Linear Equations in One and Two Variables - Linear Equations in Two Variables, Class 9, Mathematics PDF Download

EQUATION

A statement of equality which contain one or more unknown quantity or variable (literals) is called an equation.

An equation has two parts. The part which is on the left side to the equality sign is known as left hand side (L.H.S) and the part which is on the right side to the equality sign is known as right hand side (R.H.S).
Consider an equation :

Variable : The unknown quantiLt.iHe.sS used in any equatioRn .Har.eS known as variables. Generally, they are denoted by the last English alphabets x, y, z etc.

Linear Equation : An equation in which the maximum power of variable is one is called a linear equation.
Ex. :- 4x + 5 = 3x + 1, 2x + 3y = 4 are linear equations.

Linear Equation in one variable : In general the equation of the form ax + b = c where, a, b and c are real numbers and a ≠ 0 is called linear equations in one variable.

Solution of linear equations in one variable : The value of the variable which when substituted in an equation, makes L.H.S. = R.H.S. is said to satisfy the equation is called a solution or a root of the equation.

Ex. : 2x + 5 = 0 ⇒ 2x = – 5 ⇒  is a solution of the equation The standard form of the linear equation in one variable is ax + b = 0, where a and b are real numbers and a ≠ 0. ax + b = 0 gives us as ax = – b or   is called root or solution of the equation ax + b = 0.

Ex. Verify that x = 4 is a solution of the equation 2x – 3 = 5.

Sol. Substituting x = 4 in the given equation, we get

L.H.S. = 2x – 3 = 2 × 4 – 3 = 8 – 3 = 5 = R.H.S.

Hence, x = 4 is a solution of the equation 2x – 3 = 5

Ex. Solve: 3x + 2 = 11

Sol. 3x + 2 = 11
⇒ 3x = 11 – 2 ⇒ 3x = 9⇒
Hence, x = 3 is the solution of the given equation.

RULES FOR SOLVING A LINEAR EQUATION IN ONE VARIABLE

Rule-I : Same quantity (number) can be added to both sides of an equation without changing the equality

Ex. Solve : x – 3 = 4

Sol. ⇒ x – 3 + 3 = 4 + 3 (equal number is added on both sides) x = 7

        [Adding 5/2 on both sides]

Thusm x = 51/2 is the solution of the given equation.

Rule-II : Same quantity (number) can be subtracted from both sides of an equation without changing the equality.

Ex. :Solve : x + 5 = 9

Sol. ⇒ x + 5 – 5 = 9 – 5 (equal number 5 is subtracted from both sides)

⇒ x = 4

Thus, x = 4 is the solution of the given equation.

Ex. solve the equation 

 [ subtract 11 both sides]

 [Multiplying both sides by 5]

Thus, x = - 164/3 is the solution of the given equation.

Rule-III : Both sides of an equation may be multiplied by the same non-zero number without changing the equality.

Ex. Solve  : x/3 = -3

Sol.    [Multiplying both sides by 3]

x = -9

Thus, x = –9 is a solution of the given eqution.

Rule-IV : Both sides of an equation may be divided by the same non-zero number without changing the equality.

Ex. Solve  : 2x = 7

Sol. 2x/2 = 7/2 [Dividing both sides by 2]

x = 7/2 

Hence, x = 7/2 is a solution of the given eqution.

Rule-V : (Transposition) If any term of an equation is taken from one side to the other side then the sign changes. This process is called transposition.

Ex. 10x – 27 = 7 – 7x

⇒ 10x + 7x = 7 + 27 (By transposition method)

  [dividing both sides by 17]

x = 2

Hence, x = 2 is a solution of the given equation.

Ex. Solve 3x – 7 = 17

Sol. We have 3x – 7 = 17
⇒ 3x = 17 + 7 [On transposing 7 to R.H.S.]
⇒ 3x = 24  ⇒ x = 24/3 = 8
Hence, x = 8 is the solution of the given equation.

Rule-VI : (Cross multiplication method) : If  then n(ax + b) = m (cx + d). This is called cross-multiplication method.

