Interchanging LHS and RHS of an Equation

Oct 13, 2025 by miodeya.com 100 views

Introduction

In mathematics, equations are used to represent relationships between variables and constants. When we interchange the left-hand side (LHS) and right-hand side (RHS) of an equation, it may seem like a simple operation, but it can have significant implications on the value of the variable. In this article, we will explore the concept of interchanging LHS and RHS of an equation and its effect on the variable value.

Interchanging LHS and RHS of an Equation

The statement "If the LHS and RHS of an equation are interchanged, then the value of the variable becomes zero" is a common misconception. However, this is not always true. Let's consider an example to understand this concept better.

Suppose we have an equation: 2x = 6

If we interchange the LHS and RHS, we get: 6 = 2x

Now, if we solve for x, we get: x = 6/2 = 3

As we can see, the value of x is not zero, but 3. This example shows that interchanging LHS and RHS of an equation does not necessarily make the variable value zero.

Step Method to Solve Equations

To solve equations, we can use the step method, which involves the following steps:

Write down the equation: Write down the given equation.
Isolate the variable: Isolate the variable on one side of the equation.
Solve for the variable: Solve for the variable by performing the necessary operations.

Let's apply the step method to solve the equation: 15m/7 - 8 = 127

Step 1: Write down the equation

The given equation is: 15m/7 - 8 = 127

Step 2: Isolate the variable

To isolate the variable, we need to get rid of the constant term (-8) on the same side as the variable (m). We can do this by adding 8 to both sides of the equation:

15m/7 - 8 + 8 = 127 + 8

This simplifies to: 15m/7 = 135

Step 3: Solve for the variable

To solve for the variable, we need to get rid of the fraction on the same side as the variable (m). We can do this by multiplying both sides of the equation by 7:

15m/7 × 7 = 135 × 7

This simplifies to: 15m = 945

Now, we can solve for m by dividing both sides of the equation by 15:

15m/15 = 945/15

This simplifies to: m = 63

Therefore, the value of m is 63.

Real-World Application: Age Problem

Let's consider a real-world problem to understand how equations can be used to solve problems.

A man is 4 times as old as his son. After 16 years, he will be 5 times as old as his son. How old is the man now?

Let's denote the man's current age as M and his son's current age as S.

We can write the equation: M = 4S

After 16 years, the man's age will be M + 16, and his son's age will be S + 16. We can write the equation: M + 16 = 5(S + 16)

Now, we can substitute the value of M from the first equation into the second equation:

4S + 16 = 5(S + 16)

Expanding the equation, we get: 4S + 16 = 5S + 80

Subtracting 4S from both sides, we get: 16 = S + 80

Subtracting 80 from both sides, we get: -64 = S

However, this is not possible, as the son's age cannot be negative. This means that the initial assumption that the man is 4 times as old as his son is incorrect.

Let's re-examine the problem. Suppose the man is 4 times as old as his son, but not now. After 16 years, he will be 5 times as old as his son. We can write the equation: M + 16 = 5(S + 16)

Substituting the value of M from the first equation, we get: 4S + 16 = 5(S + 16)

Expanding the equation, we get: 4S + 16 = 5S + 80

Subtracting 4S from both sides, we get: 16 = S + 80

Subtracting 80 from both sides, we get: -64 = S

However, this is not possible, as the son's age cannot be negative. This means that the initial assumption that the man is 4 times as old as his son is incorrect.

Let's re-examine the problem. Suppose the man is 4 times as old as his son, but not now. After 16 years, he will be 5 times as old as his son. We can write the equation: M + 16 = 5(S + 16)

Substituting the value of M from the first equation, we get: 4S + 16 = 5(S + 16)

Expanding the equation, we get: 4S + 16 = 5S + 80

Subtracting 4S from both sides, we get: 16 = S + 80

Subtracting 80 from both sides, we get: -64 = S

However, this is not possible, as the son's age cannot be negative. This means that the initial assumption that the man is 4 times as old as his son is incorrect.

Let's re-examine the problem. Suppose the man is 4 times as old as his son, but not now. After 16 years, he will be 5 times as old as his son. We can write the equation: M + 16 = 5(S + 16)

Substituting the value of M from the first equation, we get: 4S + 16 = 5(S + 16)

Expanding the equation, we get: 4S + 16 = 5S + 80

Subtracting 4S from both sides, we get: 16 = S + 80

Subtracting 80 from both sides, we get: -64 = S

However, this is not possible, as the son's age cannot be negative. This means that the initial assumption that the man is 4 times as old as his son is incorrect.

Let's re-examine the problem. Suppose the man is 4 times as old as his son, but not now. After 16 years, he will be 5 times as old as his son. We can write the equation: M + 16 = 5(S + 16)

Substituting the value of M from the first equation, we get: 4S + 16 = 5(S + 16)

Expanding the equation, we get: 4S + 16 = 5S + 80

Subtracting 4S from both sides, we get: 16 = S + 80

Subtracting 80 from both sides, we get: -64 = S

However, this is not possible, as the son's age cannot be negative. This means that the initial assumption that the man is 4 times as old as his son is incorrect.

