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**Number problems** are a category of word problems that provide verbal clues about one or more unknown numbers, and the goal is to use those clues to set up and solve an algebraic equation. Unlike most real-world word problems — which involve contexts such as prices, distances, or quantities of physical objects — number problems are abstract: they deal with numbers themselves rather than with a practical scenario. Because of this simplicity, number problems serve as an excellent first application of the seven-step problem-solving strategy. They allow students to focus on the core skills of identifying operation keywords (such as "difference," "sum," and "twice"), translating verbal clues into an equation, solving the equation, and verifying the result — all without the added complexity of interpreting a real-world context.

Number Problems

Apply the seven-step problem-solving strategy to a number problem involving a single subtraction operation.

**Problem:** The difference of a number and six is $$13$$. Find the number.

1. **Read** the problem and confirm all words are familiar.
2. **Identify** what to find: the number.
3. **Name** the unknown: Let $$n$$ = the number.
4. **Translate** using the keyword *difference*, which signals subtraction. Restate the problem as a single sentence — "The difference of the number and $$6$$ is $$13$$" — and convert it to the equation:

$$n - 6 = 13$$

5. **Solve** by adding $$6$$ to both sides (Addition Property of Equality):

$$n - 6 + 6 = 13 + 6$$

$$n = 19$$

6. **Check**: The difference of $$19$$ and $$6$$ is $$19 - 6 = 13$$. $$\checkmark$$
7. **Answer**: The number is $$19$$.

This example demonstrates a one-step number problem: the word "difference" maps directly to subtraction, and the resulting equation $$n - 6 = 13$$ requires only one inverse operation (adding $$6$$) to isolate the variable.

Solving a Number Problem Using "Difference"

Apply the seven-step problem-solving strategy to a number problem whose translation involves two operations — multiplication and addition — producing a two-step equation.

**Problem:** The sum of twice a number and seven is $$15$$. Find the number.

1. **Read** the problem and confirm understanding of all terms.
2. **Identify** what to find: the number.
3. **Name** the unknown: Let $$n$$ = the number.
4. **Translate** by recognizing two keywords: *twice* signals multiplication by $$2$$, and *sum* signals addition. Restate the problem as "The sum of twice $$n$$ and $$7$$ is $$15$$," which converts to:

$$2n + 7 = 15$$

5. **Solve** using two inverse operations. First, subtract $$7$$ from both sides:

$$2n + 7 - 7 = 15 - 7$$

$$2n = 8$$

Then divide both sides by $$2$$:

$$\frac{2n}{2} = \frac{8}{2}$$

$$n = 4$$

6. **Check**: Is the sum of twice $$4$$ and $$7$$ equal to $$15$$?

$$2 \cdot 4 + 7 \stackrel{?}{=} 15$$

$$15 = 15 \checkmark$$

7. **Answer**: The number is $$4$$.

Unlike one-step number problems where a single keyword produces a single-operation equation, this problem combines two keywords ("twice" and "sum") to generate $$2n + 7 = 15$$, which requires both subtraction and division to solve. As comfort grows, intermediate steps may be condensed — but writing out every step is perfectly fine while building confidence.

Solving a Two-Step Number Problem Using "Sum" and "Twice"

Apply the seven-step problem-solving strategy to a number problem involving **two unknown numbers** that are related by a "more than" phrase and whose sum is given.

**Problem:** One number is five more than another. The sum of the numbers is $$ 21 $$. Find the numbers.

1. **Read** the problem.
2. **Identify** what to find: two numbers.
3. **Name** the unknowns using a single variable. Let $$ n $$ = the first number. Because one number is "five more than" the other, the second number is $$ n + 5 $$.
4. **Translate** into an equation: The sum of the two numbers is $$ 21 $$, so: $$ n + (n + 5) = 21 $$.
5. **Solve** the equation. First, combine the like terms $$ n + n = 2n $$: $$ 2n + 5 = 21 $$. Subtract $$ 5 $$ from both sides: $$ 2n = 16 $$. Divide both sides by $$ 2 $$: $$ n = 8 $$. Find the second number: $$ n + 5 = 8 + 5 = 13 $$.
6. **Check**: Is $$ 13 $$ five more than $$ 8 $$? $$ 8 + 5 = 13 $$ $$ \checkmark $$. Is their sum $$ 21 $$? $$ 8 + 13 = 21 $$ $$ \checkmark $$.
7. **Answer**: The numbers are $$ 8 $$ and $$ 13 $$.

