Algebra Concepts: How to Solve an Equation with One Unknown Value
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Algebra Concepts: How to Solve an Equation with One Unknown Value

The article discusses the concepts and rules for solving equations with one unknown value.

It often happens that kids are taught to solve problems of mathematics with a set of rules. Without actually understanding the concepts behind the process of solution, they learn to solve the equations by mugging up those rules. And thus, there are a lot of loop holes in the concepts - people understand how to solve an equation, but they don't understand why the equation is solved with the specified steps. This article (and few others which will be published later) is an effort to fill the holes in basic concepts.

Let's consider an extremely simple equation,

x + 5 = 8

First of all it's important to understand the meaning of this equation. The above equation basically mean that 'x' is a value to which when 5 would be added, the result would be 8. So, to find out the value of 'x', we should deduct 5 from 8, resulting in 3. Let's cross check, by replacing 'x' by 3 in the LHS of the equation.

3 + 5 will be equal to 8, which is equal to RHS of the equation.

Now let's consider the rules of solving an equation with one variable. We try to keep the variable at one side of the equation, taking all the the constants to the other side. To do that, we have to use the complement of the mathematical operation by which the constant is interacting with the variable. The complement of addition is subtraction, and vice versa. Similarly the complement of multiplication is division, and vice versa.

In solving the above equation, we have logically followed these rules. Keeping x at one side, we simply moved 5 to the other side, by using complementary operation subtraction of addition.

x = 8 - 5

x = 3

Let's consider another equation, a little more complex than our previous one,

3x = 24 + x

The meaning of this equation is, 'x' is a value which when multiplied by 3, is equal to the sum of 'x' and 24.

Now, although we've one unknown in the equation, its at both the sides of the equation. To take x from RHS to LHS, we'll have to use the complementary operation, thus making the equation,

3x - x = 24

2x = 24

Now the equation means, x when multiplied by 2 is equal to 24. To take the constant 2 to RHS we'll use the division operation, complement of multiplication.

x = 24 / 2

x = 12

Let's cross check the solution, replacing the value of x by 12 in LHS we get,

3x = 36

In RHS, we get 24 + 12 = 36

Since, LHS is equal to RHS, thus our solution is correct.

Understanding the equations at each stage like this is very important for developing the correct concepts of mathematics.

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Comments (1)

Very simplistic and detailed approach for understanding and solving equations... Many individuals with math phobia will really appreciate this. Thanks for posting. Voted up

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