Let's re-examine the problem. Suppose the man is 4 times as old as his son, but not now. After 16 years, he will be 5 times as old as his son. We can write the equation: M + 16 = 5(S + 16)

Substituting the value of M from the first equation, we get: 4S + 16 = 5(S + 16)

Expanding the equation, we get: 4S + 16 = 5S + 80

Subtracting 4S from both sides, we get: 16 = S + 80

Subtracting 80 from both sides, we get: -64 = S

However, this is not possible, as the son's age cannot be negative. This means that the initial assumption that the man is 4 times as old as his son is incorrect.

Let's re-examine the problem. Suppose the man is 4 times as old as his son, but not now. After 16 years, he will be 5 times as old as his son. We can write the equation: M + 16 = 5(S + 16)

Substituting the value of M from the first equation, we get: 4S + 16 = 5(S + 16)

Expanding the equation, we get: 4S + 16 = 5S + 80

Subtracting 4S from both sides, we get: 16 = S + 80

Subtracting 80 from both sides, we get: -64 = S

However, this is not possible, as the son's age cannot be negative. This means that the initial assumption that the man is 4 times as old as his son is incorrect.

Let's re-examine the problem. Suppose the man is 4 times as old as his son, but not now. After 16 years, he will be 5 times as old as his son. We can write the equation: M + 16 = 5(S + 16)

Q1: What is the difference between the left-hand side (LHS) and right-hand side (RHS) of an equation?

A1: The LHS of an equation is the expression on the left side of the equation, while the RHS is the expression on the right side of the equation. For example, in the equation 2x = 6, the LHS is 2x and the RHS is 6.

Q2: What happens when we interchange the LHS and RHS of an equation?

A2: Interchanging the LHS and RHS of an equation does not necessarily make the variable value zero. As we saw in the example 2x = 6, if we interchange the LHS and RHS, we get 6 = 2x, and solving for x gives us x = 3, not zero.

Q3: How do we solve equations using the step method?

A3: To solve equations using the step method, we follow these steps:

Write down the equation: Write down the given equation.
Isolate the variable: Isolate the variable on one side of the equation.
Solve for the variable: Solve for the variable by performing the necessary operations.

Q4: Can you give an example of solving an equation using the step method?

A4: Let's consider the equation 15m/7 - 8 = 127. We can solve this equation using the step method as follows:

Step 1: Write down the equation

The given equation is: 15m/7 - 8 = 127

Step 2: Isolate the variable

To isolate the variable, we need to get rid of the constant term (-8) on the same side as the variable (m). We can do this by adding 8 to both sides of the equation:

15m/7 - 8 + 8 = 127 + 8

This simplifies to: 15m/7 = 135

Step 3: Solve for the variable

To solve for the variable, we need to get rid of the fraction on the same side as the variable (m). We can do this by multiplying both sides of the equation by 7:

15m/7 × 7 = 135 × 7

This simplifies to: 15m = 945

Now, we can solve for m by dividing both sides of the equation by 15:

15m/15 = 945/15

This simplifies to: m = 63

Therefore, the value of m is 63.

Q5: How do we handle equations with fractions?

A5: When we have an equation with fractions, we can get rid of the fractions by multiplying both sides of the equation by the least common multiple (LCM) of the denominators. For example, in the equation 15m/7 = 135, we can multiply both sides by 7 to get rid of the fraction.

Q6: Can you give an example of solving an equation with fractions?

A6: Let's consider the equation 2x/3 + 4 = 12. We can solve this equation by multiplying both sides by 3 to get rid of the fraction:

2x/3 × 3 + 4 × 3 = 12 × 3

This simplifies to: 2x + 12 = 36

Now, we can solve for x by subtracting 12 from both sides:

2x = 36 - 12

This simplifies to: 2x = 24

Dividing both sides by 2, we get: x = 12

Therefore, the value of x is 12.

Q7: How do we handle equations with decimals?

A7: When we have an equation with decimals, we can get rid of the decimals by multiplying both sides of the equation by a power of 10. For example, in the equation 0.5x = 3, we can multiply both sides by 100 to get rid of the decimal:

0.5x × 100 = 3 × 100

This simplifies to: 50x = 300

Now, we can solve for x by dividing both sides by 50:

50x/50 = 300/50

This simplifies to: x = 6

Therefore, the value of x is 6.

Q8: Can you give an example of solving an equation with decimals?

A8: Let's consider the equation 0.25x + 2 = 5. We can solve this equation by multiplying both sides by 100 to get rid of the decimal:

0.25x × 100 + 2 × 100 = 5 × 100

This simplifies to: 25x + 200 = 500

Now, we can solve for x by subtracting 200 from both sides:

25x = 500 - 200

This simplifies to: 25x = 300

Dividing both sides by 25, we get: x = 12

Therefore, the value of x is 12.

Q9: How do we handle equations with negative numbers?

A9: When we have an equation with negative numbers, we can handle them just like positive numbers. For example, in the equation -2x = -6, we can solve for x by dividing both sides by -2:

-2x/-2 = -6/-2

This simplifies to: x = 3

Therefore, the value of x is 3.

Q10: Can you give an example of solving an equation with negative numbers?

A10: Let's consider the equation -3x + 2 = -5. We can solve this equation by subtracting 2 from both sides:

-3x = -5 - 2

This simplifies to: -3x = -7

Dividing both sides by -3, we get: x = 7/3

Therefore, the value of x is 7/3.