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Just to add to the discussion, here is the solution using the substitution method.
Solve x+y-2=0 for x. x=-y+2
Substitute -y+2 for the x in the equation 2x-3y+1=0.
Distribute the 2: -2y+4-3y+1=0
Substitute the value of 1 into either equation for the y value. I will use x+y-2=0.
The solution is the point (1,1).
The system can be solved either using elimination method, or substitution method.
We'll choose elimination method to remove y variable. For this reason we'll multiply the 2nd equation by 3:
3x + 3y - 6 = 0
We'll add this equation to the 1st equation of the system:
3x + 3y - 6 + 2x - 3y + 1 = 0
We'll eliminate variable y:
5x - 5 = 0
We'll isolate 5x to the left side, adding 5 both sides:
5x = 5
x = 1
We'll replace x by 1 within the second equation, to determine y:
1 + y - 2 = 0
y = 2 - 1
y = 1
The solution of the system of independent equations is (1 ; 1).
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