# Eureka Math Algebra 2 Module 1 Lesson 36 Answer Key

## Engage NY Eureka Math Algebra 2 Module 1 Lesson 36 Answer Key

### Eureka Math Algebra 2 Module 1 Lesson 36 Opening Exercise Answer Key

Find all solutions to each of the systems of equations below using any method.
2x – 4y = – 1                  y = x2 – 2                 x2 + y2 = 1
3x – 6y = 4                     y = 2x – 5                 x2 + y2 = 4
All three systems have no real number solutions, which is evident from the non-Intersecting graphs in each. Instead of graphing the systems, students may have used an analytic approach such as the approach outlined in the Discussion below.

Discussion (10 minutes)
Ask students to explain their reasoning for each of the three systems in the Opening Exercise with both approaches shown for each part, allowing six students the opportunity to present their solutions to the class. It is important to go through both the analytical and graphical approaches for each system so that students draw the connection between graphs that do not intersect and systems that have no analytic solution. Be sure to display the graph of each system of equations as students are led through this discussion.

Part (a):
→ Looking at the graphs of the equations in the first system, 2x – 4y = – 1 and 3x – 6y = 4, how can we tell that the system has no solution?
The two lines never intersect.
The two lines are parallel.

→ Using an algebraic approach, how can we tell that there is no solution?
If we multiply both sides of the top equation by 3 and the bottom equation by 2, we see that an equivalent system can be written.
6x – 12y = – 3
6x – 12y = 8
Subtracting the first equation from the second results in the false number sentence
0 = 11.
Thus, there are no real numbers x and y that satisfy both equations.

→ The graphs of these equations are lines. What happens if we put them in slope-intercept form?
Rewriting both linear equations in slope-intercept form, the system from part (a) can be written as
y = $$\frac{1}{2}$$x + $$\frac{1}{4}$$
y = $$\frac{1}{2}$$x – $$\frac{2}{3}$$.
From what we know about graphing lines, the lines associated to these equations have the same slope and different y-intercepts, so they will be parallel. Since parallel lines do not intersect, the lines have no points in common and, therefore, this system has no solution.

Part (b):
→ Looking at the graphs of the equations in the second system, y = x2 – 2 and = 2x – 5, how can we tell that the system has no solution?
The line and the parabola never intersect.

→ Can we confirm, algebraically, that the system in part (b) has no real solution?
Yes. Since y = x2 – 2 and y = 2x – 5, we must have x2 – 2 = 2x – 5, which is equivalent to the quadratic equation x2 – 2x + 3 = 0. Solving forx using the quadratic formula, we get

Since the square root of a negative real number is not a real number, there is no real number x that satisfies the equation x2 – 2x + 3 = 0; therefore, there is no point in the plane with coordinates (x, y) that satisfies both equations in the original system.

Part (c):
→ Looking at the graphs of the equations in the final system, x2 + y2 = 1 and x2 + y2 = 4, how can we tell that there the system has no solution?
The circles are concentric, meaning that they have the same center and different radii. Thus, they never intersect, and there are no points that lie on both circles.

→ Can we algebraically confirm that the system in part (c) has no solution?
Yes. If we try to solve this system, we could subtract the first equation from the second, giving the false number sentence 0 = 3. Since this statement is false, we know that there are no values of x andy that satisfy both equations simultaneously; thus, the system has no solution.

At this point, ask students to summarize in writing or with a partner what they have learned so far. Use this brief exercise as an opportunity to check for understanding.

### Eureka Math Algebra 2 Module 1 Lesson 36 Exercise Answer Key

Exercise 1.
Are there any real number solutions to the system y = 4 and x2 + y2 = 2? Support your findings both analytically and graphically.
x2 + (4)2 = 2
x2 + 16 = 2
x2 = – 14
Since x2 is non-negative for all real numbers x, there are no real numbers x so that x2 = – 14. Then, there is no pair of real numbers (x, y) that solves the system consisting of the line y = 4 and the circle x2 + y2 = 2. Thus, the line y = 4 does not intersect the circle x2 + y2 = 2 in the real plane. This is confirmed graphically as follows.

