# Eureka Math Algebra 1 Module 1 Lesson 13 Answer Key

## Engage NY Eureka Math Algebra 1 Module 1 Lesson 13 Answer Key

### Eureka Math Algebra 1 Module 1 Lesson 13 Exercise Answer Key

Exercise 1.
Describe the property used to convert the equation from one line to the next:
x(1-x)+2x-4=8x-24-x2
x-x2+2x-4=8x-24-x2 ______
x-x2+2x-4=8x-24-x2 Distributive property
x+2x-4=8x-24 _______
x+2x-4=8x-24 Added x2 to both sides of the equation
3x-4=8x-24 _________
3x-4=8x-24 collected like terms
3x+20=8x _________
3x+20=8x Added 24 to both sides of the equation
20=5x ________
20=5x Subtracted 3x from both sides of the equation

In each of the steps above, we applied a property of real numbers and/or equations to create a new equation.

a. Why are we sure that the initial equation x(1-x)+2x-4=8x-24-x2 and the final equation 20=5x have the same solution set?
We established last class that making use of the commutative, associative, and distributive properties and properties of equality to rewrite an equation does not change the solution set of the equation.

b. What is the common solution set to all these equations?
x=4

→ Do we know for certain that x=4 is the solution to every equation shown? Explain why.
→ Have students verify this by testing the solution in a couple of the equations.

Exercise 2.
Solve the equation for x. For each step, describe the operation used to convert the equation.
3x-[8-3(x-1)]=x+19
3x-[8-3(x-1)]=x+19
3x-(8-3x+3)=x+19 Distributive property
3x-(11-3x)=x+19 Commutative property/collected like terms
3x-11+3x=x+19 Distributive property
6x-11=x+19 Commutative property/collected like terms
5x-11=19 Subtracted x from both sides
5x=30 Added 11 to both sides
x=6 Divided both sides by 5

Exercise 3.
Solve each equation for x. For each step, describe the operation used to convert the equation.
a. 7x-[4x-3(x-1)]=x+12
7x-(4x-3x+3)=x+12 Distributive property
7x-(x+3)=x+12 Collected like terms
7x-x-3=x+12 Distributive property
6x-3=x+12 Collected like terms
5x-3=12 Subtracted x from both sides
5x=15 Added 3 to both sides
x=3 Divided both sides by 5
{3}

b. 2[2(3-5x)+4]=5[2(3-3x)+2]
2(6-10x+4)=5(6-6x+2) Distributive property
2(10-10x)=5(8-6x) Commutative property/collected like terms
20-20x=40-30x Distributive property
20+10x=40 Added 30x to both sides
10x=20 Subtracted 20 from both sides
x=2 Divided both sides by 10
{2}

c. $$\frac{1}{2}$$ (18-5x)=$$\frac{1}{3}$$(6-4x)
3(18-5x)=2(6-4x) Multiplied both sides by 6
54-15x=12-8x Distributive property
54=12+7x Added 15x to both sides
42=7x Subtracted 12 from both sides
6=x Divided both sides by 7
{6}

Note with the class that students may have different approaches that arrived at the same answer.
Ask students how they handled the fraction in part (c).
→ Was it easier to use the distributive property first or multiply both sides by 6 first?

Exercise 4.
Consider the equations x+1=4 and (x+1)2=16.
a. Verify that x=3 is a solution to both equations.
3+1=4
(3+1)2=16

b. Find a second solution to the second equation.
x=-5

c. Based on your results, what effect does squaring both sides of an equation appear to have on the solution set?
Answers will vary. The new equation seems to retain the original solution and add a second solution.

Exercise 5.
Consider the equations x-2=6-x and (x-2)2=(6-x)2.
a. Did squaring both sides of the equation affect the solution sets?
No. x=4 is the only solution to both equations.

b. Based on your results, does your answer to part (c) of the previous question need to be modified?
The new equation retains the original solution and may add a second solution.

