Solving an Equation

Solving an Equation is a way of finding the value of an unknown variable that the equation contains. An equation is said to be balanced if it has an ‘equal to’ sign. Thus, this equation represents that it has two quantities equal on both sides. The two sides of the equation are LHS (Left hand side) and RHS (Right hand side). Class 7 students can easily solve simple linear equations by reading this article here.

For example, x – 4 = 5 is an equation. It represents that x – 4 (LHS) is equal to 5 (RHS). x is an unknown quantity or variable here. Hence, we need to solve this equation here to get the value of x.

Solving an equation means finding the solution of the equation. A solution is the value of an unknown variable that can be used to verify the given equation. Let us learn in this article how to solve an equation with examples.

How to Solve an Equation?

What is an equation? An equation is a condition on a variable such that two expressions in the variable should have equal value. The value of the variable for which the equation is satisfied is called the solution of the equation.

To solve any equation we need to perform arithmetic operations, to separate the variable, such as:

• Adding the same number on both sides
• Subtracting same number on both sides
• Multiplying with the same number on both sides
• Dividing by same number on both the sides

Based on these operations, we can isolate the given variable in the equation and solve any equations. The solution so obtained will justify the equation. Let us solve some examples to understand better.

Solved Examples

Example 1: Solve 3 + x = 4

Solution: Given, the equation is;

3 + x = 4

We can see, on the Left hand side, the variable x is present. Thus, we need to make the variable ‘x’ alone on LHS. Thus, by subtracting the 3 from LHS and RHS we get;

3 + x – 3 = 4 – 3

x = 1

Hence, the solution is x = 1.

Verification:

For x = 1

Taking LHS and proving it equal to RHS.

LHS = 3 + x = 3 + 1 = 4 = RHS

Hence, verified.

Example: 2: Solve x – 9 = 0.

Solution: Given, the equation is;

x – 9 = 0

On the left hand side of the equation, the variable x is present, for which we need to find the solution. Thus, adding 9 on both sides of the equation, we get;

x – 9 + 9 = 0 + 9

x = 9

Hence, the solution is x = 9.

Verification: To verify our solution, we have to put the value obtained for x into the given equation.

Taking LHS and proving it equal to RHS.

LHS = x – 9 = 9 – 9 = 0 = RHS

Hence, verified.

Example 3: Solve 3x = 21.

Solution: Given, the equation is:

3x = 21

Since, x is multiplied by 3 in the left hand side of the equation, thus, we have to perform a division here to remove 3. Thus, dividing both sides by 3, we get;

(3x)/3 = 21/3

x = 7 (Since, 3 × 7 = 21)

Hence, the required solution is x = 7.

Verification: Put x = 7 on the LHS of the given equation.

LHS = 3x = 3 × 7 = 21 = RHS

Thus, verified.

Example 4: Solve the following equation: x/5 = 2.

Solution: Given, the equation is:

x/5 = 2

In this equation, the LHS side has the variable x which is divided by 5. Thus, we need to make the variable alone on the left hand side.

So, multiplying the equation by 5 on both sides, we get;

(x/5) × 5 = 2 × 5

x = 10

Hence, the required solution is x = 10.

Verification:

Put x = 10 in the LHS of the equation x/5 = 2, to get the value equal to RHS.

LHS = x/5 = 10/5 = 2 = RHS

Hence, verified.

Example 5: Solve 2x + 6 = 12 and verify your answer.

Solution: Given ,the equation is:

2x + 6 = 12

Subtract 6 on both sides of the equation.

2x + 6 – 6 = 12 – 6

2x = 6

Now, divide both sides by 2.

(2x/2) = 6/2

x = 3

Hence, the required solution is 6.

Verification:

Taking the LHS, we have

LHS = 2x + 6

Now putting the value of x = 6, we get;

LHS = 2 (3) + 6

= 6 + 6

= 12

= RHS

Hence, verified.

Solving an Equation with More Than One Solution

In your higher classes, you will learn to solve equations with two or more solutions. Let us take an example of a simple equation.

Example: (x – 2) (x – 5) = 0

In this case, the variable x will have two solutions.

X – 2 = 0

X = 2

And

X – 5 = 0

X = 5

Thus, the two solutions are x = 2 and x = 5.

An equation having a degree of variable equal to 2 (variable raised to the power such as x²) is called a quadratic equation. We generally use the quadratic formula to solve such equations. Also, sometimes we can use the factorisation method to solve the equation with two solutions.

Solving Equations With Fractions

If an equation has fractions, then we have to make the fractions as like fractions first and then solve it for variables. To convert fractions into like fractions means we need to make the denominators of the fractions the same, by using the LCM method. Let us solve an example to understand it.

Example: Solve ⅛ x +½ = ¼.

Solution: Given, the equation is:

⅛ x +½ = ¼

As we can see, in the given equation, both left and right hand sides have fractions present. We need to solve the equation for x.

Since, all the three fractions i.e., ⅛, ½ and ¼ are unlike (denominators are not same)

Thus, finding the LCM (Least common multiple) of denominators we have;

LCM (2, 4, 8) = 8

Now multiply both sides of the equation by 8, to get;

8 (⅛ x +½) = 8 (¼)

8.(⅛ x) + 8.(½) = 8.¼

X + 4 = 2

Now, we subtract 4 from both sides, to get;

X + 4 – 4 = 2 – 4

X = -2

Hence, the required solution is x = -2.

Verification:Let us check if the solution obtained is correct or not.

The equation is ⅛ x +½ = ¼.

Put x = -2, on the LHS.

LHS = ⅛ x +½

= ⅛ (-2) + ½

= -¼ +½

=-¼ + 2/4

= (-1+2)/4

= ¼

= RHS

Word Problem On Solving an Equation

Problem: The sum of three times a number and 11 is 32. Find the number.

Solution: Let the number be x.

As per the given equation,

Sum of three times a number and 11 = 32

3x + 11 = 32

Now we need to solve the above equation for x.

Subtract the equation from 11 on both sides, to get;

3x + 11 – 11 = 32 – 11

3x = 21

Now divide the equation by 3 on both sides.

3x/3 = 21/3

x = 7

Hence, the required number is 7.