Into Math Grade 8 Module 7 Lesson 5 Answer Key Examine Special Systems

We included HMH Into Math Grade 8 Answer Key PDF Module 7 Lesson 5 Examine Special Systems to make students experts in learning maths.

HMH Into Math Grade 8 Module 7 Lesson 5 Answer Key Examine Special Systems

I Can identify the number of solutions to a system of linear equations in any form.

Spark Your Learning

What values would give you no solutions to the system of equations. How do you know?
y = 2x + 5
y = HMH Into Math Grade 8 Module 7 Lesson 5 Answer Key Examine Special Systems 1x + HMH Into Math Grade 8 Module 7 Lesson 5 Answer Key Examine Special Systems 1

HMH Into Math Grade 8 Module 7 Lesson 5 Answer Key Examine Special Systems 2

HMH Into Math Grade 8 Module 7 Lesson 5 Answer Key Examine Special Systems 3
Answer:
Given that the equation is
y = 2x + 5
The equation for the no solutions to the system of equations.
y = 2x + 6
Subtract the above equations then we get
0 = 1
The system has no solution.

Turn and Talk What values would make the system of equations have an infinite number of solutions?
Answer:
Given that the equation is
y = 2x + 5
The equation for the no solutions to the system of equations.
y = 2x + 5
Subtract the above equations then we get
0 = 0
The system has an infinite number of solutions.

Build Understanding

Just as when solving a linear equation, the solution set for a system of two linear equations can have one solution, no solution, or infinitely many solutions.
A system of two linear equations has no solution if the graphs of the two lines never intersect because the lines are parallel.
HMH Into Math Grade 8 Module 7 Lesson 5 Answer Key Examine Special Systems 4
1. On a map, a street that runs in a straight line can be represented by a linear equation.
A. Solve the system of equations represented by Oak Street and Pine Street.
Answer:
The equation for Oak street = x – y = 5
Pine Street = y = x + 5
x – y = 5
y = x + 5
y – x = 5
Solving both the equations
x – y = 5
-x + y = 5
Thus it has no solution

B. What was the result? What does it tell you about the number of solutions to the system?
Answer: It has no solution.

C. What are the slopes of the two streets? the y-intercepts?
_______________
Answer:
y = mx + x
x – y = 5
y = x – 5
Slope = 1 and y-intercept = -5
-x + y = 5
y = 5 + x
y = x + 5
Slope = 1 and y-intercept = 5

A system of two linear equations has infinitely many solutions if the graphs of the two lines are concurrent and therefore intersect at infinitely many points.

D. Solve the system of equations represented by South Birch and North Birch.
__________________
Answer:
South Birch: 3x + y = 2
North Birch: y = -3x + 2
3x + y = 2
Both the equations are same.
Thus the system has infinite solution.

E. What was the result? What does this tell you about the number of solutions to the system?
__________________
Answer: The result is 0.
The system of equations has infinite solutions.

F. What are the slopes of the two streets? the y-intercepts?
__________________
Answer:
South Birch: 3x + y = 2 ⇒ y = -3x + 2
North Birch: 3x + y = 2 ⇒ y = -3x + 2
Slope = -3
y-intercept = 2

Turn and Talk Compare the answers for Part C and F. Explain the similarities and differences between a system with no solutions and infinitely many solutions.

Step It Out

2. The system HMH Into Math Grade 8 Module 7 Lesson 5 Answer Key Examine Special Systems 5 has infinitely many solutions.
A. Graph the system on your own paper. What do you notice?
_____________________
Answer: The equations are same.

HMH Into Math Grade 8 Module 7 Lesson 5 Answer Key Examine Special Systems 6

B. What do you notice about the pair of equations?
________________________
Answer:
Both the equations are same and meet at the same point.
C. Fill in the blanks to use elimination to solve the system:
HMH Into Math Grade 8 Module 7 Lesson 5 Answer Key Examine Special Systems 7
Answer:
HMH-Into-Math-Grade-8-Module-7-Lesson-5-Answer-Key-Examine-Special-Systems-7

D. How does the final equation relate to the system having infinitely many solutions?
________________________
Answer: The final equation relates to the system having infinitely many solutions is 0 = 0.