Exercise 6.
Consider the equation x3+2=2x2+x.
a. Verify that x=1, x=-1, and x=2 are each solutions to this equation.
(1)3+2=2(1)2+1 True
(-1)3+2=2(-1)2+(-1) True
(2)3+2=2(2)2+2 True

b. Bonzo decides to apply the action “ignore the exponents” on each side of the equation. He gets
x+2=2x+x. In solving this equation, what does he obtain? What seems to be the problem with his technique?
x=1 The problem is that he only finds one of the three solutions to the equation.

c. What would Bonzo obtain if he applied his “method” to the equation x2+4x+2=x4? Is it a solution to the original equation?
x=-$$\frac{1}{2}$$ No. It is not a solution to the original equation.

Exercise 7.
Consider the equation x-3=5.
a. Multiply both sides of the equation by a constant, and show that the solution set did not change.
7(x-3)=7(5)
7(8-3)=7(5)
7(5)=7(5)

Now, multiply both sides by x.
x(x-3)=5x

b. Show that x=8 is still a solution to the new equation.
8(8-3)=5(8)
8(5)=5(8)

c. Show that x=0 is also a solution to the new equation.
0(0-3)=5(0)

Now, multiply both sides by the factor x-1.
(x-1)x(x-3)=5x(x-1)

d. Show that x=8 is still a solution to the new equation.
(8-1)(8)(8-3)=5(8)(8-1)
(7)(8)(5)=5(8)(7)

e. Show that x=1 is also a solution to the new equation.
(1-1)(1)(1-3)=5(1)(1-1)
0(1)(-2)=5(1)(0)
0=0

f. Based on your results, what effect does multiplying both sides of an equation by a constant have on the solution set of the new equation?
Multiplying by a constant does not change the solution set.

g. Based on your results, what effect does multiplying both sides of an equation by a variable factor have on the solution set of the new equation?
Multiplying by a variable factor could produce additional solution(s) to the solution set.

### Eureka Math Algebra 1 Module 1 Lesson 13 Problem Set Answer Key

Question 1.
Solve each equation for x. For each step, describe the operation used to convert the equation. How do you know that the initial equation and the final equation have the same solution set?
Steps will vary as in the Exit Ticket and exercises.

a. $$\frac{1}{5}$$ [10-5(x-2)]=$$\frac{1}{10}$$ (x+1)
Solution set is {39/11}.

b. x(5+x)=x2+3x+1
Solution set is {$$\frac{1}{2}$$}.

c. 2x(x2-2)+7x=9x+2x3
Solution set is {0}.

Question 2.
Consider the equation x+1=2.
Students should write the new equations and the solution sets:

a. Find the solution set.
Solution set is {1}.

b. Multiply both sides by x+1, and find the solution set of the new equation.
New solution set is {±1}.

c. Multiply both sides of the original equation by x, and find the solution set of the new equation.
New solution set is {0,1}.

Question 3.
Solve the equation x+1=2x for x. Square both sides of the equation, and verify that your solution satisfies this new equation. Show that –$$\frac{1}{3}$$ satisfies the new equation but not the original equation.
The solution of x+1=2x is x=1. The equation obtained by squaring is (x+1)2=4x2.
Let x=1 in the new equation. (1+1)2=4(1)2 is true, so x=1 is still a solution.
Let x=-$$\frac{1}{3}$$ in the new equation. (-$$\frac{1}{3}$$+1)2=4(-$$\frac{1}{3}$$)2 is true, so x=-$$\frac{1}{3}$$ is also a solution to the new equation.

Question 4.
Consider the equation x3=27.
a. What is the solution set?
Solution set is {3}.

b. Does multiplying both sides by x change the solution set?
Yes.

c. Does multiplying both sides by x2 change the solution set?
Yes.

Question 5.
Consider the equation x4=16.
a. What is the solution set?
Solution set is {-2,2}.

b. Does multiplying both sides by x change the solution set?
Yes.

c. Does multiplying both sides by x2 change the solution set?
Yes.

### Eureka Math Algebra 1 Module 1 Lesson 13 Exit Ticket Answer Key

Question 1.
Solve the equation for x. For each step, describe the operation and/or properties used to convert the equation.
5(2x-4)-11=4+3x
Solution set is {5}.

Question 2.
Consider the equation x+4=3x+2.
a. Show that adding x+2 to both sides of the equation does not change the solution set.
x+4=3x+2
4=2x+2
2=2x
1=x
x+4+x+2=3x+2+x+2
2x+6=4x+4
2=2x
1=x

b. Show that multiplying both sides of the equation by x+2 adds a second solution of x=-2 to the solution set.