Ex.  Solve :

Sol. 8(x + 1) = – 3(x – 3) (by cross-multiplication)

8x + 8 = – 3x + 9

8x + 3x = 9 – 8 (by transposition)

11x = 1 ⇒ x = 

Hence, x =  is a solution of the given equation.

VARIABLE ON BOTH SIDES OF THE EQUATION

We need to simplify both sides of the equation and get the variable on one side, so as to find its value and solve the equation.

Ex. Solve : 10a – 28 = 6 – 7a

Sol. Bring –7a to L.H.S.

So, 10a + 7a – 28 = 6

Take –28 to R.H.S.

So, 10a + 7a = 6 + 28

⇒ 17a = 34

⇒ a = 34/17 = 2

Ex. Which of the following are linear equations ?

1. 15 + 3x = 7
2.  
3. x² + 1 = 2x
4.  x² – x + 7 = x² + 3x – 1
5. x + 2x² + x3 = 2x² + 7

Sol. 1. The given equation is 15 + 3x = 7 ⇒ 3x = 7 – 15 ⇒ 3x = – 8 ⇒ 3x + 8 = 0
⇒ It is a linear equation.

2. The given equation is ⇒ 5 + 3x = 4x² ⇒ 4x² – 3x – 5 = 0

As it contains a quadratic polynomial, it is not a linear equation.

3. The given equation x² + 1 = 2x is not a linear equation as it contains a quadratic polynomial.

4. The given equation is x² – x + 7 = x² + 3x – 1 ⇒ – x – 3x = – 7 – 1 ⇒  4x = 8 ⇒ 4x – 8 = 0

which is a linear equation.

5. The given equation is x + 2x² + x³ = 2x² + 7
The given equation can be written as x³ + 2x² – 2x² + x – 7 = 0 ⇒ x3 + x – 7 = 0

As the L.H.S. is a cubic polynomial so the given equation is not a linear equation.

LINEAR EQUATIONS IN TWO VARIABLES

An equation of the form ax + by + c = 0 or ax + by = c. where a,b,c are real numbers a ≠ 0, b ≠ 0 and x & y are variables, is called a linear equation in two variables.

Ex. : x + y = 176, 1.25 + 3t = 5, p + 4q = 7 are the example of linear equation in two variables.

SOLUTION OF LINEAR EQUATION IN TWO VARIABLES

Any values of x and y which satisfies the equation ax + by + c = 0, is called a solution of it.

Ex. Consider the equation 3x + 2y = 14 ... (1)

(i) On putting x = 2 and y = 4 in (1), we get

L.H.S. = 3 × 2 + 2 × 4 = 14 i.e. 14 = 14

L.H.S. = R.H.S. Therefore, (2, 4) is a solution of equation (1).

(ii) On putting x = – 2 and y = 4 in (1), we get

3 × (–2) + 2 × 4 ≠ 14 i.e. L.H.S. ≠ R.H.S. Therefore, (–2, 4) is not a solution of equation (1).

Ex. Show that (x = 1, y = 1) as well as (x = 2, y = 5) is a solution of 4x – y – 3 = 0.

Sol. If we put x = 1 and y = 1 in the given equation., we have

we have L.H.S = 4 × 1 – 1 – 3 = 0 = R.H.S.

so, x = 1, y = 1 is a solution of 4x – y – 3 = 0

If we put x = 2, y = 5 in the equation 4x – y – 3 = 0, we have

L.H.S. = 4 × 2 – 5 – 3 = 0 = R.H.S.

So, x = 2, y = 5 is a solution of the equation 4x – y – 3 = 0

Ex. The cost of a notebook is twice the cost of a pen. Write a linear equation in two variable to represent this statement.

Sol. Let the cost of a note book be Rs. x and that of a pen be Rs. y.

∵ The cost of a notebook is twice the cost of a pen.

∴ x = 2y

So, x – 2y = 0, is the required equation.

Ex. In a one-day international cricket match between India and Australia played in Kolkata, Dhoni and Yuvraj together scored 198 runs. Express this information in the form of an equation.

Sol. Let the number of runs scored by Dhoni be x and the runs scored by Yuvraj be y.

∵ Dhoni and Yuvraj together scored 198 runs.

So, x + y = 198 ⇒ x + y – 198 = 0 is the required equation.