   Introduction to Linear Equations

In the earlier classes, you have come across several algebraic expressions and equations. Some examples of expressions we have so far worked with are:

5x, 2x – 3, 3x + y, 2xy + 5, xyz + x + y + z, x2 + 1, y + y2

Some examples of equations are: You would remember that equations use the equality (=) sign; it is missing in expressions.

Of these given expressions, many have more than one variable. For example, 2xy + 5 has two variables. We however, restrict to expressions with only one variable when we form equations. Moreover, the expressions we use to form equations are linear. This means that the highest power of the variable appearing in the expression is 1.

These are linear expressions: These are not linear expressions: (since highest power of variable > 1)

Here we will deal with equations with linear expressions in one variable only. Such equations are known as linear equations in one variable. The simple equations which you studied in the earlier classes were all of this type.

Let us briefly revise what we know:

•   An algebraic equation is an equality involving variables. It has an equality sign.The expression on the left of the equality sign is the Left Hand Side (LHS). The expression on the right of the equality sign is the Right Hand Side (RHS). • In an equation the values of the expressions on the LHS and RHS are equal. This happens to be true only for certain values of the variable. These values are the solutions of the equation. • How to find the solution of an equation?

We assume that the two sides of the equation are balanced. We perform the same mathematical operations on both
sides of the equation, so that the balance is not disturbed. A few such steps give the solution.

## Solving Equations which have Linear Expressions on one Side and Numbers on the other Side

Let us recall the technique of solving equations with some examples. Observe the solutions; they can be any rational number.

### Example 1

Find the solution of 2x – 3 = 7

### Solution

#### Step 1 Add 3 to both sides. (The balance is not disturbed)

Or #### Step 2 Next divide both sides by 2. Or (required solution)

Slove  2y+9 =4

### Solution

Transposing 9 ro RHS Or Dividing both sides by 2, (solution)

To check the answer: LHS  = (as required)

Do you notice that the solution is a rational number? In Class VII, the equations

we solved did not have such solutions.

### Example 3.

Solve ### Solution

Transposing to the RHS, we get Or Multiply both sides by 3, Or (solution)

Check: LHS (as required)

Do you now see that the coefficient of a variable in an equation need not be an integer?

### Example 4

Solve ### Solution

We Have or (transposing to R H S)

or or (dividing both sides by – 7)

or or (solution)

Check :  LHS (as required)

Cite this Simulator:

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