## Engage NY Eureka Math Algebra 2 Module 1 Lesson 23 Answer Key

### Eureka Math Algebra 2 Module 1 Lesson 23 Example Answer Key

Example 1.

x | \( \frac{x+1}{x} \) | \( \frac{x+2}{x+1} \) |

0.5 | ||

1 | ||

1.5 | ||

2 | ||

5 | ||

10 | ||

10 |

Answer:

x | \( \frac{x+1}{x} \) | \( \frac{x+2}{x+1} \) |

0.5 | 3 | 1.6667 |

1 | 2 | 1.5000 |

1.5 | 1.6667 | 1.4000 |

2 | 1.5 | 1.3333 |

5 | 1.2 | 1.1667 |

10 | 1.1 | 1.0909 |

10 | 1.01 | 1.0099 |

### Eureka Math Algebra 2 Module 1 Lesson 23 Opening Exercise Answer Key

Use the slips of paper you have been given to create visual arguments for whether \(\frac{1}{3}\) or \(\frac{3}{8}\) is larger.

Answer:

Ask students to make visual arguments as to whether \(\frac{1}{3}\) or \(\frac{3}{8}\) is larger. Use the following as either scaffolding for struggling students or as an example of student work.

→ We can use the following area models to represent the fraction \(\frac{1}{3}\) and \(\frac{3}{8}\) as we did in Lesson 22.

→ We see that these visual representations of the fractions give us strong evidence that \(\frac{1}{3}\) < \(\frac{3}{8}\).

→ Discuss with your neighbor another way to make a comparison between the two fractions and why we might not want to always rely on visual representations.

Students should suggest finding decimal approximations of fractions and converting the fractions to equivalent fractions with common denominators. Reasons for not using visual representations may include the difficulty with fractions with large denominators.

Once students have had a chance to discuss alternative methods, ask them to choose one of the two methods to verify that the visual representations above are accurate.

Decimal approximations: We have \(\frac{1}{3}\) ≈ 0.333 and \(\frac{3}{8}\) = 0.3 75; thus, \(\frac{1}{3}\) < \(\frac{3}{8}\).

Common denominators: We have \(\frac{1}{3}\) = \(\frac{8}{24}\) and \(\frac{3}{8}\) = \(\frac{9}{24}\). Since \(\frac{8}{24}\) < \(\frac{9}{24}\), we know that \(\frac{1}{3}\) < \(\frac{3}{8}\).

### Eureka Math Algebra 2 Module 1 Lesson 23 Exercise Answer Key

We will start by working with positive integers. Let m and n be positive integers. Work through the following exercises with a partner.

Exercise 1.

Fill out the following table.

n | \( \frac{1}{n} \) |

1 | |

2 | |

2 | |

4 | |

5 | |

6 |

Answer:

n | \( \frac{1}{n} \) |

1 | 1 |

2 | \( \frac{1}{2} \) |

2 | \( \frac{1}{3} \) |

4 | \( \frac{1}{4} \) |

5 | \( \frac{1}{5} \) |

6 | \( \frac{1}{6} \) |

Exercise 2.

Do you expect \(\frac{1}{n}\) to be larger or smaller than \(\frac{1}{n+1}\)? Do you expect \(\frac{1}{n}\) to be larger or smaller than \(\frac{1}{n+2}\)? Explain why.

Answer:

From the table, as n increases, \(\frac{1}{n}\) decreases. This means that since 1 + n > n, we will hove \(\frac{1}{n+1}\) < \(\frac{1}{n}\). That is \(\frac{1}{n}>\frac{1}{n+1}>\frac{1}{n+2}\).

Exercise 3.

Compare the rational expressions \(\frac{1}{n}\), \(\frac{1}{n+1}\), and \(\frac{1}{n+2}\) for n = 1, 2, and 3. Do your results support your conjecture from Exercise 2? Revise your conjecture if necessary.

Answer:

Exercise 4.

From your work In Exercises 1 and 2, generalize how \(\frac{1}{n}\) compares to \(\frac{1}{n+m}\), where m and n are positive integers.

Answer:

Since m is a positive integer being added to n, the denominator will increase, which will decrease the value of the rational expression overall. That is, \(\frac{1}{n}\) > \(\frac{1}{n+m}\) for positive integers m and n.

Exercise 5.

