How do you solve \[2x + y = 5\] and \[x - y = 1?\]

Hint: In this question we have given a linear equation in two variables for which we need to find the value of both \[x\] and \[y\] using additional property.

Complete step-by-step solution:
In this case we have a perfect scenario because the \[y\] terms are additive inverses. When the two equations are added, the \[y\] terms will add to 0.
\[2x + y = 5\;\] …. (A)
\[x - y = 1\;\] …. (B)
On adding equations, A and B we will get the following equation:
A + B: \[3x = 6\] …. (C)
So, the final value of \[x\] will be 2.
\[x = 2\]
Substituting \[x = 2\] in equation (B) we will get the value of \[y\] as follows:
\[2 - y = 1\]
\[2 - 1 = y\]
\[y = 1\]

Thus, we get final values as \[x = 2\;\;\;and{\text{ }}y = 1\].

Note: Well, it would be easier when we have few equations and variables. If we have 2 equations and 2 variables it is ok; when we get to 3 equations and 3 variables it becomes more complicated, it is still possible, but we have more work to do. The number of substitutions increases together with the possibility to make mistakes.
In general, in linear equations, we need n equations if we have n variables. Let us have 3 equations and 3 variables \[x\] , \[y\] and \[z\] . Now pick up an equation with \[x\] and segregate it to say \[x\] in terms of \[y\] , \[z\] . When we put this value of \[x\] in two other equations we get two equations in \[y\] and \[z\] .
We can now find \[y\] in terms of \[z\] say using the second equation and when we put in the third equation, we get the value of \[z\] . Once \[z\] is known, it is easy to find \[y\] and then \[x\] .
More than 3 equations and 3 variables and it gets almost impossible and other methods would be better like the slope intercept form of equation or the distributive properties of linear equations.