## Engage NY Eureka Math Algebra 1 Module 1 Lesson 19 Answer Key

### Eureka Math Algebra 1 Module 1 Lesson 19 Example Answer Key

Example 1.

Rearranging Familiar Formulas

The area A of a rectangle is 25 in^{2}. The formula for area is A=lw.

→ If the width w is 10 inches, what is the length l?

Answer:

l=\(\frac{5}{2}\)

→ If the width w is 15 inches, what is the length l?

Answer:

l=\(\frac{5}{3}\)

→ Rearrange the area formula to solve for l.

Answer:

A=lw

\(\frac{A}{w}\)=\(\frac{lw}{w}\)

Answer:

\(\frac{A}{w}\)=l or l=\(\frac{A}{w}\)

→ Verify that the area formula, solved for l, will give the same results for l as having solved for l in the original

area formula.

Answer:

A=lw

l=\(\frac{A}{w}\)=\(\frac{25}{10}\)=\(\frac{5}{2}\)

l=\(\frac{A}{w}\)=\(\frac{25}{15}\)=\(\frac{5}{3}\)

Walk students through the solution to this problem. Have them write the reasons for each step in the equation-solving process on their papers. Much of the work students will do in future classes will involve rearranging formulas to highlight a variable of interest. Begin to set the stage that solving for a variable before plugging in values is often easier than solving after substituting the values, especially when the numbers are not user friendly. If time permits, give them A=10.356 and w=5 \(\frac{3}{11}\), and ask them to solve for the length.

Example 2.

Comparing Equations with One Variable to Those with More Than One Variable

Demonstrate how to reverse the distributive property (factoring) to solve for x. Start by solving the related equation OR solve both equations at the same time, one line at a time. Encourage students to make notes justifying their reasoning on each step. Continue emphasizing that the process is the same because variables are just numbers whose values have yet to be assigned.

Example 2.

Comparing Equations with One Variable to Those with More Than One Variable

Answer:

### Eureka Math Algebra 1 Module 1 Lesson 19 Exercise Answer Key

Exercises 1–3

Solve each equation for x. For part (c), remember a variable symbol, like a, b, and c, represents a number.

a. 2x-6=10

Answer:

x=8

b. -3x-3=-12

Answer:

x=3

c. ax-b=c

Answer:

ax-b=c

ax=b+c

x=\(\frac{b+c}{a}\)

Exercises 2–3

Exercise 2.

Compare your work in parts (a) through (c) above. Did you have to do anything differently to solve for x in part (c)?

Answer:

The process to solve all three equations is the same.

Exercise 3.

Solve the equation ax-b=c for a. The variable symbols x, b, and c represent numbers.

Answer:

Solving for a is the same process as solving for x. a=\(\frac{b+c}{x}\)

Debrief student responses to Exercises 2 and 3 as a whole class. Make sure to emphasize the points below.

→ Variables are placeholders for numbers and as such have the same properties.

→ When solving an equation with several variables, you use the same properties and reasoning as with single-variable equations.

→ The equation in Exercise 3 holds as long as x does not equal 0 (division by 0 is undefined). Consider your result from Exercise 1(c). Does this equation hold for all values of the variables involved?

→ No. It only holds if a≠0.

Exercise 4.

Have students work in small groups or with a partner. Solving these exercises two ways will help students to further understand that rearranging a formula with variables involves the same reasoning as solving an equation for a single variable.

Exercises 4–5

Exercise 4.

Solve each problem two ways. First, substitute the given values, and solve for the given variable. Then, solve for the given variable, and substitute the given values.

a. The perimeter formula for a rectangle is p=2(l+w), where p represents the perimeter, l represents the length, and w represents the width. Calculate l when p=70 and w=15.

Answer:

Sample responses:

Substitute and solve. 70=2(l+15), l=20

Solve for the variable first: l=\(\frac{p}{2}\)-w

b. The area formula for a triangle is A=\(\frac{1}{2}\) bh, where A represents the area, b represents the length of the base, and h represents the height. Calculate b when A=100 and h=20.

Answer:

b=\(\frac{2A}{h}\), b=10

Have one or two students present their solutions to the entire class.

Exercise 5.

