Solve \[x + 2y = 5\]; \[y = - 2x - 2\]

Hint:
We will solve the above problem by graphical method. We will substitute different values of \[y\] in both the given equations to get the corresponding value of \[x\]. Then using these points we will draw the required graph. From the graph, we will find the intersection point which will give the solution of the system of linear equations.

Complete step by step solution:
The given equations form a non-homogeneous system of linear equations in two variables.
To solve this system of linear equations, we will adopt the graphical method.
We will find two points on each line and draw their graphs using these points. Then, we will check if the lines meet at any point. If they do, then the system of equations is consistent and has a unique solution.
The given non-homogeneous system of linear equations is:
\[x + 2y = 5\]………………………………….\[\left( 1 \right)\]
\[y = - 2x - 2\]……………………………….\[\left( 2 \right)\]
Now we will consider the equation \[\left( 1 \right)\].
Substituting \[y = 0\] in equation \[\left( 1 \right)\], we get
\[\begin{array}{l}x + 2\left( 0 \right) = 5\\ \Rightarrow x = 5\end{array}\]
So, we get one point as \[A(5,0)\].
Now, substituting \[y = 1\] in equation \[\left( 1 \right)\], we get
\[\begin{array}{l}x + 2\left( 1 \right) = 5\\ \Rightarrow x + 2 = 5\end{array}\]
Subtracting 2 from both sides, we get
\[ \Rightarrow x = 5 - 2 = 3\]
 Here, we get \[x = 3\]. Therefore, the next point is \[B(3,1)\].
Now we will consider the equation \[\left( 2 \right)\].
Substituting \[y = 0\] in equation \[\left( 2 \right)\], we get
\[\begin{array}{l}\left( 0 \right) = - 2x - 2\\ \Rightarrow - 2x = 2\end{array}\]
Dividing both sides by \[ - 2\], we get
\[ \Rightarrow x = - \dfrac{2}{2} = - 1\]
Here, we get \[x = - 1\]. So, we get the point as \[C( - 1,0)\].
Substituting \[y = 2\] in equation \[\left( 2 \right)\], we get
\[\begin{array}{l}2 = - 2x - 2\\ \Rightarrow - 2x = 4\end{array}\]
Dividing both sides by \[ - 2\], we get
\[ \Rightarrow x = - \dfrac{4}{2} = - 2\]
As \[x = - 2\], so the next point is \[D( - 2,2)\].
Let us plot the points \[A(5,0)\], \[B(3,1)\], \[C( - 1,0)\], and \[D( - 2,2)\] on the graph.

From the above graph, we observe that the lines meet at the point \[E( - 3,4)\]. Therefore, the solution of the system of linear equations is \[x = - 3,y = 4\].

Note:
We can also solve the above problem by the substitution method.
From equation \[\left( 1 \right)\], we have \[x + 2y = 5\]. Let us take \[2y\] to the RHS. We get
\[x = 5 - 2y\]………………………\[\left( 3 \right)\]
Let us substitute this equation in equation \[\left( 2 \right)\]. We get
\[y = - 2(5 - 2y) - 2\]
Multiplying \[ - 2\] with \[5 - 2y\], we get
\[ \Rightarrow y = - 10 + 4y - 2\]
Rearranging like terms on either side of the equation, we have
\[\begin{array}{l} \Rightarrow 4y - y = 10 + 2\\ \Rightarrow 3y = 12\end{array}\]
Dividing both sides by 3, we get
\[ \Rightarrow y = 4\]
Substituting \[y = 4\] in equation \[\left( 3 \right)\], we have
\[x = 5 - 2(4) = - 3\]
Therefore, the solution of the given equations is \[x = - 3,y = 4\].