Engage NY Eureka Math 8th Grade Module 7 Lesson 9 Answer Key

Opening Exercise
a. Compute the decimal expansions of $$\frac{5}{6}$$ and $$\frac{7}{9}$$.
$$\frac{5}{6}$$ = 0.8333… and $$\frac{7}{9}$$ = 0.7777…

b. What is $$\frac{5}{6}$$ + $$\frac{7}{9}$$ as a fraction? What is the decimal expansion of this fraction?
$$\frac{5}{6}$$ + $$\frac{7}{9}$$ = $$\frac{15 + 14}{18}$$ = $$\frac{29}{18}$$
$$\frac{29}{18}$$ = 1.61111″…”

c. What is $$\frac{5}{6}$$ × $$\frac{7}{9}$$ as a fraction? According to a calculator, what is the decimal expansion of the answer?
$$\frac{5}{6}$$ × $$\frac{7}{9}$$ = $$\frac{35}{54}$$ = 0.648 148 148 1″…”

d. If you were given just the decimal expansions of $$\frac{5}{6}$$ and $$\frac{7}{9}$$, without knowing which fractions produced them, do you think you could easily add the two decimals to find the decimal expansion of their sum? Could you easily multiply the two decimals to find the decimal expansion of their product?
No. To add 0.8333… and 0.777…, we need to start by adding together their rightmost digits. But these decimals are infinitely long, and there are no rightmost digits. It is not clear how we can start the addition.
Thinking about how to multiply the two decimals, 0.8333… × 0.77777…, is even more confusing!

Exercise 1.
Two irrational numbers x and y have infinite decimal expansions that begin 0.67035267… for x and 0.84991341… for y.
a. Explain why 0.670 is an approximation for x with an error of less than one thousandth. Explain why 0.849 is an approximation for y with an error of less than one thousandth.
The difference between 0.670 and 0.67035267… is 0.00035267…, which is less than 0.001, a thousandth.
The difference between 0.849 and 0.84991341… is 0.00091341…, which is less than 0.001, a thousandth.

b. Using the approximations given in part (a), what is an approximate value for x + y, for x × y, and for x2 + 7y2?
x + y is approximately 1.519 because 0.670 + 0.849 = 1.519.
x × y is approximately 0.56883 because 0.670 × 0.849 = 0.56883.
x2 + 7y2 is approximately 5.494507 because (0.670)2 + 7(0.849)2 = 5.494 507.

c. Repeat part (b), but use approximations for x and y that have errors less than $$\frac{1}{10^{5}}$$.
We want the error in the approximation to be less than 0.00001.
If we approximate x by truncating to five decimal places, that is, as 0.67035, then the error is 0.00000267…, which is indeed less than 0.00001.
Truncating y to five decimal places, that is, as 0.84991, gives an error of 0.00000341…, which is indeed less than 0.00001.
Now:
x + y is approximately 1.52026 because 0.67035 + 0.84991 = 1.52026.
x × y is approximately 0.5697371685 because 0.67035 × 0.84991 = 0.5697371685.
x2 + 7y2 is approximately 5.505798179 because (0.67035)2 + 7(0.84991)2 = 5.505798179.

Exercise 2.
Two real numbers have decimal expansions that begin with the following:
x = 0.1538461…
y = 0.3076923…
a. Using approximations for x and y that are accurate within a measure of $$\frac{1}{10^{3}}$$, find approximate values for x + y and y-2x.
Using x ≈ 0.153 and y ≈ 0.307, we obtain x + y ≈ 0.460 and y-2x ≈ 0.001.

b. Using approximations for x and y that are accurate within a measure of $$\frac{1}{10^{7}}$$, find approximate values for x + y and y-2x.
Using x ≈ 0.1538461 and y ≈ 0.3076923, we obtain x + y ≈ 0.4615384 and y-2x ≈ 0.0000001.

