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Linear Equations in One Variable
• Last Updated : 27 Oct, 2020

Linear equation is an algebraic equation that is a representation of the straight line. Linear equations are composed of variables and constants. These equations are of first-order, that is, the highest power of any of the involved variables i.e. 1. It can also be considered as a polynomial of degree 1. Linear equations containing only one variable are called homogeneous equations. The corresponding variable is called the homogeneous variables.
For instance,

• x + 2y = 3 is a linear equation in two variables.
• x + y + z = 8 is in three variables.
• x + y2 = 1 is not a linear equation because the highest power of y is 2.

### Standard Form of Linear Equation in One Variable

A linear equation in one variable can be expressed in the form of ax+b = 0, where x is the variable and a and b are the constants involved. These constants (a and b) should be non-zero real numbers. These types of equations have only one possible solution for the value of the variable.

### Steps for Solving Linear Equations in One Variable

The following steps are performed to solve linear equations in one variable:

Step 1: In case the integers a and b are fractional numbers, LCM has to be taken to clear them.

Step 2: The constants are taken to the right side of the equation.

Step 3: All the terms involving the variable are isolated to the left hand side of the equation, to evaluate the value of the variable.

Step 4: The solution is verified.

Examples of Linear Equation in One Variable:

The following are some of the examples of linear equations in one variable:

21x = 55

5/4 + 1/2 x = 1

9y – 4 = 8

5/4 (z-3) = 0

If we analyze these examples, we have only one variable, and the highest power of this variable in any term is 1.
This algebraic equation can be evaluated by taking all the terms involving the variables on the left-hand side (L.H.S) and constants on the right side (R.H.S), to solve for the corresponding variable value.

### Sample Problems on Linear Equations

Example 1: Solve for y, 8y – 4 = 0

Solution:

Solving for value of y,

Adding 4 to both sides of the equation ,

⇒ 8y -4 + 4 = 4

⇒ 8y = 4

Dividing both sides of equation by 8

⇒ y = 4/8

Simplifying the equation ,

⇒ y = 1/2

Example 2: Solve the equation in x, 3x +10 = 55

Solution:

Taking constants to RHS,

3x = 45

⇒ x = 15

Example 3: Solve the equation in x, 4/5x -5 = 15

Solution:

Taking constants to RHS,

4/5x = 20

⇒x = 100/4

⇒x = 25

### Sample Word Problems

Problem 1: There are two numbers, one equal to 5/4 and other equal to 1/2 times some number x. The sum of these two numbers is 1. Find x.

Solution:

Since, the sum of both the numbers is 1, we have

5/4 + 1/2 x = 1

Transposing all the constants to R.H.S of the equation (transposing is an operation of shifting the operands to the other side by reversing the sign of the operand upon taking to other side)

⇒ 1/2x = 1 – 5/4

⇒  1/2x = -1/4

⇒ 2 * (1/2x) = 2 * -1/4

Multiplying both sides of the equation with 2 we get ,

⇒ x = -1/2

Problem 2: The height of the rectangle if 4m less than the base of the rectangle. The perimeter of the rectangle is 32 m. Find the length and height of the rectangle.

Solution:

Perimeter P of the rectangle = 32 m

Let base of the rectangle be x metres.

Therefore, the height of rectangle is x-4 m.

Perimeter of the rectangle is equal to sum of all sides.

Since, opposite sides of the rectangle are equal we have,

2 (x) + 2(x-4) = 32

⇒ 2x + 2x – 8 = 32

⇒ 4x – 8 = 32

Adding 8 on both sides of the equation,

⇒ 4x = 40

⇒ x = 10

Therefore, base of rectangle = 10 m.

And height = 6 m

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