This example illustrates a key technique for two-unknown number problems: express both quantities through a single variable using the stated relationship ("five more than" becomes $$ n + 5 $$), then write an equation from the remaining fact (the sum). The resulting equation involves combining like terms before using inverse operations to isolate the variable.

Solving a Two-Number Problem Using "Sum" and "More Than"

Apply the seven-step problem-solving strategy to a two-unknown number problem where the given sum is negative and the relationship uses "less than," producing negative solutions.  **Problem:** The sum of two numbers is $$-15$$. One number is nine less than the other. Find the numbers.  1. **Read** the problem. 2. **Identify** what to find: two numbers. 3. **Name** the unknowns. Let $$n$$ = the first number. Because one number is "nine less than" the other, the second number is $$n - 9$$. 4. **Translate** into an equation: The sum of the two numbers is $$-15$$, so:  $$n + (n - 9) = -15$$  5. **Solve** the equation. Combine like terms:  $$2n - 9 = -15$$  Add $$9$$ to both sides:  $$2n = -6$$  Divide both sides by $$2$$:  $$n = -3$$  Find the second number: $$n - 9 = -3 - 9 = -12$$.  6. **Check**: Is $$-12$$ nine less than $$-3$$? $$-3 - 9 = -12$$ $$\checkmark$$. Is their sum $$-15$$? $$-3 + (-12) = -15$$ $$\checkmark$$. 7. **Answer**: The numbers are $$-3$$ and $$-12$$.  This example extends the two-unknown number problem pattern to situations involving negative numbers. The phrase "nine less than" translates to subtraction ($$n - 9$$), and the negative sum means the equation $$2n - 9 = -15$$ requires adding $$9$$ rather than subtracting. The verification step is especially important here, because working with negative numbers increases the likelihood of sign errors.

Solving a Two-Number Problem with Negative Results

Apply the seven-step problem-solving strategy to a number problem that combines **multiplication and addition** in the relationship between two unknowns, producing a three-step equation.

**Problem:** One number is ten more than twice another. Their sum is one. Find the numbers.

1. **Read** the problem.
2. **Identify** what to find: two numbers.
3. **Name** the unknowns using a single variable. Let $$x$$ = the first number. Because one number is "ten more than twice" the other, the phrase combines two operations: *twice* signals multiplication by $$2$$, and *ten more than* signals adding $$10$$. The second number is therefore $$2x + 10$$.
4. **Translate** into an equation: Their sum is $$1$$, so:

$$x + (2x + 10) = 1$$

5. **Solve** the equation. First, combine like terms $$x + 2x = 3x$$:

$$3x + 10 = 1$$

Subtract $$10$$ from both sides:

$$3x = -9$$

Divide both sides by $$3$$:

$$x = -3$$

Find the second number: $$2(-3) + 10 = -6 + 10 = 4$$.

6. **Check**: Is $$4$$ ten more than twice $$-3$$?

$$2(-3) + 10 \stackrel{?}{=} 4$$

$$-6 + 10 \stackrel{?}{=} 4$$

$$4 = 4 \checkmark$$

Is their sum $$1$$?

$$-3 + 4 \stackrel{?}{=} 1$$

$$1 = 1 \checkmark$$

7. **Answer**: The numbers are $$-3$$ and $$4$$.

This example extends earlier two-number problems in two ways. First, the relationship between the unknowns involves **two operations** — multiplication followed by addition — rather than a simple additive phrase like "five more than." The expression $$2x + 10$$ captures both operations in a single algebraic expression. Second, the resulting equation $$3x + 10 = 1$$ requires **three algebraic steps** (combining like terms, subtracting, and dividing) rather than two, because combining $$x$$ and $$2x$$ into $$3x$$ is needed before the inverse operations can isolate the variable.

Solving a Two-Number Problem Using "Sum" and "More Than Twice"

A supervisor's report states that the number of units produced in Shift A was 'twice' the number produced in Shift B. Which operation is used to represent 'twice' in an equation?

