# Linear Equations in One Variable

What is an equation? An equation is a statement in which two algebraic expressions are equal. Linear Equations of one variable is the equation which involves only one variable with degree one. General form is ax + b = 0, where a is not equal to 0.
Let us discuss the situation that " A farmer traveled a distance of 61 km in 9 hrs. He traveled partly on foot at an average of 4 km / hr and partly on bicycle at an average speed of 9 km/hr. Find the distance traveled by him on foot ". If we look at this situation it seems to be complicated. But we can easily solve this type of situations framing linear equations in few seconds.

## Definition

An equation statement involving one variable with non zero coefficient and degree one is called linear equation in one variable. The general form of linear equation in one variable is usually written as ax + b = 0, where a and b are real numbers and "a" is not equal to 0.
Here, Degree of x = 1
For example: 3x + 10 = 0, x = 1, 3 + y = 5 are all linear equations with variable one.
The value of the variable which when substituted for the variable in the equation, makes its two sides equal, is called a solution or root of the equation.

## How to Solve

Rules for solving linear equations:
The equality of a linear equation is not changed,
Rule 1 : When the same number is added to both sides of the equation.
Rule 2:  When the same number is subtracted from both sides of the equation.
Rule 3 : When both sides of the equation are multiplied by the same non-zero numbers.
Rule 4:  When both sides of the equation are divided by the same non-zero number.

## Examples

There are different techniques to solve such problem. Some of the examples are given below for better understanding.

### Solving Linear Equations with Variables on One Side

Example 1:
Solve ( 5x - 7) = 8
Solution: We have 5x - 7 = 8
=>5x = 8 + 7 = 15 [ when we transpose -7 to the other side it becomes + 7 ]
=> x = 15 / 5 = 3, Hence the solution

### Solving Linear Equations with Variables and Fractions on both Sides

Example 2: Solve  $\frac{6x - 7}{3x + 1}\: = \: \frac{2x+1}{x+5}$
Solution: We have  $\frac{6x - 7}{3x + 1}\: = \: \frac{2x+1}{x+5}$
=>  ( 6x - 7 ) ( x + 5) = ( 2x + 1) ( 3x + 1)  [ by cross multiplication ]
=>  6x2 +23 x - 35 = 6 x2 + 5x + 1  [ Using FOIL ]
=>  6 x2 - 6 x2 + 23 x - 5 x - 35  = 6x2 - 6 x2 + 5x + 1 - 5 x [adding -6x2 - 5x on both sides ]
=> 18 x - 35 = 1
=>18 x - 35 + 35  = 1 + 35  [ adding 35 on both sides ]
=> 18 x = 36        [ dividing both sides by 18 ]
=>  x = 36/18 = 2
The solution to the given equation is x = 2.

### Solving Linear Equations with Variables on both Sides

Example 3: Verify that x = 2 is a solution of the linear equation 2x + 7 = 13 - x
Solution: Substituting x = 2 in 2x + 7 = 13 - x, we get
2 ( 2) + 7 = 13 - 2
=>4 + 7 = 11
=> 11 = 11
=> x = 2 is the solution of the given equation.

## Word Problems

Method of solving liner equations in one variable word problems we should follow the following steps.
Step 1: Read the problem carefully and find out what is given and what is unknown.
Step 2: Represent the unknown quantity by any variable, say x.
Step 3: Frame an equation in x, as per the conditions given in the problem.
Step 4 : Solve for x.
Example : A chemist has one solution containing 50% acid and a second one containing 25% acid. How much of each should be mixed  to make 10 litres of a Rs. 40 acid solution?
Solution: It is given that the total number of litres of the mixture = 10 L
Let us fill up the following table.
 Solution 1 Solution 2 Concentration 50% 25% No. of litres x ( 10 - x)
Quantity of acid in 50% of the solution = 50% x = 0.5 x
Quantity of acid in 25% of the solution = 25% of ( 10 - x) = 0.25 ( 10 - x)
Quantity of acid in 40% of the mixture  = 40 % of 10 = 0.40 $\times$ 10 = 4
Hence we have the equation,
0.5 x + 0.25 ( 10 - x ) = 4
=> 0.5 x + 0.25(10) - 0.25 x = 4
=> 0. 5 x - 0.25x + 2.5 = 4
=> 0.25 x= 1.5
=> x = 1.5/0.25 = 6 litres
Therefore the scientist should mix 6 litres of 50% solution and 4 litres of 25 % solution.

## Practice Problems

Practice below problems and test your knowledge.
Problem 1:
Verify that x = 3 is not a solution of the equation, 3x - 5 = 4 + x
Problem 2: Find the value of x, 2x + 6 = 0
Problem 3: Solve for x; 5x + 4 = 4x - 10