Pair of linear equations, mainly we work with a set of two equations with two variables which can be expressed as ax + by + c = 0, where a, b and c are real numbers and a and b are not zero. The value of x and y is a pair of solution for such equations. If we substitute the values back to the given equations, it satisfy the system. We can represent linear equations in ways ways called algebraically and geometrically. Solving linear equations in one variable involves basic algebraic and equality properties. We shall discuss the more examples of linear equations in two variables in this section.

When we consider the linear equations in two variables, their solutions will satisfy any one of the following conditions.

1. Only one solution (unique solutions)

2. No solution

3. Infinitely many solutions satisfying the given equation.

To know more about the behavior of solution set, click here.

To know more about the behavior of solution set, click here.

Example 1

3x + 3y = 11Â Â Â Â Â --------(1)

6x + 6y = 13Â Â Â --------Â (2)

Multiplying (1) by number 2, we get,Â

2 ( 3x+ 3y ) = 2(11)

6x + 6y = 22Â -------------------(3)

Subtracting , (2) fromÂ (3), we get,Â Â

Â

0 = 9, which is not true.

**Example 2: **Solve -2x + y - 2 = 0 and x - y - 1 = 0

**Solution**: -2x + y = 2 ...equation (1)

**Solution**: Consider the system of given equations

A system of linear equations contains two or more equations. The solution of such a system is the ordered pair that is a solution to both equations.

x - y = -2 ...equation (2)

Solve equation (2) for x and substitute the value in equation (1)

Â x = y - 2

Again, -2 ( y - 2) + y = 2

-2y + 4 + y = 2

- y + 4 = 2

-y = -2

or y = 2

Put value of y in equation (2) to get the value of x.

x - y = -2

x - 2 = -2

x = -2 + 2

or x = -

So, the solution set for the given equations is (0, 2).

**Example 3**: By using the method of graphing solve:

x - y = 6

-2x + 2y = 1

x - y = 6

-2x + 2y = 1

Now put y = 0, 1, 2Â inÂ x - y = 6 (isolate x) we get

Â yÂ | Â xÂ |

Â 0 | Â 6 |

Â 1 | Â 7 |

Â 2 | Â 8 |

Consider the second equation and isolate y again

-2x + 2y = 1

y = $\frac{1}{2}$ + x

Now put x = 0, 1, 2Â inÂ -2x + 2y = 1

Â xÂ | Â yÂ |

Â 0 | Â 0.5 |

Â 1 | Â 1.5 |

Â 2 | Â 2.5 |

From the graph, we can see that both the lines are parallel to each other. Since lines are parallel to each other they have different intercepts and no chance to meet any point. Hence there will be no solution to the given system of equations.

x - 2y = 8 and x - 3y = 10