In a professional business or logistics environment, verbal reports must often be translated into algebraic equations. Match each common descriptive phrase used in these number problems to the mathematical operation it represents.

When a professional is translating a verbal report into a mathematical model, they follow a specific process. Arrange the following steps for solving a number problem in the correct order.

In a professional mathematics training context, 'number problems' are categorized as abstract because they focus on the relationship between numbers themselves rather than using physical units like currency, weight, or distance.

State the definition of a 'number problem' as presented in your training and identify the primary reason why these abstract problems are used as an initial step before introducing real-world scenarios.

Defining Number Problems and Their Role

When translating a professional report into an algebraic model, terms like 'sum,' 'difference,' and 'twice' are specifically known as ________ keywords because they indicate which mathematical process to use.

Based on the course materials for 'Number Problems,' identify the specific characteristic that distinguishes these problems from 'real-world' applications. Furthermore, recall and list the four core skills that students focus on when solving this category of problems as part of their initial training.

Foundational Strategy in Algebraic Training

As part of your professional development in algebraic modeling, you are asked to contribute to a training summary regarding problem-solving foundations. Write a brief essay that addresses the following: 
1. Define what a 'number problem' is in the context of algebraic equations.
2. List three specific operation keywords (such as 'sum' or 'twice') that are commonly used to translate verbal clues into mathematical symbols.
3. Explain the primary reason why these problems are classified as 'abstract' rather than 'practical' or 'real-world' scenarios.

Foundational Characteristics of Number Problems

In a professional algebra training module, 'number problems' are used as the primary introductory application for problem-solving because they allow the learner to focus on which specific core skill?

In a foundational algebra course, what is the primary purpose of the 'verifying the result' step when solving a number problem?

Apply the seven-step problem-solving strategy to a number problem whose translation involves two operations — multiplication by a specific number and addition — producing a two-step equation.  **Problem:** The sum of seven times a number and eight is $$36$$. Find the number.  1. **Read** the problem and confirm understanding of all terms. 2. **Identify** what to find: the number. 3. **Name** the unknown: Let $$n$$ = the number. 4. **Translate** by recognizing two keywords: *times* signals multiplication (here, by $$7$$), and *sum* signals addition. Restate the problem as "The sum of $$7$$ times $$n$$ and $$8$$ is $$36$$," which converts to:  $$7n + 8 = 36$$  5. **Solve** using two inverse operations. First, subtract $$8$$ from both sides:  $$7n + 8 - 8 = 36 - 8$$  $$7n = 28$$  Then divide both sides by $$7$$:  $$\frac{7n}{7} = \frac{28}{7}$$  $$n = 4$$  6. **Check**: Is the sum of seven times $$4$$ plus $$8$$ equal to $$36$$?  $$7 \cdot 4 + 8 \stackrel{?}{=} 36$$  $$28 + 8 \stackrel{?}{=} 36$$  $$36 = 36 \checkmark$$  7. **Answer**: The number is $$4$$.  This problem combining "times" and "sum" generates the equation $$7n + 8 = 36$$, which requires both subtraction and division to solve.

Solving a Two-Step Number Problem Using "Sum" and "Times"

Apply the seven-step problem-solving strategy to a real-world word problem that asks to find an unknown past value based on a known present value and a related comparative phrase.

**Problem:** According to the National Automobile Dealers Association, the average cost of a car in 2014 was \$28,400. This was \$1,600 less than six times the cost in 1975. What was the average cost of a car in 1975?

1. **Read** the problem carefully. The goal is to find a past cost using a present cost.
2. **Identify** what to find: the average cost of a car in 1975.
3. **Name** the unknown: Let $$ c $$ = the average cost of a car in 1975. The phrase "\$1,600 less than six times the cost in 1975" tells us that the cost in 2014 can be expressed as $$ 6c - 1{,}600 $$.
4. **Translate** into an equation: The average cost in 2014 was \$28,400, so: $$ 6c - 1{,}600 = 28{,}400 $$
5. **Solve** the equation. Combine the two operations by first adding 1,600 to both sides:
$$ 6c = 30{,}000 $$
Divide both sides by 6:
$$ c = 5{,}000 $$
6. **Check**: Is \$1,600 less than six times \$5,000 equal to \$28,400?
$$ 6(5{,}000) - 1{,}600 \stackrel{?}{=} 28{,}400 $$
$$ 30{,}000 - 1{,}600 \stackrel{?}{=} 28{,}400 $$
$$ 28{,}400 = 28{,}400 \checkmark $$
7. **Answer**: The average cost of a car in 1975 was \$5,000.