c. We now reveal that x = $$\frac{2}{13}$$ and y = $$\frac{4}{13}$$. How accurate is your approximate value to y-2x from part (a)? From part (b)?
The error in part (a) is 0.001. The error in part (b) is 0.0000001.

d. Compute the first seven decimal places of $$\frac{6}{13}$$. How accurate is your approximate value to x + y from part (a)? From part (b)?
$$\frac{6}{13}$$ = 0.4615384…
The error in part (a) is 0.4615384…-0.460 = 0.0015384…, which is less than 0.01.
Our approximate answer in part (b) and the exact answer match in the first seven decimal places. There is likely a mismatch from the eighth decimal place onward. This means that the error is no larger than 0.0000001, or $$\frac{1}{10^{7}}$$.

Eureka Math Grade 8 Module 7 Lesson 9 Problem Set Answer Key

Question 1.
Two irrational numbers x and y have infinite decimal expansions that begin 0.3338117… for x and 0.9769112… for y.
a. Explain why 0.33 is an approximation for x with an error of less than one hundredth. Explain why 0.97 is an approximation for y with an error of less than one hundredth.
The difference between 0.33 and 0.3338117… is 0.0038117…, which is less than 0.01, a hundredth.
The difference between 0.97 and 0.9769112… is 0.0069112…, which is less than 0.01, a hundredth.

b. Using the approximations given in part (a), what is an approximate value for 2x(y + 1)?
2x(y + 1) is approximately 1.3002 because 2 × 0.33 × 1.97 = 1.3002.

c. Repeat part (b), but use approximations for x and y that have errors less than $$\frac{1}{10^{6}}$$.
We want the error in the approximation to be less than 0.000001.
If we approximate x by truncating to six decimal places, that is, as 0.333811, then the error is 0.0000007…, which is indeed less than 0.000001.
Truncating y to six decimal places, that is, as 0.976911, gives an error of 0.0000002…, which is indeed less than 0.000001.
Now:
2x(y + 1) is approximately 1.319829276, which is a rounding of 2 × 0.333811 × 1.976911.

Question 2.
Two real numbers have decimal expansions that begin with the following:
x = 0.70588…
y = 0.23529…
a. Using approximations for x and y that are accurate within a measure of $$\frac{1}{10^{2}}$$, find approximate values for x + 1.25y and x/y.
Using x ≈ 0.70 and y ≈ 0.23, we obtain x + 1.25y ≈ 0.9875 and x/y ≈ 3.0434….

b. Using approximations for x and y that are accurate within a measure of $$\frac{1}{10^{4}}$$, find approximate values for x + 1.25y and $$\frac{x}{y}$$.
Using x ≈ 0.7058 and y ≈ 0.2352, we obtain x + 1.25y ≈ 0.9998 and $$\frac{x}{y}$$ ≈ 3.000850….

c. We now reveal that x and y are rational numbers with the property that each of the values x + 1.25y and $$\frac{x}{y}$$ is a whole number. Make a guess as to what whole numbers these values are, and use your guesses to find what fractions x and y might be.
It looks like x + 1.25y = 1 and $$\frac{x}{y}$$ = 3. Thus, we guess x = 3y and so 3y + 1.25y = 1, that is, 4.25y = 1, so y = $$\frac{1}{4.25}$$ = $$\frac{100}{425}$$ = $$\frac{4}{17}$$ and x = 3y = $$\frac{12}{17}$$.

Eureka Math Grade 8 Module 7 Lesson 9 Exit Ticket Answer Key

Question 1.
Suppose x = $$\frac{2}{3}$$ = 0.6666… and y = $$\frac{5}{9}$$ = 0.5555….
a. Using 0.666 as an approximation for x and 0.555 as an approximation for y, find an approximate value for x + y.
x + y = $$\frac{2}{3}$$ + $$\frac{5}{9}$$ = $$\frac{11}{9}$$ = 1 + $$\frac{2}{9}$$ = 1.22222…
c. Use approximations for x and y, each accurate to within an error of $$\frac{1}{10^{5}}$$, to estimate a value of the product x × y.