Which of the following equations could be the result of using the comparison method to solve the system shown?
x - 4y - 1 = 0
x + 5y - 4 = 0
A 4y + 1 = 4 - 5y
B -4y + 1 = 4 - 5y
C 4y + 1 = 5y - 4
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(x-4y-1)=0 x+5y-4=0 by solving the above both equations ie; equation 1-equation2 we get x-4y-1=x+5y-4 since RHS1= RHS2 we has to equate LHS1=LHS2 -4y -1= =5y-4 taking - common on both sides of equations we get 4y+1=4-5y hence option A is correct.
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