Subtract \(2\): \(\frac{11}{4}x = 66\). 5. Multiply by \(\frac{4}{11}\): \(x = 24\). 6. The result is meaningful because \(24\) is a whole number and one-fourth of \(24\) is \(6\), also a whole number.

Answer

a) The original flock had \(24\) sheep. b) The apprentice is not correct in this case. Both the original number, \(24\), and one-fourth of it, \(6\), are whole numbers.

Model situations with equations and inequalities 73

Two rectangular flower beds have the same length. Bed A is \(3\,\text{ft}\) wide, and Bed B is \(5\,\text{ft}\) wide. Bed B has \(14\,\text{ft}^2\) more area than Bed A. a) Use an equation to find the common length. b) Bed C has the same length and a width equal to the sum of the other two widths, \(8\,\text{ft}\). Find the area of Bed C. c) Explain why the area of Bed C equals the sum of the areas of Beds A and B.

Hints

- Compare the two area expressions using their common length. - Find how much more area is added for each foot of length. - Notice that \(8 = 3 + 5\). - Use the distributive property to explain the area relationship.

Solution

1. Let \(L\) feet be the common length. The difference in area is \(5L - 3L = 14\). 2. Simplify: \(2L = 14\), so \(L = 7\). 3. The areas of Beds A and B are \(7 \cdot 3 = 21\,\text{ft}^2\) and \(7 \cdot 5 = 35\,\text{ft}^2\). 4. Bed C has area \(7 \cdot 8 = 56\,\text{ft}^2\). 5. Because all three beds have length \(L\), the distributive property gives \(L(3 + 5) = 3L + 5L\). Therefore, Bed C's area is the sum of the other two areas.

Answer

a) The common length is \(7\,\text{ft}\). b) The area of Bed C is \(56\,\text{ft}^2\). c) Since the common length is a shared factor, \(L(3 + 5) = 3L + 5L\), so the combined width produces the sum of the two areas.

Model situations with equations and inequalities 74

A store sells large wooden crates for \(\$12.00\) and small crates for \(\$4.50\). A customer spends \(\$153.00\) and buys three times as many small crates as large crates. a) Use an equation to find how many crates of each size the customer bought. b) The customer bought \(24\) crates in all. What would the total cost have been if the customer had instead bought twice as many large crates as small crates?

Hints

- Express the number of small crates in terms of the number of large crates. - Write an equation for the original total cost. - For part b), keep the total number of crates unchanged. - Translate the new ratio into two quantities that add to \(24\).

Solution

1. Let \(x\) be the number of large crates. Then the number of small crates is \(3x\). 2. Write \(12x + 4.50(3x) = 153\). 3. Combine like terms: \(12x + 13.5x = 153\), so \(25.5x = 153\). 4. Divide by \(25.5\): \(x = 6\). The customer bought \(6\) large crates and \(18\) small crates. 5. For part b), let \(y\) be the number of small crates. Then there are \(2y\) large crates, and \(y + 2y = 24\). 6. Solve: \(3y = 24\), so \(y = 8\). The new quantities are \(8\) small and \(16\) large crates. 7. The new cost is \(16 \cdot 12 + 8 \cdot 4.50 = 192 + 36 = 228\) dollars.

Answer

a) The customer bought \(6\) large crates and \(18\) small crates. b) The new total cost would be \(\$228.00\).

Model situations with equations and inequalities 75

A theater group buys \(6\) identical tickets. Each ticket costs \(x\) dollars plus a \(\$2.75\) booking fee. The total charge is \(\$76.50\). a) Write an equation that models the purchase. b) Find the ticket price \(x\) before the booking fee. c) If the ticket price before the fee were \(\$1.50\) higher, how much would the total charge increase? Explain using the structure of your model.

Hints

- First express the complete cost of one ticket, including the fee. - The same complete cost is charged six times. - Solve by undoing the multiplication before undoing the added fee. - For part c), consider how a change in one ticket's base price is repeated across all six tickets.

Solution

1. Each complete ticket charge is \(x+2.75\), and there are \(6\) identical charges, so \(6(x+2.75)=76.50\). 2. Divide by \(6\): \(x+2.75=12.75\). 3. Subtract \(2.75\): \(x=10.00\). 4. Increasing \(x\) by \(\$1.50\) increases each of the \(6\) ticket charges by \(\$1.50\), so the total increases by \(6\cdot1.50=\$9.00\).

Answer

a) \(6(x+2.75)=76.50\) b) \(x=\$10.00\) c) The total charge would increase by \(\$9.00\).

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