Question

Show graphically that each one of the following systems of equations is inconsistent (i.e. has no solution):
$2y-x=\dfrac{9}{2}$
$6y-3x=21$

Answer 1

Hint: First draw the graph of the two given equations. To draw the graph of a straight line, we need at least two points. So, choose one of the equations and substitute x = 0, determine y, then substitute y = 0, determine x. Now, apply the same process for the second equation. Plot the graph of the two equations using the points obtained. Now, if the graphs of these equations are parallel but not coincident to each other then the system of equations is inconsistent. In case the graphs are coincident or overlapped then there will be infinitely many solutions.

Complete step-by-step answer:

Let us assume the two equations as:
$\begin{align}
  & 2y-x=\dfrac{9}{2}....................(i) \\
 & 6y-3x=21................(ii) \\
\end{align}$
Considering equation (i),
$2y-x=\dfrac{9}{2}$
Substituting x = 0, we get,
$\begin{align}
  & 2y=\dfrac{9}{2} \\
 & \Rightarrow y=\dfrac{9}{4} \\
\end{align}$
Substituting y = 0, we get,
$x=\dfrac{-9}{2}$
Therefore, the two points are: $\text{ }A\left( \dfrac{-9}{2},0 \right)\text{ and }B\left( 0,\dfrac{9}{4} \right)$.
Considering equation (ii),
$6y-3x=21$
Substituting x = 0, we get,
$\begin{align}
  & 6y=21 \\
 & \Rightarrow y=\dfrac{7}{2} \\
\end{align}$
Substituting y = 0, we get,
$\begin{align}
  & -3x=21 \\
 & \Rightarrow x=-7 \\
\end{align}$
Therefore, the two points are: $C\left( -7,0 \right)\text{ and }D\left( 0,\dfrac{7}{2} \right)$.
Therefore, the graph of the two functions can be plotted as:

Clearly, we can see that the two lines are parallel to each other and do not intersect at any point. Hence, the system of equations is inconsistent (i.e. has no solution).

Note: There is another method to check the consistency or inconsistency of a system of equations without the use of graphs. To apply this method, write the two equations in the form: ${{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0$ and ${{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0$ respectively. Now, consider the ratios: $\dfrac{{{a}_{1}}}{{{a}_{2}}},\dfrac{{{b}_{1}}}{{{b}_{2}}}\text{ and }\dfrac{{{c}_{1}}}{{{c}_{2}}}$. From here, three cases can arise:
Case (i): If $\dfrac{{{a}_{1}}}{{{a}_{2}}}\ne \dfrac{{{b}_{1}}}{{{b}_{2}}}$ then the system has unique solution.
Case (ii): If $\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}\ne \dfrac{{{c}_{1}}}{{{c}_{2}}}$ then the system has no solution.
Case (iii): If $\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{{{c}_{1}}}{{{c}_{2}}}$ then the system has infinitely many solutions.
As we can see that the above question has a situation of case (ii). Therefore, the system of equations has no solution.