Substitution and Elimination Method - Pair of Linear Equations in Two Variables, CBSE, Class 10, Mat | Extra Documents, Videos & Tests for Class 10 PDF Download

ALGEBRAIC METHOD OF SOLVING A PAIR OF LINEAR EQUATIONS IN TWO VARIABLES
Some times, graphical method does not give an accurate answer. While reading the co-ordinate of a point on a graph paper we are likely to make an error. So we require some precise method to obtain accurate result. The algebraic methods are given below :
(i) Method of elimination by substitution.
(ii) Method of elimination by equating the coefficients.
(iii) Method of cross multiplication.

ALGEBRAIC SOLUTION BY SUBSTITUTION METHOD
To solve a pair of linear equations in two variables x and y by substitution method, we follow the following steps:
Step-I : Write the given equations
a₁x + b₁y + c₁ = 0 ...(i)
and a₂x + b₂y + c₂ = 0 ...(ii)
Step-II : Choose one of the two equations and express y in terms of x (or x in terms of y), i.e., express, one variable in terms of the other.
Step-III : Substitute this value of y obtained in step-II, in the other equation to get a linear equation in x.
Step-IV : Solve the linear equation obtained in step-III and get the value of x.
Step-V : Substitute this value of x in the relation obtained in step-II and find the value of y.
 

Ex.6 Solve for x and y : 4x + 3y = 24, 3y – 2x = 6.
 Sol. 4x + 3y = 24 ...(i)
3y – 2x = 6 ...(ii)
From equation (i), we get

Substituting in equation (ii), we get

⇒ 24 – 4x – 2x = 6
⇒ – 6x = – 24 + 6
⇒ 6x = 18
⇒ x = 3
Substituting x = 3 in (iii), we get

Hence, x = 3, y = 4.

ALGEBRAIC SOLUTION BY ELIMINATION METHOD
To solve a pair of linear equations in two variables x and y by elimination method, we follow the following steps:
Step-I : Write the given equations
a₁x + b₁y + c₁ = 0 ...(i)
and a₂x + b₂y + c₂ = 0 ...(ii)
Step-II : Multiply the given equations by suitable numbers so that the coefficient of one of the variables are numerically equal.
Step-III : If the numerically equal coefficients are opposite in sign, then add the new equations otherwise subtract.
Step-IV : Solve the linear equations in one variable obtained in step-III and get the value of one variable.
Step-V : Substitute this value of the variable obtained in step-IV in any of the two equations and find the value of the other variable.
 

Ex.7 Solve the following pair of linear equations by elimination method : 3x + 4y = 10 and 2x – 2y = 2.
 Sol. We have, 3x + 4y = 10 ...(i)
and 2x – 2y = 2 ...(ii)
Multiplying (ii) by 2, we get 4x – 4y = 4 ...(iii)
Adding (i) and (iii), we get 7x = 14 ⇒ x = 2
Putting x = 2 in equation (ii), we get 2 × 2 – 2y = 2⇒ 2y = 4 – 2  y = 1
Hence, the solution is x = 2 and y = 1

Ex.8 Solve : ax + by = c, bx + ay = 1 + c
 Sol. ax + by = c ...(i)
bx + ay = 1 + c ...(ii)
Adding (i) and (ii), we get
(a + b) x + (a + b) y = 2c + 1

Subtracting (iv) from (iii) we get

ALGEBRAIC SOLUTION BY CROSS-MULTIPLICATION METHOD
Consider the system of linear equations
a₁x + b₁y + c₁ = 0 ...(i)
a₂x + b₂y + c₂ = 0 ...(ii)
To solve it by cross multiplication method, we follow the following steps :
Step-I : Write the coefficients as follows :

The arrows between the two numbers indicate that they are to be multiplied. The products with upward arrows are to be subtracted from the products with downward arrows.
To apply above formula, all the terms must be in left to the equal sign in the system of equations –
Now, by above mentioned rule, equation (i) reduces to

Case-I : If a₁b₂ – a₂b₁ ⇒0  x and y have some finite values, with unique solution for the system of equations.
Case-II : If a₁b₂ – a₂b₁ = 0 ⇒a₁/a₂= b₁/b₂
Here two cases arise :

......(ii)

So (i) and (ii) are dependent,so there are infinite number of solutions.

So system of equations is inconsistent.

Ex.9 Solve by cross-multiplicaiton method : x + 2y + 1 = 0 and 2x – 3y – 12 = 0
 Sol. We have,x + 2y + 1 = 0 and 2x – 3y – 12 = 0
By cross-multiplication method, we have

Hence the solution is x = 3 and y = –2.

 Ex.10

Sol. 

 ...(i)

....(ii)

7u - 2v = 5 .....(iii)
8u + 7v = 15 ....(iv)

Multiply (iii) by 7 and (iv) by 2 and add