Exercise 2.
Does the line y = x intersect the parabola y = – x2? If so, how many times and where? Draw graphs on the same set of axes.

x = – x2
x + x2 = 0
x(1 + x) = o
x = o or x = – 1
If x = 0, then y = – x2 = 0, and if x = – 1, then y = – x2 = – (-1)2 = -1. calculator to shot
The line y = x intersects the parabola y = -x2 at two distinct points: (0, 0) and (-1,-1).

Exercise 3.
Does the line y = – x intersect the circle x2 + y2 = 1? If so, how many times and where? Draw graphs on the same set of axes.

x2 + (- x)2 = 1
2x2 = 1
x2 = $$\frac{1}{2}$$
x = –$$\frac{\sqrt{2}}{2}$$ or x = $$\frac{\sqrt{2}}{2}$$
The hne y = – x intersects the circle x + y = 1 at two distinct points: $$\left(-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)$$ and $$\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)$$

Exercise 4.
Does the line y = 5 intersect the parabola y = 4 – x2? Why or why not? Draw the graphs on the same set of axes.

5 = 4 – x2
1 = – x2
x2 = – 1
A squared real number cannot be negative, so the line y = 5 does not intersect the parabola y = 4 – x2.

### Eureka Math Algebra 2 Module 1 Lesson 36 Problem Set Answer Key

Question 1.
For each part, solve the system of linear equations, or show that no real solution exists. Graphically support your answer.
a. 4x + 2y = 9
x + y = 3

Multiply the second equation by 4.
4x + 2y = 9
4x + 4y = 12
Subtract the first equation from the second equation.
2y = 3
Then y = $$\frac{3}{2}$$.
Substitute $$\frac{3}{2}$$ for y in the original second equation.
x + $$\frac{3}{2}$$ = 3
Then x = $$\frac{3}{2}$$.
The lines from the system intersect at the point $$\left(\frac{3}{2}, \frac{3}{2}\right)$$,

b. 2x – 8y = 9
3x – 12y = 0

Multiply the first equation by 3 and the second equation by 2 on both sides.
6x – 24y = 27
6x – 24y = 0
Subtracting the new second equation from the first equation gives the false number sentences 27 = 0. Thus, there is no solution to the system. The graph of the system appropriately shows two parallel lines.

Question 2.
Solve the following system of equations, or show that no real solution exists. Graphically confirm your answer.
3x2 + 3y2 = 6
x – y = 3

We can factor out 3 from the top equation and isola te y in the bottom equation to give us a better idea of what the graphs of the equations in the system look like. The first equation represents a circle centered at the origin with radius √2, and the second equation represents the line y = x – 3.
3x2 + 3(x – 3)2 = 6
x2 + (x – 3)2 = 2
x2 + (x2 – 6x + 9) = 2
2x2 – 6x + 7 = 0
We solve for x using the quadratic formula:
a = 2, b = – 6, c = 7
x = $$\frac{-(-6) \pm \sqrt{(-6)^{2}-4(2 \cdot 7)}}{2 \cdot 2}$$
x = $$\frac{6 \pm \sqrt{36-56}}{4}$$
The solutions would be $$\frac{6+\sqrt{-20}}{4}$$ and $$\frac{6-\sqrt{-20}}{4}$$
Since both solutions for x contain a square root of a negative number, no real solution x exists; so the system has no solution (x, y) where x and y are real numbers.

Question 3.
Find the value of k so that the graph of the following system of equations has no solution.
3x – 2y – 12 = 0
kx + 6y – 10 = 0
First, we rewrite the linear equations in the system in slope-intercept form.
y = $$\frac{3}{2}$$x – 6
y = –$$\frac{k}{6}$$x + $$\frac{10}{6}$$
There is no solution to this system when the lines are parallel. Two lines are parallel when they share the same slope and have different y-intercepts. Here, the first line has slope $$\frac{3}{2}$$ and y-in tercept – 6, and the second line has slope –$$\frac{k}{6}$$ and y-intercept $$\frac{10}{6}$$. The lines have different y-intercepts and will be parallel when $$-\frac{k}{6}=\frac{3}{2}$$
$$\frac{3}{2}=-\frac{k}{6}$$
2k = – 18
k = – 9
Thus, there is no solution only when k = – 9.