Will your conjecture change or stay the same If the numerator is 2 instead of 1? Make a conjecture about what happens when the numerator is held constant, but the denominator increases for positive numbers.

Answer:

It will stay the same because this would be the same as multiplying the in equality by 2, and multiplication by a positive number does not change the direction of the inequality. If the numerator is held constant and the denominator increases, you are dividing by a larger number, so you get a smaller number overall

### Eureka Math Algebra 2 Module 1 Lesson 23 Problem Set Answer Key

Question 1.

For parts (a) – (d), rewrite each rational expression as an equivalent rational expression so that all expressions have a common denominator.

a. \(\frac{3}{5}, \frac{9}{10}, \frac{7}{15}, \frac{7}{21}\)

Answer:

\(\frac{18}{30}, \frac{27}{30}, \frac{14}{30}, \frac{10}{30}\)

b. \(\frac{m}{s d}, \frac{s}{d m}, \frac{d}{m s}\)

Answer:

c. \(\frac{1}{(2-x)^{2}}, \frac{3}{(2 x-5)(2-x)}\)

Answer:

d. \(\frac{3}{x^{2}-x}, \frac{5}{x}, \frac{2 x+2}{2 x^{2}-2}\)

Answer:

Question 2.

If x is a positive number, for which values of x is x < \(\frac{1}{x}\)?

Answer:

Before we can compare two rational expressions, we need to express them as equivalent expressions with a common denominator. Since x ≠ 0, we have x = \(\frac{x^{2}}{x}\) Then x < \(\frac{1}{x}\) exactly when \(\frac{x^{2}}{x}\) < \(\frac{1}{x}\) which happens when

x^{2} < 1. The only positive real number values of x that satisfy x^{2} < 1 are 0 < x < 1. Question 3. Can we determine if \(\frac{y}{y-1}>\frac{y+1}{y}\) for all values y > 1? Provide evidence to support your answer.

Answer:

Question 4.

For positive x, determine when the following rational expressions have negative denominators.

a. \(\frac{3}{5}\)

Answer:

Never; 5 is never less than 0.

b. \(\frac{x}{5-2x}\)

Answer:

5 – 2x < 0 when 5 < 2x, which is equivalent to \(\frac{5}{2}\) < x.

c. \(\frac{x+3}{x^{2}+4 x+8}\)

Answer:

For any real number x, x^{2} + 4x + 8 is never negative. One way to see this is that x^{2} + 4x + 8 = (x + 2)^{2} + 4, which is the sum of two positive numbers.

d. \(\frac{3 x^{2}}{(x-5)(x+3)(2 x+3)}\)

Answer:

For positive x, x + 3 and 2x + 3 are always positive. The number x – 5 is negative when x < 5, so the denominator is negative when x < 5.

Question 5.

Consider the rational expressions \(\frac{x}{x-2}\) and \(\frac{x}{x-4}\).

a. Evaluate each expression for x = 6.

Answer:

If x = 6, then \(\frac{x}{x-2}\) = 1.5 and \(\frac{x}{x-4}\) = 3.

b. Evaluate each expression for x = 3.

Answer:

If x = 3, then \(\frac{x}{x-2}\) = 3 and \(\frac{x}{x-4}\) = – 3.

c. Can you conclude that \(\frac{x}{x-2}\) < \(\frac{x}{x-4}\) for all positive values of x? Explain how you know. Answer: No, because \(\frac{x}{x-2}\) > \(\frac{x}{x-4}\) when x = 3, it is not true that \(\frac{x}{x-2}\) < \(\frac{x}{x-4}\) for every positive value of x.

d. Extension: Raphael claims that the calculation below shows that \(\frac{x}{x-2}\) < \(\frac{x}{x-4}\) for all values of x, where x ≠ 2 and x ≠ 4. Where is the error In the calculation?

Starting with the rational expressions \(\frac{x}{x-2}\) and \(\frac{x}{x-4}\), we need to first find equivalent rational expressions with a common denominator. The common denominator we will use is (x – 4)(x – 2). We then have

Since x^{2} – 4x < x^{2} – 2x for x > 0, we can divide each expression by (x – 4)(x – 2). We then have \(\frac{x(x-4)}{(x-4)(x-2)}\) < \(\frac{x(x-2)}{(x-4)(x-2)}\), and we can conclude that \(\frac{x}{x-2}\) < \(\frac{x}{x-4}\) for all positive values of x.