3. The system HMH Into Math Grade 8 Module 7 Lesson 5 Answer Key Examine Special Systems 8 has no solution.

A. Graph the system on your own paper. What do you notice?
Answer:
HMH Into Math Grade 8 Module 7 Lesson 5 Answer Key Examine Special Systems_3A

B. What do you notice about the pair of equations?
Answer:
We notice that the pair of equations are parallel and non-intersecting equations.

C. Fill in the blanks to solve the system by substitution:
HMH Into Math Grade 8 Module 7 Lesson 5 Answer Key Examine Special Systems 10
Answer:
2x + y = 3 → y = -2x + 3
4x + 2y = 8
4x + 2(-2x + 3) – 8
4x -4x + 6 – 8 = -2

D. How does the final equation relate to the system having no solutions?

Turn and Talk If you solve the equations in Task 3 for y, what equations do you get? How is that related to the graph of the two lines?

4. How many solutions does each system of equations have?

A.
6x + 8y = 20
9x + 12y = 30
6x + 8y = 20 can be simplified by a factor of ___, resulting in the equation ______
9x + 12y = 30 can be simplified by a factor of ______, resulting in the equation ____
The given equations are (equivalent / unique). Therefore the system has ____ solution(s).
Answer:
6x + 8y = 20 ⇒ 3x + 4y = 10 (Common factor is 2)
9x + 12y = 30 ⇒ 3x + 4y = 10 (Common factor is 3)
Both the equations are equivalent. Therefore the system has no solution.

B.
5x + 2y = 7
15x + 6y = 24
All terms in 15x + 6y = 24 can be divided by ____. resulting in the equation ____.
The expression 5x + 2y (can / cannot) equal both 7 and ___.
Therefore the system has ______ solution(s)
Answer:
5x + 2y = 7
15x + 6y = 24
All terms in 15x + 6y = 24 can be divided by 3.
5x + 2y = 8
The expression 5x + 2y cannot equal both 7 and 8.
Therefore the system has no solution.

Check Understanding

Question 1.
Rory bought a pencil and a folder for $0.35. Emily bought 3 pencils and 3 folders for $1.05. The system of equations shown represents this,
x + y = 35
3x + 3y = 105
HMH Into Math Grade 8 Module 7 Lesson 5 Answer Key Examine Special Systems 11
Could each of them have paid the same price for pencils and for folders?
Answer:
Given that the equation is
x + y = 35 is an equation 1
3x + 3y = 105 is an equation 2
Multiply equation 1 with 3 then we get
3x + 3y = 105 is an equation 3
Subtract equation 2 from equation 3 then we get
0 = 0
Rory and Emily had paid the same price for pencils and for folders.

Use the information to answer Problems 2-4.

Consider the system of equations
6x + 4y = 14
9x + 6y = C

Question 2.
Which value(s) of C represent a system with infinitely many solutions?
Answer:
Given that the equations is
6x + 4y = 14 is an equation 1
9x + 6y = C is an equation 2
Multiply equation 1 with 9 then we get
54x + 36y = 126 is an equation 3
Multiply equation 2 with 6 then we get
54x + 36y = 6c is an equation 4
If c = 21 then 6(21) = 126
Subtract equation 3 from 4 then we get
0 = 0
The system has infinite solution.

Question 3.
Which value(s) of C represent a system with no solution?
Answer:
If c = 0 then the system with no solution.
Given that the equations is
6x + 4y = 14 is an equation 1
9x + 6y = C is an equation 2
Multiply equation 1 with 9 then we get
54x + 36y = 126 is an equation 3
Multiply equation 2 with 6 then we get
54x + 36y = 6c is an equation 4
c = 0 then 6(0) = 0
54x + 36y = 0 is an equation 5
Subtract equation 5 from equation 3 then we get
0 = 126.
The system has no solution.

Question 4.
Is there any value of C that would result in the system having one solution? Why or why not?
Answer:

On Your Own

Question 5.
Stacy and Bridget are wrapping presents. Their times are represented by the system shown, where x is the wrapping rate in minutes per box and y is the packing rate in minutes per bag.
HMH Into Math Grade 8 Module 7 Lesson 5 Answer Key Examine Special Systems 12
6x + 3y = 36
2x + y = 12
How many solutions does the system have? Explain.
Answer:
Given that,
The equations are
6x + 3y = 36 is an equation 1
2x + y = 12 is an equation 2
Multiply equation with 3 then we get.
6x = 3y = 36 is an equation 3
Subtract equation 3 from equation 1 then we get
0 = 0
The system as many infinite solutions.