The next set of exercises increases slightly in difficulty. Instead of substituting, students solve for the requested variable. Have students continue to work in groups or with a partner. If the class seems to be getting stuck, solve part of one exercise as a whole class, and then have them go back to working with their partner or group.

Have students present their results to the entire class. Look for valid solution methods that arrive at the same answer using a slightly different process to isolate the variable. For parts (b–ii), students may need a reminder to use the square root to “undo” the square of a number. They learned about square roots and solving simple quadratic equations in Grade 8.

Exercise 5.

Rearrange each formula to solve for the specified variable. Assume no variable is equal to 0.

a. Given A=P(1+rt),

i. Solve for P.

Answer:

P=\(\frac{A}{1+rt}\)

ii. Solve for t.

Answer:

t=(\(\frac{A}{P}\)-1)÷r

b. Given K=\(\frac{1}{2}\) mv^{2},

i. Solve for m.

Answer:

m=\(\frac{2k}{v^{2}}\)

ii. Solve for v.

Answer:

v=±\(\sqrt{\frac{2 K}{m}}\)

### Eureka Math Algebra 1 Module 1 Lesson 19 Problem Set Answer Key

For Problems 1–8, solve for x.

Question 1.

ax+3b=2f

Answer:

x= \(\frac{2 f-3 b}{a}\)

Question 2.

rx+h=sx-k

Answer:

\(\frac{h+k}{s-r}\)

Question 3.

3px=2q(r-5x)

Answer:

x = \(\frac{2 q r}{3 p+10 q}\)

Question 4.

\(\frac{x+b}{4}\) = c

Answer:

x = 4c – b

Question 5.

\(\frac{x}{5}\)-7=2q

Answer:

x = 10q + 35

Question 6.

\(\frac{x}{6}\)–\(\frac{x}{7}\)=ab

Answer:

x = 42ab

Question 7.

\(\frac{x}{m}\)–\(\frac{x}{n}\)=\(\frac{1}{p}\)

Answer:

x = \(\frac{n m}{(n-m) p}\)

Question 8.

\(\frac{3 a x+2 b}{c}\) = 4d

Answer:

x = \(\frac{4 c d-2 b}{3 a}\)

Question 9.

Solve for m.

t=\(\frac{ms}{m+n}\)

Question 10.

\(\frac{1}{u}\) + \(\frac{1}{v}\) = \(\frac{1}{f}\)

Answer:

u = \(\frac{vf}{v-f}\)

Question 11.

Solve for s.

A=s^{2}

s = ± \(\sqrt{A}\)

Question 12.

Solve for h.

V=πr^{2} h

Answer:

h = \(\frac{V}{\pi r^{2}}\)

Question 13.

Solve for m.

T=4\(\sqrt{m}\)

Answer:

m = \(\frac{T^{2}}{16}\)

Question 14.

F=G \(\frac{m n}{d^{2}}\)

Answer:

d = ±\(\sqrt{\frac{G m n}{F}}\)

Question 15.

Solve for y.

ax+by=c

Answer:

y = \(\frac{c-a x}{b}\)

Question 16.

Solve for b_{1}.

A=\(\frac{1}{2}\) h(b_{1}+b_{2})

Answer:

b_{1} = \(\frac{2A}{h}\) – b_{2}

Question 17.

The science teacher wrote three equations on a board that relate velocity, v, distance traveled, d, and the time to travel the distance, t, on the board.

v=\(\frac{d}{t}\)

t=\(\frac{d}{v}\)

d=vt

Would you need to memorize all three equations, or could you just memorize one? Explain your reasoning.

Answer:

You could just memorize d=vt since the other two equations are obtained from this one by solving for v and t.

### Eureka Math Algebra 1 Module 1 Lesson 19 Exit Ticket Answer Key

Given the formula =\(\frac{1+a}{1-a}\),

Question 1.

Solve for a when x=12.

Answer:

12=\(\frac{1+a}{1-a}\)

12(1-a)=1+a

12-12a=1+a

11=13a

\(\frac{11}{13}\)=a

Question 2.

Rearrange the formula to solve for a.

Answer:

x=\(\frac{1+a}{1-a}\)

x(1-a)=1+a

x-xa=1+a

x-1=a+xa

x-1=a(1+x)

\(\frac{x-1}{1+x}\)=a