The equations are those where the given expression is equated to a constant or an expression. An equation will involve one or more variables. A linear equation is one of the most important topics in algebra that exists in mostly all the classes from high school to university level. In order to learn about mathematics advanced topics, one should be master in all the algebra concepts. We have come up with the meaningful strategy to solve linear equations to help students to overcome difficulties in solving linear equation problems. Let us discuss about the steps to solve linear equations with one variable and two variables.

Type |
General Form |
Number of Solutions | Examples |

Linear equations in one variable | ax + b = 0 |
Only one solution i.e. x = -b/a | 2x - 3 = 0 |

Linear equations in two variables | System of two equations $a_1$ x + $b_1$y = $c_1$ and $a_2$ x + $b_2$ y = $c_2$ |
a. One Solutions b. No solution Or c. Infinitely many solutions *solution depends on the given equations | 3x + 5y = 10 |

1. Apply FOIL or BODMAS and simplify the expressions by combining the like terms on both sides of the equations. If the given linear equation is not of the form px + q = 0.

2. Add the additive inverse of the constant term on both sides of the equation so that the constant term is eliminated from the left side of the equation.

3. To find the value of the variable. (i.e), multiply both sides by the multiplicative inverse of the coefficient of x and simplify to get the value of x.

We can use cross multiplication method when we have single fraction on either side of the equation.

2. Use any of the methods to solve both the equations: Substitution Method, Elimination Method, Cross multiplication, Graphical Method or Matrix Method.

3. Get your answer and verify it.

Applying Distributive property,4x - 12 + 7 = 20 - 5x + 20

=> 4x - 5 = 40 - 5x

[ combining the like terms on the left side and right side of the equation]

=> 4x + 5x - 5 = 40 - 5x + 5x [ adding 5x on both sides ]

=> 9x -5 = 40

=> 9x - 5 + 5 = 40 + 5 [ adding 5 on both sides ]

=> 9x = 45

=> $\frac{1}{9}$ (9x) = $\frac{1}{9}$ (45) [Multiply both the sides by 1/5]

=> x = 5

Let us first number the equations .

3x + 2y = 10 ------------(1)

6x + 4 y = 20 -----------(2)

Multiplying Equation (1) by 2, we get,

2 ( 3x + 2y ) = 2 ( 10)

6x + 4y = 20 ------------(3)

Subtracting (2) from (3), we get

0 = 0, which is true.

This is true for any values of x.

Since the values of y depends on the values of x, there will be infinite number of values for x and the corresponding values of y.

Hence there will be infinite number of ordered pairs (x,y) common to both the lines ( equations ).