Answer:
The solution to the given systems of linear equation is {3, -1} lies in IV Quadrant.
Step-by-step explanation:
Given:
x-3y = 6 ........( 1 )
x+y=2 ........( 2 )
To find :
x = ?
y = ?
Solution:
Let solve the above equation to get the solutions.
Eliminate X by subtracting the two equations equation 1 minus equation 2
![(x-3y) - (x+y) = 6 - 2\\-4y = 4\\\therefore y=-1](https://tex.z-dn.net/?f=%28x-3y%29%20-%20%28x%2By%29%20%3D%206%20-%202%5C%5C-4y%20%3D%204%5C%5C%5Ctherefore%20y%3D-1)
Now substitute for y = -1 in equation 2 we get
![x+(-1)=2\\\therefore x = 2+1\\\therefore x = 3](https://tex.z-dn.net/?f=x%2B%28-1%29%3D2%5C%5C%5Ctherefore%20x%20%3D%202%2B1%5C%5C%5Ctherefore%20x%20%3D%203)
Therefore, solution set is {3, -1}
In
I Quadrant {x,y} = {+,+}......both the coordinates are positive.
II Quadrant {x,y} = {-,+}.....X coordinate is negative and Y coordinate is positive.
III Quadrant {x,y} = {-,-}......both the coordinates are negative.
IV Quadrant {x,y} = {+,-}.....X coordinate is positive and Y coordinate is negative. {3,-1}
Point A on the graph is the solution.