Question #be783

2 Answers
Mar 15, 2017

There are possibilities for solutions for B, and C, but not A.

Explanation:

#A.# If one equation is a multiple of the other then the slope of a line defined by the first equation will be equal to that of the second equation with the same y-intercept at #(2,0)#. but this will be the same line in both cases, so it cannot be a solution to a system of equations.

#B.# It is possible for no solutions if the computation of the equations results in a vertical line parallel to the y-axis. This solution is undefined.

#C.# Since we have been given only the y-intercept, then it is possible for infinitely many solutions because there can be infinitely many slopes of a straight line intersecting the x-axis at #(2,0)#.

Mar 15, 2017

A. Yes #-# B. No #-# C. Yes

Explanation:

I am assuming that #y#-intercept is at #(0,2)# and not #(2,0)#.

A. If both equations in a system of linear equations have #y#-intercepts at #(2,0)#, there is a possibility of one solution, if slopes of the two equations are different. Example give below with equations #y=x+2# and #y=-3x+2#.

graph{(y-x-2)(y+3x-2)=0 [-10, 10, -5, 5]}

#B.# It is not possible to have no solutions as at least #(2,0)#, is common to both equations.

#C.# If both equations in a system of linear equations have #y#-intercepts at #(2,0)# and slopes are also equal, the two equations (say #y=x+2# and #3y=3x+6#) are identical and all the points on one are also n other equation and hence infinite solutions including #(0,2)#.