Question

Question #be783

Answer 1

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)`.

Answer 2

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)`.