Linear Equation in One Variable

Linear Equation in One Variable
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Equations can be classified on the basis of numbers and powers of variables they contain.

Different types of equations require different approach to solve them. Therefore, it is important to understand how they are classified.

Types of Equation

According to the number of variables

Equation in one variable

These types of equations have one variable only.

e.g., + 3 = 8

5 − = 9

2n+5 = 4n

Equation in two variables

These types of equations have two variables.

e.g., + 3 = 8y

5x − = 9

2xy + 5 = 4x

According to the degree of an equation

Degree of Equation

For an equation in one variable, the highest power of the variable is called the degree of the equation.

e.g.,

x+ 3 = 8,

Highest power of is 1

∴ Degree of the equation is one

5 −y2= 9y,

Highest power of y is 2

∴ Degree of the equation is two.

For equation with more than one variable, sum of the powers of variables in each term is compared, the largest power is called the degree of the equation.

e.g.,

+ 3 = 8y,
Sum of powers of variables in terms — 1, 0, 1
largest sum is one
∴ Degree of the equation is one

5x2 − = 9,
Sum of powers of variables in terms — 2, 1, 0
largest sum is two
∴ Degree of the equation is two.

4xy+5=4x,
Sum of powers of variables in  terms — 2, 0, 1
largest sum is two
∴ Degree of the equation is two

2xy2 −5x=10,
Sum of powers of variables in  terms — 3, 1, 0
largest sum is three
∴ Degree of the equation is three

Linear equation

An equation of 1st degree is called linear equation.
e.g.,
+ 3 = 8,
7x + 3 = 2y

Quadratic equation

An equation of 2nd degree is called Quadratic equation.
5x2 − = 9,
3y2  = 27

Linear Equation in One Variable

An equation of first degree with one variable is called a Linear equation in one variable.

A Linear Equation has one variable whose highest power is 1.
e.g., 2+ 3 = 7,
2y = 4y−9

Reducing non-linear equations to linear form

Some non-linear equations in one variable can be bought to linear form by multiplying both sides of the equation with a suitable expression.

(Ex2.6-Q1)

(8x-3)/3x = 2
⇒ 8x−3 = 2×3x

This equation is not in a linear form to begin with as the highest power of x is 1/2, but when we transposed 3x in the denominator from LHS to RHS, it became linear.
8x−3 = 2×3x
⇒ 8x−3 = 6x
⇒ 8x−6x = 3
⇒ 2x = 3
⇒ x = 3/2

(Ex2.6-Q4)
(7y+ 4)/(y+2) =−4/3
⇒ (7y + 4)3 = −4(y+2)     (cross multiplication)
⇒ 21y + 12 = −4y−8
⇒ 21y+4=−8−12
⇒ 25y = −20
⇒ = −20/25
⇒ = −4/5

Cross multiplication

Transposing denominators from both side simultaneously in single step is called cross multiplication.
\frac{a}{b} = \frac{c}{d}

2018-07-07-22-33-27_deco-1-1.jpg

ad = bc

Application of Linear equations in one variable

Different types of problems on numbers, ages, perimeters, combination of currency notes, and so on can be solved using linear equations.

Try These

Q1) The ages of Hari and Harry are in the ratio 5:7. Four years from now the ratio of their ages will be 3:4. Find their present ages.

Q2) I have a total of Rs 300 in coins of denomination Re 1, Rs 2 and Rs 5. The number of Rs 2 coins is 3 times the number of Rs 5 coins. The total number of coins is 160. How many coins of each denomination are with me?

Ans

A1) Let, the present age of Hari and Harry be 5x and 7x respectively,

Four years from now
Age of Hari will be 5+ 4 
Age of Hari will be 7+ 4 

A/Q

(5+ 4) : (7+ 4 )= 3 : 4

⇒ (5x+4)/(7x+4) = 3/4
⇒ (5+ 4)4 = 3(7+ 4)
⇒ 20x + 16 = 21x + 12
⇒ 20x−21x  = 12−16
⇒ −1x = −4
x = 4

∴ Present age of Hari = 5x = 5×4 = 20 years
and Present age of Harry = 7x = 7×4 = 28 years

A2) Let, the number of coins of Rs 5 be x,
then, the number of coins of Rs 2 be 3x,
the number of coins of Re 1 be 160 − (x+3x),A/Q5⋅x + 2⋅3x + 1⋅(160 − 4x) = 300
⇒ 5x + 6x + 160 − 4x = 300
⇒ 7x = 300 − 160
⇒ 7x = 140
⇒ x = 140/7 = 20∴ the number of coins of Rs 5 = x = 20,the number of coins of Rs 2 = 3x = 3×20 = 60,the number of coins of Re 1
= 160 − 4x = 160 − 4×20
= 160 − 80 = 80.

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