Answer:

The error in logic in this calculation is that the denominator (x – 4) (x – 2) is not always a positive number for all positive values of x. In fact, if 2 < x < 4, then (x – 4)(x – 2) < 0. Thus, even though x^{2} – 4x < x^{2} – 2x when x > 0, the inequality \(\frac{x^{2}-4 x}{(x-4)(x-2)}\) < \(\frac{x^{2}-2 x}{(x-4)(x-2)}\) is not valid for every positive value of x.

Question 6.

Consider the populations of two cities within the same state where the large city’s population is P, and the small city’s population is Q. For each of the following pairs, state which of the expressions has a larger value. Explain your reasoning in the context of the populations. a. P + Q and P

Answer:

The value of P + Q is larger than P. The expression P + Q represents the total population of the two cities, and P represents the population of the larger city. Since these quantities are populations of cities, we can assume they are greater than zero.

b. \(\frac{P}{P+Q}\) and \(\frac{Q}{P+Q}\)

Answer:

The value of \(\frac{P}{P+Q}\) is larger. As stated in part (a), P + Q represents the total population of the two cities. Hence, \(\frac{P}{P+Q}\) and \(\frac{Q}{P+Q}\) represent each city’s respective fraction of the total population. Since \(\frac{P}{P+Q}\) > \(\frac{Q}{P+Q}\).

c. 2Q and P + Q

Answer:

The value of P + Q is larger than the value of 2Q. The population of the smaller of the two cities is represented by Q, so 2Q represents a population twice the size of the smaller city, but P > Q so P + Q > Q + Q and thus P + Q > 2Q.

d. \(\frac{P}{Q}\) and \(\frac{Q}{P}\)

Answer:

The value of \(\frac{P}{Q}\) is larger. These expressions represent the ratio between the populations of the cities. For instance, the larger city is \(\frac{P}{Q}\) times larger than the smaller city. Since P > Q, \(\frac{P}{Q}\)> 1 > \(\frac{Q}{P}\), we can say that there are \(\frac{P}{Q}\) people in the larger city for every one person in the smaller city.

e. \(\frac{P}{P+Q}\) and \(\frac{1}{2}\)

Answer:

The value \(\frac{P}{P+Q}\) is larger. Since P is the population of the larger city, the first city represents more than half

of the total.

f. \(\frac{P+Q}{P}\) and P – Q

Answer:

The value of P – Q is larger. The expression P – Q represents the difference in population between the two cities. The expression \(\frac{P+Q}{P}\) can represent the ratio of how much larger the total is compared to the population of the larger city, but we know that P represents more than half of the total; therefore, \(\frac{P+Q}{P}\) cannot be larger than 2. Without the context, we could not say that P – Q is larger than 2, but in the context of the problem, since P is the population of a large city, and Q is the population of a small city, P – Q > 2.

Thus, P – Q > \(\frac{P+Q}{P}\)

g. \(\frac{P+Q}{2}\) and \(\frac{P+Q}{Q}\)

Answer:

The value of \(\frac{P+Q}{2}\) is larger. The sum divided by the number of cities represents the average population of the two cities and will be significantly higher than the ratio represented by \(\frac{P+Q}{Q}\). Alternatively, Q is much larger than 2, so \(\frac{P+Q}{2}\) < \(\frac{P+Q}{Q}\).

h. \(\frac{1}{P}\) and \(\frac{1}{Q}\)

Answer:

The value of \(\frac{1}{Q}\) is larger. The expression \(\frac{1}{Q}\) represents the proportion of the population of the second city a single citizen represents. Similarly for \(\frac{1}{P}\), since the second city has a smaller population, each individual represents a larger proportion of the whole than in the first city.

### Eureka Math Algebra 2 Module 1 Lesson 23 Exit Ticket Answer Key

Use the specified methods to compare the following rational expressions: \(\frac{x+1}{x^{2}}\) and \(\frac{1}{x}\).

Question 1.

Fill out the table of values.

Answer:

Question 2.

Graph y = \(\frac{x+1}{x^{2}}\) and y = \(\frac{1}{x}\) for positive values of x.

Answer:

Question 3.

Find the common denominator, and compare numerators for positive values of x.

Answer:

The common denominator of x and x^{2} is x^{2}.

For any value of x, x^{2} is positive. Since

x + 1 > x,

we then have,