# Solving Linear Equations

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.

## What are Linear Equations

The equations are said to be linear if the degree of each term is 1. Solution to the linear equations of one or more variable is very much useful in solving for any unknown quantity in any complicated real life problems. This is very much useful in finding the total cost, sales and profit of the multiple products produced in a factory. Linear equations involves, one variable, two variables, three variables and so on. Algebraic equations with degree ‘1’ are usually termed as Linear Equations. Slope intercept form and two point form are examples of general formulas of linear equations.

 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 Solutionsb. No solution Orc. Infinitely many solutions*solution depends on the given equations 3x + 5y = 10

## Steps for Solving Linear Equations

Steps for solving linear equations with one variable:

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.

Cross Multiplication: When $\frac{a}{b}$  =  $\frac{c}{d}$ , then a x d = b x c.
We can use cross multiplication method when we have single fraction on either side of the equation.

Steps for solving linear equations with two variables:
1. Write the equations in the general form i.e. ax + by + c = 0.
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.

## Examples

Example 1: Solve 4 ( x - 3) + 7 = 5( 4 - x) + 20
Solution: We have 4 ( x- 3) + 7 = 5 ( 4 - x) + 20
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

Example 2: Find the solution (s) ( if any ) for the equations  3x + 2y = 10 and 6x + 4y = 20
Solution: Let us solve the two equations by elimination method.

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 ).

## Practice Questions

Solve the following problems:
Question 1. Solve 5x + 27y = -8 and x+y = 10
Question 2. Find the value of x. $\frac{4x+5}{4}$  = $\frac{4x-7}{5}$
Question 3. Solve 3 ( 8x - 5) + 20 = 7 ( 3 - x) + 2( x-3)