Use the information to answer Problems 6-7.

A system of two equations includes the equation 2x + 6y = 8. The other equation in the system was smudged but shows 3x + HMH Into Math Grade 8 Module 7 Lesson 5 Answer Key Examine Special Systems 13 = 12.

Question 6.
Reason Is it possible for the system to have infinitely many solutions? If so, what could be the smudged term? If not, explain your reasoning.
Answer:
The missing term in the question is 9y.
The equations is
2x + 6y = 8 is an equation 1
3x + 9y = 12 is an equation 2
Multiply equation 1 with 3 then we get
6x + 18y = 24 is an equation 3
Multiply equation 2 with 2 then we get
6x + 18y = 24 is an equation 4
Subtract equation 4 from equation 3 then we get
0 = 0
The system has infinitely many solutions.

Question 7.
Reason Is it possible for the system to have no solutions? If so, what could be the smudged term? If not, explain your reasoning.
Answer:
2x + 6y = 8 —- × 3
3x + HMH Into Math Grade 8 Module 7 Lesson 5 Answer Key Examine Special Systems 13 = 12 — × 2
6x + 18y = 24
6x + 2HMH Into Math Grade 8 Module 7 Lesson 5 Answer Key Examine Special Systems 13 = 24
According to the equation
2HMH Into Math Grade 8 Module 7 Lesson 5 Answer Key Examine Special Systems 13 = 18y
HMH Into Math Grade 8 Module 7 Lesson 5 Answer Key Examine Special Systems 13 = 18/2
HMH Into Math Grade 8 Module 7 Lesson 5 Answer Key Examine Special Systems 13 = 9y
Thus a system of two equations have the solutions.

Question 8.
Open Ended Provide an equation that, when combined with the equation 7x – y = 5, could form a system with the given number of solutions.
A. infinitely many solutions ___________
Answer:

B. no solutions ___________
Answer:

For Problems 9-12, state the number of solutions to the system: one, none, or infinitely many.

Question 9.
3x + 4y = 7
3x + 4y = 9
Answer:
Given that the equations are
3x + 4y = 7 is an equation 1
3x + 4y = 9 is an equation 2
Subtracting equation 1 from equation 2
0 = 2
Therefore the number of solutions to the system is no solution

Question 10.
x + y = 2
y – x = 12
Answer:
Given that the equations are
x + y = 2 is an equation 1
y – x = 12 is an equation 2
Rewrite the equation 2
-x + y = 12 is an equation 3
Subtract equation 1 from equation 3
-2x = 10
x = 10/-2
x = -5
Substitute x = -5 in equation 1
-5 + y = 2
y = 2 – 5
y = -3
The solution is (-5, -3)
Therefore the number of solutions to the system is one.

Question 11.
y = 3x – 5
y = -3x – 5
Answer:
Given that the equations are
y = 3x – 5
y = -3x – 5
Rewrite the above equations
y – 3x = -5 is an equation 1
y + 3x = -5 is an equation 2
Subtract equation 2 from equation 1
-6x = 0
x = 0/-6
x = 0
Substitute equation x = 0 in equation 1
y = 3(0) -5
y = -5
The solution is (0, -5)
Therefore the number of solutions to the system is one.

Question 12.
y = 5 + 2x
y = 2x + 5
Answer:
Given that the equations are
y = 5 + 2x
y = 2x + 5
Rewrite the above equations
y – 2x = 5 is an equation 1
y – 2x = 5 is an equation 2
Subtract equation 2 from equation 1 then we get
0 = 0
The system as infinitely manysolutions.

Question 13.
Rosie spends 2 hours building each model plane x and 6 hours building each model boaty. She has a total of 40 hours to spend each week. Yuri spends 1 hour building each model plane and 3 hours building each model boat, but has only 20 hours to spend each week. The system represents their total work times.
HMH Into Math Grade 8 Module 7 Lesson 5 Answer Key Examine Special Systems 14
2x + 6y = 40
x + 3y = 20
How many solutions does the system have? Explain.
Answer:
Given that,
The equations are
2x + 6y = 40 is an equation 1
x + 3y = 20 is an equation 2
Multiply equation 2 with 2 then we get
2x + 6y = 40 is an equation 3
Subtract equation 3 from equation 2 then we get
0 = 0
The system as infinite solutions.

