Solving equations of degree 1 : LINEAR

Degree 1 equations or linear equations are equations in one or more variable such that degree of each variable is one. Let us consider some special cases of linear equations :

One variable
Such equations will always have a solution. General form is \(ax=b\) and solution is \(x=(b/a)\)

One equation in Two variables
This is not enough to determine x and y uniquely. There can be infinitely many solutions.

Two equations in Two variables
If you have a linear equation in 2 variables, you need at least 2 equations to solve for both variables. The general form is :
\(ax + by = c\)
\(dx + ey = f\)

If \((a/d) = (b/e) = (c/f)\) then there are infinite solutions. Any point satisfying one equation will always satisfy the second

If \((a/d) = (b/e) \neq (c/f)\) then there is no such x and y which will satisfy both equations. No solution

In all other cases, solving the equations is straight forward, multiply eq (2) by a/d and subtract from (1).

More than two equations in Two variables
Pick any 2 equations and try to solve them :
Case 1 : No solution --> Then there is no solution for bigger set
Case 2 : Unique solution --> Substitute in other equations to see if the solution works for all others
Case 3 : Infinite solutions --> Out of the 2 equations you picked, replace any one with an un-picked equation and repeat.

More than 2 variables
This is not a case that will be encountered often on the GMAT. But in general for n variables you will need at least n equations to get a unique solution. Sometimes you can assign unique values to a subset of variables using less than n equations using a small trick. For example consider the equations :
\(x + 2y + 5z = 20\)
\(x + 4y + 10z = 40\)
In this case you can treat \(2y+z\) as a single variable to get :
\(x + (2y+5z) = 20\)
\(x + 2*(2y+5z) = 40\)
These can be solved to get x=0 and 2y+5z=20

There is a common misconception that you need n equations to solve n variables. This is not true.