In this type of equations, the expressions which are involved in the formation of the equation are made up of one variable. I.e. the highest power of the variables used in the equations is 1. The solution to this linear equation can be any rational number. This equation may consist expressions which are linear on both sides of the equal to sign.

Just like numbers, we can also transpose the variables from one side of the equation to the other side. The simplification of the equations which was formed by expressions and this can be done by bringing the equation into a linear form by equating the expression by multiplication using suitable techniques. Utilization of linear equation can be seen in diverse scenarios such as problems on numbers, perimeter, ages, currency, and even algebra has linear equations applications.

## Solving Linear Equations

### Performing Mathematical Operations on Equations

When we are doing mathematical operations on a linear equation, we should do it on both sides of the equality otherwise the equality won’t hold true.

Suppose,Â 4*x* + 3 = 3*x* +7 is a linear equation. If we want to subtract 3 from the given equation, then we do it on both sides of the equality, so that the equality holds true.

4x+3âˆ’3=3x+7âˆ’3

â‡’4x=3x+4

Similarly, if we want to multiply or divide the equation, we multiply or divide all the terms on the left side of the equality and to the right side of the equality by the given number.

Note: we can not multiply or divide the equation by 0.

### Solving Equations with Linear ExpressionÂ on one side and numbers on the other Side

Suppose we have to find the solution of 2xâˆ’3=7, where the linear expression is on the left-hand side, and numbers on the right-hand side.

Step 1: Transpose all the constant terms from the left-hand side to the right-hand side.

2x=7+3=10â‡’2x=10

Step 2: Divide both sides of the equation by the coefficient of the variable.

In the above equation 2*x* is on the left-hand side. The coefficient of 2*x* is 2.

On dividing the equation by two, We get:

\(\frac{1}{2}\) Ã—Â 2xÂ = \(\frac{1}{2}\)Â Ã—Â 10

â‡’x=\(\frac{10}{2}\)Â =Â 5, Which is the required solution.

#### For More Information On Solving An Equation, Watch The Below Video.

### Solving Equations with variables on both sides

Suppose we have to solve 3*x* – 3 = *x* + 2. In this equation, there are variables on both sides of the equation.

Step 1: Transpose all the terms with a variable from the right-hand side to the left-hand side of the equation and all the constants from the left-hand side to the right-hand side of the equation.

3xâˆ’x=2+3

â‡’2x=5

Step 2: Divide both sides of the equation by the coefficient of the variable.

\(\frac{1}{2}\)Ã—2x=\(\frac{1}{2}\)Ã—5

â‡’x=\(\frac{5}{2}\)

To know more about Solving Linear Equations, visit here.

### Applications (Word Problems)

Sum of two numbers is 74. One of the numbers is 10 more than the other. What are the numbers?

Let one of the numbers be *x.*

Then the other number is *x* + 10.

Given that the sum of the two numbers is 74.

So, x+(x+10)=74

â‡’2x+10=74

â‡’2x=74âˆ’10=64

â‡’x=\(\frac{64}{2}\)=32

One of the number is 32 and the other number is 42.

## Questions

### Equations Reducible to the Linear Form

Solve: \(\frac{x+1}{2x+3}\)=\(\frac{3}{8}\)

Multiplying both sides with 2*xÂ *+ 3

â‡’\(\frac{x+1}{2x+3}\)Ã—(2x+3)=\(\frac{3}{8}\)Ã—(2x+3)

â‡’x+1=\(\frac{3(2x+3)}{8}\)

Multiplying both sides with 8

â‡’8(x+1)=3(2x+3)

â‡’8x+8=6x+9

â‡’8x=6x+9âˆ’8

â‡’8x=6x+1

â‡’8xâˆ’6x=1

â‡’x=\(\frac{1}{2}\)

### Reducing Equations to Simpler Form

Simplify the equation \(\frac{6x+1}{3}\)+1=\(\frac{x-3}{6}\).

\(\frac{6x+1}{3}\)+1=\(\frac{x-3}{6}\)

â‡’\(\frac{6(6x+1)}{3}\)+6Ã—1=\(\frac{6(x-3)}{6}\)Â Â (Multiplying both sides by 6)

â‡’2(6x+1)+6=(xâˆ’3)

â‡’12x+2+6=xâˆ’3Â Â Â Â Â Â Â Â Â Â Â (opening the brackets)

â‡’12x+8=xâˆ’3

â‡’12xâˆ’x+8=âˆ’3

â‡’11x+8=âˆ’3

â‡’11x=âˆ’3âˆ’8

â‡’11x=âˆ’11

â‡’x=âˆ’1 (required solution)

LHS: \(\frac{6(-1)+1}{3}\)+1=\(\frac{-6+1}{3}\)+1=\(\frac{-5}{3}\)+\(\frac{3}{3}\)=\(\frac{-2}{3}\)

RHS: \(\frac{(-1)-3}{6}\)Â =\(\frac{-4}{6}\)Â =\(\frac{-2}{3}\) LHS = RHS

## Introduction to Linear Equations in One Variable

### Variables and Constants

A **constant** is a value or number that never changes in an expression and it’s constantly the same.

A **variable** is a letter representing some unknown value. Its value is not fixed, it can take any value. On the other hand, the value of a constant is fixed.

For example, in the expression 4x+7, 4 and 7 are the constants and *x* is a variable.

To know more about Variables and Constants, visit here.

### Algebraic Equation

The statement of **equality **of two **algebraic expressions** is an **algebraic equation**. It is of the form P=Q, where P and Q are algebraic expressions.

6*x* + 5 and 5*x* + 3Â Â are algebraic expressions. On equating the algebraic expressions we get an algebraic equation.

6*x* + 5 = 5*x* + 3Â is an algebraic equation.

To know more about Algebraic Equation, visit here.

### Linear Equations in One Variable

A **linear equation** is an algebraic equation in which each term is either a **constant** or the product of a **constant **and a **single variable**, where the highest power of the variable is **one**.

If the linear equation has only a **single variable** then it is called a **linear equation** in one variable.

For example, 7*x* + 4 = 5*x* + 8 is a linear equation in one variable.

To know more about Linear Equations in One Variable, visit here.

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