Question 14.
Reason Devante and Jim each draw a rectangle. The perimeter of Devante’s rectangle is 60 centimeters. The sum of the length and width of Jim’s rectangle is 15 centimeters. The system shown represents the dimensions of their rectangles.
2L + 2W = 60
L + W= 15
Do the rectangles have the same dimensions? Explain.
Answer:
Given that
2L + 2W = 60
L + W= 15
2(L + W) = 60
2(15) = 60
The two rectangles have the same dimensions.

For Problems 15—20, solve each system of equations.

Question 15.
x + 2y = 7
4x + 8y = 28
Answer:
Given that,
The equations are
x + 2y = 7 is an equation 1
4x + 8y = 28 is an equation 2
Multiply equation 1 with 4 then we get
4(x + 2y) = 4(7)
4x + 8y = 28 is an equation 3
Subtract equation 2 from equation 3 then we get
0 = 0
The system has infinitely many solutions.

Question 16.
5x – 2y = 19
5x + 2y = 11
Answer:
Given that the equations are
5x – 2y = 19 is an equation 1
5x + 2y = 11 is an equation 2
Subtract equation 2 from equation 1 then
-4y = 8
y = 8/-4
y = -2
substitute y = -4 in equation 1 then
5x – 2(-2) = 19
5x + 4 = 19
5x = 19 – 4
5x = 5
x = 5/5
x = 1
The solution is (1, -2)
Therefore the system has only one solution.

Question 17.
6x – 4y = 2
-9x + 6y = 3
Answer:
Given that the equations are
6x – 4y = 2 is an equation 1
-9x + 6y = 3 is an equation 2
Multiply equation 1 with -9 then we get
-30x + 36y = -18 is an equation 3
Multiply equation 2 with 6 then we get
-30x + 36y = 18 is an equation 4
Subtract equation 4 from equation 3 then we get
0 = -36
The system has no solution.

Question 18.
-4x + 2y = 12
y = 2x + 6
Answer:
Given that,
The equations are
-4x + 2y = 12 is an equation 1
y = 2x + 6 is an equation 2
Rewrite the equation 2
2x – y = -6
The multiply the equation with -2
-2(2x – y) = -2(-6)
-4x + 2y = 12 is an equation 3
Subtract equation 2 from equation 3 then we get
0 = 0
The system has infinitely many solutions.

Question 19.
12x – 18y = 9
2x – 3y = 1
Answer:
Given that,
The equations are
12x – 18y = 9 is an equation 1
2x – 3y = 1 is an equation 2
Multiply the equation 2 with 6
6(2x – 3y) = 6(1)
12x – 18y = 6 is an equation 3
Subtract equation 3 from equation 1
Then we get
0 = -3
The system has no solution.

Question 20.
x = 2y – 10
4y = x + 12
Answer:
Given that the equations are
x = 2y – 10
4y = x + 12
Rewrite the above equations
x – 2y = -10 is an equation 1
-x + 4y = 2 is an equation 2
Multiply the equation 1 with -2
-2(x – 2y) = -2(-10)
-2x +4y = 20 is an equation 3
Subtract equation 2 from equation 3 then we get it
-x = 18
x = -18
Substitute x = -18 in equation 1
-18 – 2y = -10
-2y = -10 + 18
-2y = 8
y = 8/-2
y = -4
The solution is (-18, -4).
the system has one solution.

I’m in a Learning Mindset!

Which special systems were the most challenging for me? Why?

Lesson 7.5 More Practice/Homework

Question 1.
Gabe bought 6 singles and 9 albums from an online music store. Jenny bought 4 singles and 6 albums at a different online music store. The system shown represents their totals for singles that cost x dollars each and albums that cost y dollars each.
6x + 9y = 96
4x+ 6y = 38

A. Solve the system. What does the solution mean?
Answer:
Given that the equations are
x + 9y = 96 is an equation 1
4x+ 6y = 38 is an equation 2
Multiply equation 1 with 4 then we get
4x + 36y = 144 is an equation 3
Subtract equation 2 from equation 3 then we get
30y = 106
y = 106/30
y = 3.53
Substitute y = 3.53 in equation 1
x + 9(3.53) = 96
x + 31.77 = 96
x = 96 – 31.77
x = 64.23
Therefore the system for singles is 64.23 dollars each and albums is 3.53 dollars each.

