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 Solving Equations & Framing Formulae « PreviousNext » Commercial Mathematics 
Author: S.LAL
Created: 14 Aug, 2010; Last Modified: 29 Apr, 2018

Algebra - 03

Linear Equations in One Variable

A linear equation is an equation which involves only linear polynomialsA linear polynomial is a polynomial of degree 1, where a polynomial is an algebraic expression of two or more terms.. Equations like 7 x + 2 = 17 and 2 x + 5 y + 3 z = 25 are linear equations.

An equation involving linear polynomials of only a single variable is a linear equation in one variable or a simple equation. Some examples are:

(i) 7 x + 2 = 17 (ii) 5 2 z+12= 1 2 z
(iii) 3 x + 2 5 = 5 2 (iv) y − 3 4 y = 3 y − 16

Linear equation in one variable has the form:

a x + b = 0 {   a , b ∈ R ;   a   ≠   0   }
.......(1)

Steps for solving simple equations

The general steps for solving any simple equation with one unknown quantity are:

  1. First, if necessary, clear all fractions.
  2. Transpose all the terms containing the unknown quantity to one side of the equation, and the known quantities to the other.
  3. Collect the terms on both sides.
  4. Divide both side by the coefficient of the unknown quantity to get the solution.

Solve the equations (i) 7 x = 12 + 4 x (ii) 2 3 x + 2 = 7 4

(i) 7 x = 12 + 4 x ⇒ 7 x − 4 x = 12 (transposing) ⇒ 3 x = 12 (collecting terms) ⇒ x = 12 3 (transferring numerator factor 3 to other side) ⇒ x = 4 (ii) 2 3 x + 2 = 7 4 ⇒ 2 x + 6 3 = 7 4 Transferring denominators 3 and 4 to the other side as numerators, ⇒ 4 ( 2 x + 6 ) = 3 × 7 (fractions cleared!!) ⇒ 8 x + 24 = 21 ⇒ 8 x = − 3 ⇒ x = − 3 8

Solve the equations (i) 3 x + 2 = 7 x − 2 2 (ii) 1 2 y + 2 7 y = 2 y + 34

(i) 3 x + 2 = 7 x − 2 2 ⇒ 3 x − 7 x = − 2 2 − 2 (transposing) ⇒ − 4 x = − 3 2 (collecting terms) ⇒ 4 x = 3 2 (multiplying both sides with -1) ⇒ x = 3 2 4 (ii) 1 2 y + 2 7 y = 2 y + 34 ⇒ 7 y + 4 y 14 = 2 y + 34 ⇒ 11 y = 14 ( 2 y + 34 ) (clearing fractions) ⇒ 11 y − 28 y = ( 14 × 34 ) ⇒ − y = ( 14 × 34 ) 17 ⇒ y = − 28

One third of a number increased by 4 is equal to 7. Find the number.

Let x be the number. Then,

1 3 x + 4 = 7 ⇒ 1 3 x = 7 − 4 = 3 ⇒ x = 3 × 3 = 9

Thus, x = 9.

The sum of two numbers is 24 and their difference is 10. Find the numbers.

Let x be the first number. Then the second number is (24 - x). So,

x − ( 24 − x ) = 10 ⇒ x − 24 + x = 10 ⇒ 2 x = 34 ⇒ x = 17

Thus, one number is 17, while the other is (24 - 17) or 7.

Note that had you begun by subtracting the first number from the second, i.e. ( 24 − x ) − x = 10 , you would have got the number 7 first, from which you get 17 by subtracting from 24.

A number consists of two digits whose sum is 8. If 18 is added to the number, its digits are interchanged. Find the number. (Aggarwal & Aggarwal, Mathematics Class IX, p. 50, Ex. 1)

Let x be ten's place digit of the number. Then, (8 - x) is the unit's place digit. (Note that even if you do the reverse with x in the unit's place, you will arrive at the same answer).

Since any number, like 27, is actually ( 10 × 2 ) + 7 in expanded form, therefore the unknown number can be expressed as:

[ ( 10 × x ) + ( 8 − x ) ] or 9 x + 8

On adding 18 to the number, the digits are interchanged. This means that the new number has (8 - x) in the ten's place and x in the unit's place, wherein the new number can be expressed as:

[ 10 ( 8 − x ) + x ] or 80 − 9 x

So we have the equation:

( 9 x + 8 ) + 18 = 80 − 9 x ⇒ 9 x + 26 = 80 − 9 x ⇒ 18 x = 54 ⇒ x = 3

Thus, with x = 3 in the ten's place and (8 - 3) in the unit's place, the number is 35.

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List of References

Aggarwal, RS & Aggarwal, V, Secondary School Mathematics for Class IX, Patna, India: Bharati Bhawan, 1999.

Bibliography

Aggarwal, RS & Aggarwal, V, Secondary School Mathematics for Class IX, Patna, India: Bharati Bhawan, 1999.
Hall, HS & Knight, SR, Elementary Algebra for Schools, metric edn, Agra, India: AK Publications, 1966.

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