Question

The figure shows the graph of two parallel lines. Which of the following pairs of equations might have such a graph?

$$\begin{gathered} \left(a\right)x-2y=3 \\ x+2y=7 \\ \left(b\right)x+y=2 \\ x+y=-1 \\ \left(c\right)x-y=-2 \\ x-y=1 \\ \left(d\right)x-y=-2 \\ 2x-2y=-4 \\ \left(e\right)x+2y=2 \\ x+2y=-1 \end{gathered}$$

Short answer

Equation$$\begin{gathered} \left(c\right)x-y=-2 \\ x-y=1 \end{gathered}$$might have such graph

Step-by-step solution

Given information 

We are given a graph and set of equations

Explanation for correct options

In the graph both the lines are parallel and hence will have equal slopes

Consider the equations

$$\begin{gathered} \left(c\right)x-y=-2 \\ x-y=1 \end{gathered}$$

On further simplifying the equation we get

$$\begin{gathered} y=x+2 \\ y=x-1 \end{gathered}$$

Both the equation has same slope might have such graph

Explanation for other explanations

Consider the equations

$$\begin{gathered} \left(a\right)x-2y=3 \\ x+2y=7 \\ \left(b\right)x+y=2 \\ x+y=-1 \\ \left(d\right)x-y=-2 \\ 2x-2y=-4 \\ \left(e\right)x+2y=2 \\ x+2y=-1 \end{gathered}$$

On simplifying the equation we get

$$\begin{gathered} \left(a\right)y=\frac{x}{2}-\frac{3}{2} \\ y=-\frac{1}{2}x+\frac{7}{2} \\ \left(b\right)y=-x+2 \\ y=-x-1 \\ \left(d\right)y=x+2 \\ y=x+2 \\ \left(e\right)y=-\frac{1}{2}x+1 \\ y=-\frac{1}{2}x+1 \end{gathered}$$

Option a) both the lines have different slopes hence the lines cannot be parallel

option d) both the lines are identical hence cannot describe the given figure

Option b) and option e) are not consistent with the given graph.

Conclusion

Equation$$\begin{gathered} \left(c\right)x-y=-2 \\ x-y=1 \end{gathered}$$might have such graph.