HMH Into Math Grade 8 Module 7 Lesson 5 Answer Key Examine Special Systems 15

B. Is it possible to know if Gabe and Jenny purchased the singles and albums for the same prices? Explain.
Answer:
No, it is not possible.
Gabe and Jenny purchased the singles and albums for different prices.

For Problems 2—4. use the system of equations shown.

3x + 5y = 4
16x + 10y = 4

Question 2.
Does the system have a solution? Explain.
Answer:
Given that,
The equations are
3x + 5y = 4 is an equation 1
16x + 10y = 4 is an equation 2
Multiply equation 1 with 2 then we get
6x + 10y = 8 is an equation 3
Subtract equation 2 from equation 3 then we get
-10x = 4
x = 4/-10
x = -0.4
Substitute x = -0.4 in equation 1
3(-0.4) + 5y = 4
-1.2 + 5y = 4
5y = 4 + 1.2
5y = 5.2
y = 5.2/5
y = 1.04
The solution is (-0.4, 1.04)
The system as one solution.

Question 3.
Critique Reasoning Manuel says that he can change the 4 in the second equation to any number and the system will have no solution. Is Manuel correct? Explain.
Answer:

Question 4.
How can one number in the second equation be changed so the system has only one solution? Explain.
Answer:

For Problems 5—10, solve the system of equations.

Question 5.
x + y = 3
x – y = -3
Answer:
Given that the equations are
x + y = 3 is an equitation 1
x – y = -3 is an equation 2.
Subtract equation 2 from equation 1 then we get
2y = 6
y = 6/2
y = 3
Substitute y = 3 in equation 1
x + 3 = 3
x = 3 -3
x = 0
The solution of the equations is (0,3)
Therefore the system as only one solution.

Question 6.
5x + y = 7
10x + 2y = 16
Answer:
Given that,
5x + y = 7
Multiply the above equation with 2
2(5x + y) = 2(7)
10x + 2y = 14 is an equation 1
10x + 2y = 16 is an equation 2
Subtract equation 2 from equation 1 then we get
0 = -2.
Therefore the system has no solutions

Question 7.
13x + 4y = 9
12x + 16y = 36
Answer:
Given that,
The equations is
13x + 4y = 9
Multiply the above equation with 4
4(13x + 4y) = 4(9)
52x + 16y = 36 is an equation 1
12x + 16y = 36 is an equation 2
Subtract equation 2 from equation 1 then we get
40x = 0
x = 0/40
x = 0
Substitute x = 0 in equation 1 then
13(0) + 4y = 9
4y = 9
y = 9/4
y = 2.25
The solution is (0,2.25)
Therefore the system as only one solution.

Question 8.
4x + 3y = 11
3x + 4y = 17
Answer:
Given that the equations are
4x + 3y = 11 is an equation 1
3x + 4y = 17 is an equation 2
Multiply equation 1 with 3 then we get
12x + 9y = 33 is an equation 3
Multiply equation 2 with 4 then we get
12x + 16y = 68 is an equation 4
Subtract equation 3 from equation 4 then we get
7y = 35
y = 35/7
y = 5
Substitute y = 5 in equation 1
4x + 3(5) = 11
4x + 15 = 11
4x = 11 – 15
4x = -4
x = -4/4
x = -1
The solution is (-1, 5)
The system has one solution.

Question 9.
3x + 12y = 9
5x + 20y = 15
Answer:
Given that the equations are
3x + 12y = 9 is an equation 1
5x + 20y = 15 is an equation 2
Multiply equation 1 with 5 then we get
15x + 60y = 45 is an equation 3
Multiply equation 2 with 3 then we get
15x + 60y = 45 is an equation 4
Subtract equation 4 from equation 3 then we get
0 = 0
The system has infinitely many solutions.

Question 10.
14x – 2y = 16
-14x + 7y = -49
Answer:
Given that,
The equations is
14x – 2y = 16 is an equation 1
-14x + 7y = -49 is an equation 2
multiply equation 1 with -1 then we get
-14x + 2y = -16 is an equation 3
Subtract equation 2 from equation 3
5y = 33
y = 33/5
y = 6.6
Substitute y = 6.6 in equation 1
14x – 2(6.6) = 16
14x = 16 + 13.2
14x = 29.2
x = 29.2/14
x = 2.08
The solution is (2.08, 6.6)
The system has on solution.

