Which quadrant does the solution of the system fail?
x+y=4
2x-y=2
Please help I don't know what to do. I don't what the answer, just want to be guided to it! Thank you (:
2 x + 2 y = 8
2 x - 1 y = 2
------------- subtract
0 x + 3 y = 6
y = 2
so
x = 2
x and y are both positive, first quadrant, upper right
Thank you so much.. It means a lot!!!!!!! :D
Is this even right
To determine the quadrant where the solution of the system fails, we need to solve the system of equations first. Here's a step-by-step guide on how to do that:
Step 1: Rearrange Equation 1
x + y = 4
Step 2: Rearrange Equation 2
2x - y = 2
Step 3: Solve for y in Equation 1
y = 4 - x
Step 4: Substitute the value of y in Equation 2
2x - (4 - x) = 2
Step 5: Simplify Equation 2
2x - 4 + x = 2
3x - 4 = 2
Step 6: Solve for x in Equation 2
3x = 6
x = 6/3
x = 2
Step 7: Substitute the value of x in Equation 1
y = 4 - 2
y = 2
So the solution to the system of equations is x = 2 and y = 2.
Now, to determine the quadrant where the solution fails, we can graph the system of equations on a coordinate plane.
Plot the points (2, 2) which is the solution to the system. Now, observe in which quadrant the point lies.
Since the point (2, 2) lies in the positive quadrant (Quadrant I), the solution does not fail in any particular quadrant.