Question #be783

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 $\left(2 , 0\right)$. 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 $\left(2 , 0\right)$.

Mar 15, 2017

A. Yes $-$ B. No $-$ C. Yes

Explanation:

I am assuming that $y$-intercept is at $\left(0 , 2\right)$ and not $\left(2 , 0\right)$.

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

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

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

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