Test Prep

Question 11.
The system of equations HMH Into Math Grade 8 Module 7 Lesson 5 Answer Key Examine Special Systems 16 has infinitely many solutions. What is the value of a?
A. 5
B. 6
C. 7
D. 8
Answer:
ax + 4y = 10 —- × 3
-9x – 6y = -15 —- × 2
3ax + 12y = 30
-18x – 12y = -30
3a – 18 = 0
3a = 18
a = 6
Thus option B is the correct answer.

Question 12.
Consider the system of equations HMH Into Math Grade 8 Module 7 Lesson 5 Answer Key Examine Special Systems 17
The system has (no / one / infinitely many) solution(s).
Answer:
Given that the equations are
y = -3/2x + 4
y = -3/2x – 4
Rewrite the above equations
y + 3/2x = 4 is an equation 1
y + 3/2x = -4 is an equation 2
Subtract equation 1 from equation 2 then we get
0 = 8
The system has no solution.

Question 13.
Mark whether each system has infinitely many solutions, no solution, or one solution.
HMH Into Math Grade 8 Module 7 Lesson 5 Answer Key Examine Special Systems 18
Answer:
1) Given that the equations is
x + 3y = 6 is an equation 1
6x + 18y = 36 is an equation 2
Multiply the equation 1 with 6 then we get
6x + 18y = 36 is an equation 3
Subtract equation 2 from equation 3 then we gey
0 = 0

2) Given that the equations is
2x – 3y = -3 is an equation 1
4x + 6y = 18 is an equation 2
Multiply equation 1 with 2 then we get
4x – 6y = -6 is an equation 3
Subtract equation 2 from equation 3
8x = 12
x = 12/8
x = 3/2
Substitute x = 3/2 in equation 1
2(3/2) – 3y = -3
7/2 – 3y = -3
-3y = -3 – 7/2
-3y = -6.5
y = 6.5/3
The solution is (3/2, 6.5/3).
3) Given that the equations are
3x + 2y = 6 is an equation 1
6x – 4y = 24 is an equation 2
Multiply equation 1 with 2 then we get
6x + 4y = 12 is an equation 3
Subtract equation 2 from equation 3 then we get
12x = 36
x = 36/12
x = 3
Substitute x = 3 in equation 1
3(3) + 2y = 6
9 + 2y = 6
2y = 6 – 9
2y = -3
y = -3/2
The solution is (3, -3/2)
4) Given that the equations are
8x – 10y = 6
-8x +10y = 6
Subtract the above equations then we get
0 = 6

5) Given that the equations are
4x + 10y = 16 is an equation 1
10x + 25y = 40 is an equation 2
Divide equation 2 with 5 Then we get
2x + 5y = 8 then multiply by 2 then we get
4x + 10y = 16 is an equation 3
Subtract equation 1 from equation 3 then we get
0 = 0
HMH-Into-Math-Grade-8-Module-7-Lesson-5-Answer-Key-Examine-Special-Systems-18

Spiral Review

Use the information to answer Problems 14-15.

The height in inches of a candle that has been burning for x hours is represented by the equation h = 16 – 2x.

Question 14.
What is the meaning of the slope in the context of the burning candle?
Answer:
Given that,
The equation is h = 16 – 2x.
We know that height is the linear function of the time
Write the above equation in the form of y = mx + c
m = slope and c = y intercept.
Let us consider y = h
y = -2x + 16
The meaning of the slope is the height of the candle is burning for 2 hours.

Question 15.
What is the meaning of the y-intercept in the context of the burning candle?
Answer:
Given that,
The equation is h = 16 – 2x.
We know that height is the linear function of the time
Write the above equation in the form of y = mx + c
m = slope and c = y intercept.
Let us consider y = h
y = -2x + 16
y intercept = 16
The height of the burning candle is 16 inches.

Question 16.
What is the solution of the system of equations? HMH Into Math Grade 8 Module 7 Lesson 5 Answer Key Examine Special Systems 19
Answer:
Given that,
x = 2y -5
x – 2

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