You're asking in two variables, yes? Then I'll give out a problem...
PROBLEM: 1) x - 2y = 3; 2) 3x + y = 1
SOLUTION: It depends upon you. You can use GRAPHING, where you would list the points which are the possible solutions to the problem - at least two points would do.
You could also use SUBSTITUTION, where you have to substitute a variable to solve the other variable.
You could also use ELIMINATION, where you could cross out a variable to solve the other one.
EQUATE:
GRAPHING:
1) Points: (3,0) and (1, -1)
2) Points: (0,1) and (1,0)
SUBSTITUTION:
[tex] \left \{ {{x - 2y = 3} \atop {3x + y = 1}} \right. \\
y = -3x + 1 \\
x -2(-3x + 1) = 3 \\
x +6x - 2 = 3 \\
7x = 3 + 2 \\
7x = 5 \\
x = \frac{5}{7} \\
\\
3 ( \frac{5}{7}) + y = 1 \\
\frac{15}{7} + y = 1 \\
y = - \frac{15}{7} + 1 \\
y = - \frac{8}{7}
[/tex]
ELIMINATION:
[tex] \left \{ {x - 2y =3} \atop {3x + y =1}} \right. \\
7x = 5 \\
x = \frac{5}{7} \\
\\
\frac{5}{7} - 2y = 3 \\
-2y = 3 - \frac{5}{7} \\
-2y = - \frac{16}{7} \\
y = \frac{7}{8}
[/tex]