This problem highlights translating comparative language into an equation. The expression $$ 6c - 1{,}600 $$ accurately captures the relationship, which directly translates into a two-step equation.

Solving a Word Problem for the Past Cost of a Car

Apply the seven-step problem-solving strategy to a real-world word problem to find a past price given a current price using multiplication and addition.

**Problem:** US Census data shows that the median price of a new home in the U.S. in November $$2014$$ was $$\$280{,}900$$. This was $$\$10{,}700$$ more than $$14$$ times the price in November $$1964$$. What was the median price of a new home in November $$1964$$?

1. **Read** the problem carefully. We are comparing a present house price to a past house price.
2. **Identify** what to find: the median price of a home in $$1964$$.
3. **Name** the unknown: Let $$p$$ = the median price of a new home in $$1964$$. The phrase "$$\$10{,}700$$ more than $$14$$ times the price in November $$1964$$" translates to the algebraic expression $$14p + 10{,}700$$.
4. **Translate** into an equation: The price in $$2014$$ was $$\$280{,}900$$, so:
$$14p + 10{,}700 = 280{,}900$$

5. **Solve** the equation. First, subtract $$10{,}700$$ from both sides:
$$14p = 270{,}200$$
Divide both sides by $$14$$:
$$p = 19{,}300$$

6. **Check**: Is $$\$10{,}700$$ more than $$14$$ times $$\$19{,}300$$ equal to $$\$280{,}900$$?
$$14(19{,}300) + 10{,}700 \stackrel{?}{=} 280{,}900$$
$$270{,}200 + 10{,}700 \stackrel{?}{=} 280{,}900$$
$$280{,}900 = 280{,}900 \checkmark$$

7. **Answer**: The median price of a new home in November $$1964$$ was $$\$19{,}300$$.

This problem applies the two-step equation framework to real-world financial data, highlighting how "more than" translates to addition and "times" translates to multiplication.

Solving a Word Problem for the Past Price of a Home

Apply the seven-step problem-solving strategy to a real-world word problem in which two people's earnings add up to a known total, and one earning is described using the phrase "less than twice" the other — combining multiplication and subtraction into a single expression.

**Problem:** A married couple together earns $$\$110{,}000$$ a year. The wife earns $$\$16{,}000$$ less than twice what her husband earns. What does the husband earn?

1. **Read** the problem carefully.
2. **Identify** what to find: the husband's annual earnings.
3. **Name** the unknown: Let $$h$$ = the amount the husband earns. The phrase "$$\$16{,}000$$ less than twice what her husband earns" combines two operations: *twice* signals multiplication by $$2$$, and *$$\$16{,}000$$ less than* signals subtraction. The wife's earnings are therefore $$2h - 16{,}000$$.
4. **Translate** into an equation: Together the husband and wife earn $$\$110{,}000$$, so:
$$h + (2h - 16{,}000) = 110{,}000$$

5. **Solve** the equation. Combine the like terms $$h + 2h = 3h$$:
$$3h - 16{,}000 = 110{,}000$$
Add $$16{,}000$$ to both sides:
$$3h = 126{,}000$$
Divide both sides by $$3$$:
$$h = 42{,}000$$
Find the wife's earnings: $$2(42{,}000) - 16{,}000 = 84{,}000 - 16{,}000 = 68{,}000$$.

6. **Check**: Does $$\$42{,}000 + \$68{,}000 = \$110{,}000$$? $$110{,}000 = 110{,}000$$ $$\checkmark$$

7. **Answer**: The husband earns $$\$42{,}000$$ a year.

This example transitions from abstract number problems to a practical financial scenario while using the same algebraic techniques. The phrase "less than twice" produces a subtraction-from-a-multiple expression ($$2h - 16{,}000$$), and the equation $$h + (2h - 16{,}000) = 110{,}000$$ requires combining like terms, adding a constant to both sides, and dividing — three steps that mirror the structure seen in earlier number problems. Both unknowns are expressed through a single variable by using the stated relationship between the couple's earnings.

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