Question 4.
Offer a geometric explanation to why the equation x2 – 6x + 10 = 0 has no real solutions.
The graph of y = x2 – 6x + 10 opens upward (since the leading coefficient is positive) and takes on its lowest value at the vertex (3, 1). Hence, it does not intersect the x-axis, and, therefore, the equation has no real solutions.

Question 5.
Without his pencil or calculator, ioey knows that 2x3 + 3x2 – 1 = 0 has at least one real solution. How does he know?
The graph of every cubic polynomialfunction intersects the x-axis at least once because the end behaviors are opposite: one end goes up and the other goes down. This means that the graph of any cubic equation y = ax3 + bx2 + cx + d must have at least one x-intercept. Thus, every cubic equation must have at least one real solution.

Question 6.
The graph of the quadratic equation y = x2 + 1 has no x-intercepts. However, Gia claims that when the graph of y = x2 + 1 is translated by a distance of 1 in a certain direction, the new (translated) graph would have exactly one x-intercept. Further, if y = x2 + 1 is translated by a distance greater than 1 in the same direction, the new (translated) graph would have exactly two x-intercepts. Support or refute Gia’s claim. If you agree with her, in which direction did she translate the original graph? Draw graphs to illustrate.

By translating the graph of y = x2 + 1 DOWN by 1 unit, the new graph has equation y = x2, which has one x-intercept at x = 0. When translating the original graph DOWN by more than 1 unit, the new graph will cross the x-axis exactly twice.

Question 7.
In the previous problem, we mentioned that the graph of y = x2 + 1 has no x-intercepts. Suppose that y = x2 + 1 is one of two equations in a system of equations and that the other equation is linear. Give an example of a linear equation such that this system has exactly one solution.
The line with equation y = 1 is tangent to y = x2 + 1 only at (0, 1); so there would be exactly one real solution to the system.
y = x2 + 1
y = 1
Another possibility is an equation of any vertical line, such as x = – 3 or x = 4, or x = a for any real
number a.

Question 8.
In prior problems, we mentioned that the graph of y = x2 + 1 has no x-intercepts. Does the graph of y = x2 + 1 intersect the graph of y = x3 + 1?
Setting these equations together, we can rearrange terms to get x3 – x2 = 0, which is an equation we can solve by factoring. We have x2 (x – 1) = 0, which has solutions at 0 and 1. Thus, the graphs of these equations intersect when x = 0 and when x = 1. When x = 0, y = 1, and when x = 1, y = 2. Thus, the two graphs intersect at the points (0, 1) and (1, 2).

The quick answer: The highest term in both equations has degree 3. The third-degree term does not cancel when setting the two equations (in terms of x) equal to each other. All cubic equations have at least one real solution, so the two graphs intersect at least at one point.

### Eureka Math Algebra 2 Module 1 Lesson 36 Exit Ticket Answer Key

Solve the following system of equations, or show that it does not have a real solution. Support your answer analytically and graphically.
y = x2 – 4
y = – (x + 5)
We distribute over the set of parentheses in the second equation and rewrite the system.
y = x2 – 4
y = – (x + 5)
The graph of the system shows a parabola and a line that do not intersect. As such, we know that the system does not have a real solution.

Algebraically,
x2 – 4 = – x- 5
x2 + x + 1 = 0.
Using the quadratic formula with a = 1, b = 1, and c = 1,
x = $$\frac{-1+\sqrt{1^{2}-4(1)(1)}}{2(1)}$$ or x = $$\frac{-1-\sqrt{1^{2}-4(1)(1)}}{2(1)}$$
which indicates that the solutions would be $$\frac{-1+\sqrt{-3}}{2}$$ and $$\frac{-1-\sqrt{-3}}{2}$$
Since the square root of a negative number is not a real number, there is no real number x that solves this equation. Thus, the system has no solution (x, y) where x and y are real numbers.

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