SOLUTION: NAVIGATION: The paths of two ships are tracked on the same coordinate system. One ship is following a path described by the equation 2x+3y=6, and the other is following a path de

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Question 60579This question is from textbook Intermediate Algebra
: NAVIGATION: The paths of two ships are tracked on the same coordinate system. One ship is following a path described by the equation 2x+3y=6, and the other is following a path described by the equation y=2/3x-3.
a. Is there a possibility of a collision?
b. What are the coordinates of the danger point?
c. Is a collision a certainty?
While the answers are in the back of the book, I do not know how to get the correct answer. Could someone help me please? Thank you, I am going to school online and I am having difficulty. Much thanks!!
This question is from textbook Intermediate Algebra

Found 2 solutions by venugopalramana, Earlsdon:
Answer by venugopalramana(3286) About Me  (Show Source):
You can put this solution on YOUR website!
NAVIGATION: The paths of two ships are tracked on the same coordinate system. One ship is following a path described by the equation 2x+3y=6, and the other is following a path described by the equation y=2/3x-3..PLEASE PUT BRACKETS.DO YOU MEAN (2/3)X - 3 = (2X/3) -3.OR
[2/(3*X)] -3 ..OR
[2/(3-X)]...EACH WILL GIVE A DIFFERENT ANSWER.
I AM TAKING THE I ALTERNATIVE

a. Is there a possibility of a collision?
2X+3Y =6........1
Y=(2X/3)-3.....2
FROM EQN.1 DIVIDING WITH 3 WE GET
(2X/3)+ Y = 2
Y = 2 - (2X/3).......3..
SO
FROM EQNS. 2 AND 3 , WE GET
2X/3 -3 = 2 -(2X/3)
(2X/3) + (2X/3) = 2+ 3 =5
4X/3 =5
X=15/4
Y = [2*15/(4*3)]-3 = (5/2)-3=(5-6)/2 = -1/2
SINCE THE 2 LINES OF TRAVEL MEET AT A POINT THERE IS A POSSIBILITY OF COLLISION

b. What are the coordinates of the danger point?
THE COORDINATES OF COLLISION /DANGER POINT ARE [15/4 , -1/2]
c. Is a collision a certainty?
CANT SAY ..DEPENDS ON STARTING POINT ; DIRECTION OF TRAVEL ALONG THE LINE AND SPEED OF TRAVEL IN THAT DIRECTION WHICH ARE ALL NOT GIVEN.
SUPPOSE THEY ARE STARTING AT A POINT AND TRAVELLING IN THE DIRECTION AWAY FROM DANGER POINT THEN THEY DO NOT COLLIDE.
EVEN IF THEY ARE TRAVELLING TOWARDS THE DANGER POINT IF ONE IS GOING FASTER AND CROSSES DANGER POINT BEFORE THE OTHER REACHES THEY WILL NOT COLLIDE.
While the answers are in the back of the book, I do not know how to get the correct answer. Could someone help me please? Thank you, I am going to school online and I am having difficulty. Much thanks!!

Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
a) To see if there is a possibility of a collision, you could graph these two linear equations that represent the paths of the ships.
First, get the first equation into the slope-intercept form:
2x+3y = 6 Solve this for y by subtracting 2x.
3y = -2x+6 Now divide both sides by 3.
y = (-2/3)x + 2
Now the graph of the two equations looks like:
graph%28300%2C200%2C-5%2C5%2C-5%2C5%2C%28-2%2F3%29x%2B2%2C%282%2F3%29x-3%29
As you can see, there is a possibility of a collision if the two ships continue along their respective paths.
b) The danger point is the intersection of the two lines and this can be found by solving this system of equations:
y = (-2/3)x + 2
y = (2/3)x - 3
Set these two equations equal to each other and solve for x.
(2/3)x-3 = (-2/3)x+2 Simplify and solve for x.
(4/3)x = 5
x = 15/4 Now substitute this into either equation and solve for y.
y = (2/3)(15/4)-3
y = (5/2) - 3
y = -1/2
The coordinates of the danger (collision) point are: (15/4, -1/2)
c) Is a collision a certainty?
No, it is not. Whether the two ships collide or not depends on their individual speeds. A collission can be avoided by appropriate adjustment of their speeds even if they do remain on the same paths as indicated on the graph.