Reducing Equations To Simpler Form

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How to Reduce Equations to Simpler Form

In order to learn how to reduce the equation to a simpler form, we need to know the type of the equation we will be dealing with. In this article, we will be dealing with linear equations in one variable. A linear equation is an algebraic equation in which each term is the product of a single variable and a constant or a constant itself.  It has only the first-order power of the variable. The general form of such an equation is Ax + B = 0 where A and B are constants (A ≠ 0). 


Types of Equation 

A Linear equation may be represented in different forms. Some of the forms are as below:

  1. Some linear equations will have only one variable on the left-hand side. For example, 

3x + 5 = 20

  1. Some linear equations with one variable will have the same variable on both sides. For example, 7x – 5 = 10 + 2x

In other cases, linear equations will have a non-linear form. Such a form of the linear equation requires reduction into simpler form in order to reach the solutions. Example of such a type of equation is: \[\frac{2x+5}{3x+4}\] = 3


Such type of an equation cannot be simply solved. It has to be tactically reduced to a simpler form in order to solve it easily.


Reducing equation is a technique of solving an intricate equation by converting it into a simpler form. All equations are not always available in the form of a simple linear equation. By applying a few easy mathematical operations on such equations one can easily draw an accurate result. Methods like cross-multiplication, division or multiplication of both sides by a specific number can convert these complicated equations into their linear forms. After recovering these non-linear equations in the linear form they can be solved and the value of the variable can be calculated easily.


Some of the important tactics to be followed in order to simplify such equations are:-

  • In some cases, Cross Multiplication method can be implemented to convert such an equation into a simple form. 

  • Cross multiply the right-hand side (R.H.S.) and the left-hand side (L.H.S.) of the equation. This can be achieved by multiplying the denominator of one side with the numerator of the other side.

For example, in the equation: \[\frac{x+1}{x-3}\] = \[\frac{7}{3}\]

On cross-multiplication we get:  3(x + 1) = 7(x - 3)

  • In the next step, use distributive law and then open the brackets. According to distributive law, one must multiply everything inside the bracket with whichever number is outside.

On applying distributive law in the above equation we get:

             (3 × x) + (3 × 1) = (7 × x) – (7 × 3)

             3x + 3 = 7x - 21

  • Combine the like terms on either side by arranging the variables on L.H.S. and constants on another side of the equation i.e. R.H.S.

            7x – 3x = 21 + 3

            4x = 24

  • On adding or subtracting the same number on both sides, the value of the equation does not change. In the same way, if we multiply or divide both sides with the same number, we can get the value of the variable without changing the resultant equation.

In this equation, if we divide both sides with 4, we are left with the variable on L.H.S. and its value in the R.H.S.

            x = 6

Let us look into a more complex problem in order to understand how to reduce equations to simpler form in maths.


Example: Reduce the equation \[\frac{6x-2}{9}\] - \[\frac{3x+5}{18}\] = \[\frac{1}{3}\] to its simplest form and find the value of x that satisfies the equation:


Solution:  According to the problem:-

 \[\frac{6x-2}{9}\] - \[\frac{3x+5}{18}\] = \[\frac{1}{3}\]

In the R.H.S., first the L.C.M. of the denominator of the two terms i.e. 9 and 18 is substituted in the equation

\[\frac{2(6x-2)-(3x+5)}{18}\] = \[\frac{1}{3}\]

Now applying distributive law, open the brackets and solve the numerator of R.H.S

\[\frac{12x-4-3x+5}{18}\] = \[\frac{1}{3}\]

On adding or subtracting the like terms of the numerator we get

\[\frac{9x+1}{18}\] = \[\frac{1}{3}\]

Now, cross multiply the numerator and the denominator of L.H.S. and R.H.S. conversely

3(9x + 1) = 18

Again apply distributive law on the R.H.S.

27x + 3 = 18

Now subtracting 3 from both sides to isolate the variable we get

27x = 18 - 3

27x = 15

On dividing both sides with 27 we get

x = \[\frac{5}{9}\] (Ans)


Did You Know

  • An equation is nothing but a condition on a particular variable.

  • The value of a variable in an equation is not fixed, obviously!

  • The expression on the left side of the = sign is referred to as LHS and that on the right side of the = sign referred to as RHS. LHS = RHS. Hence the name. 

  • The value(s) for which LHS=RHS are the answer(s) of the equation. 

FAQ (Frequently Asked Questions)

Q1. Anita and Maria are Two Sisters. Their Present Ages are in the Ratio of 5:7. After Four Years the Ages of the Ratio of their Ages will be 3:4. Find the Present Ages of Anita and Maria.

Ans

Let x be the common ratios between their ages -

Let the present age of Anita = 5x

Let the present age of Maria = 7x

The age of Anita after 4 years = 5x + 4

The age of Maria after 4 years = 7x + 4

Thus, according to the problem:-

(5x+4)/(7x+4) = 3/4   

Now on cross multiplication:

4(5x + 4) = 3(7x + 4)

On applying distributive law:

20x + 16 = 21x + 12

Now rearranging the like terms on both the sides:

21x - 20x = 16 - 12

x = 4

Thus the present age of Anita is = 5x = 5 x 4 = 20 years

The present age of Maria is = 7x = 7 x 4 = 28 years

Q2. What are the Different Forms of Linear Equations? State an Example for Each.

Ans: The different forms of linear equations are represented as below:


Types of Linear Equations

General Equation

Example

Slope intercept form

y = mx + c

Y = 2x - 5

Point–slope form

y – y1 = m(x – x1 )

4y – 9 = 7(x + 4)

General Form

Ax + By + C = 0

8x + 5y + 3 = 0

Intercept form

x/x₀ + y/y₀ = 1

x/5 + y/10 = 1

A Function

Here y = f(x)

Thus, f(x) = x + C

f(x) = 2x - 9

An Identity Function

f(x) = x

f(x) = 5/8x

A Constant Function

f(x) = C

f(x) = 15