Algebraic Method of Solution of a Pair of Linear Equations

In graphs, it is not convenient to plot points representing decimals, fractions and irrational numbers.
Therefore, the graphical method is not suitable when the point representing the solution of the linear equations has non-integral coordinates like roots (√5, 3√ 2), decimals (–2.7, 5.2), fractions (3/7, 7/11), etc.
There is every possibility of making mistakes while reading such coordinates leading to error in the result.

There are several alternate algebraic methods that are less time consuming and more accurate.

Substitution Method

We express the value of one variable in terms of the other variable in the first equation and then substitute it to the second equation or vice versa to solve the pair of linear equations.

Step 1 : Find the value of one variable, say y in terms of the other variable, i.e., x from either equation, whichever is convenient.

Step 2 : Substitute this value of y in the other equation, and reduce it to an equation in one variable, i.e., in terms of x, which can be solved.
Sometimes, we can get statements with no variable. If the statement is true, we can conclude that the pair of linear equations has infinitely many solutions.
If the statement is false, then the pair of linear equations is inconsistent.

Step 3 : Substitute the value of x (or y) obtained in Step 2 in the equation used in Step 1 to obtain the value of the other variable.

Consistent & Unique Solution

Let's solve following pair of linear equations
x + y = 14 ...(i)
x – y = 4 ...(ii)

↬From (i),
y = 14 – x 

↬Substituting y = 14 – x in (ii)
x – (14 – x) = 4
⇒ x – 14 + x = 4
⇒ 2x = 4 + 14
⇒ x = 18/2
⇒ x = 9

↬Substituting x = 9 in y = 14 – x,
y = 14 – 9
⇒ y = 5

∴ The Solution of the given pair of equations is (9,5)

Verification
Substituting the values of x & y in both equations
Eq(i), LHS = 9+5 = 14 = RHS
Eq(ii), LHS = 9−5 = 4 = RHS
∵ Both equations are satisfied, the solution is correct.

Consistent & Infinite Solution

Let's solve following pair of linear equations
3x − y = 3 ...(i)
9x – 3y = 9 ...(ii)

From (i),
y = 3x – 3 

Substituting y = 3x – 3 in (ii)
9x – 3(3x – 3) = 9
⇒ 9x – 9x + 9 = 9
⇒ 9 = 9

This statement is true for all values of y. However, we do not get a specific value of y as a solution. Therefore, we cannot obtain a specific value of x.
This situation has arisen because both the given equations are equivalent.

∴ The the given pair of equations have infinitely many solutions.

Inconsistent Solution

Let's solve following pair of linear equations
2x + 3y = 3
4x + 6y = 5

From (i),
y = (3 – 2x)/3

Substituting y = (3 – 2x)/3 in (ii)
4x + 6(3 – 2x)/3 = 5
⇒ 4x + 2(3 – 2x) = 5
⇒ 4x + 6 – 4x = 5
⇒ 6  = 5

Which is a false statement.
∴ The equations do not have a common solution.

Elimination Method

Step 1 : First multiply both the equations by some suitable non-zero constants to make the coefficients of one variable (either x or y) numerically equal.

Step 2 : Then add or subtract one equation from the other so that one variable gets eliminated. If you get an equation in one variable, go to Step 3.

If in Step 2, we obtain a true statement involving no variable, then the original pair of equations has infinitely many solutions.
If in Step 2, we obtain a false statement involving no variable, then the original pair of equations has no solution, i.e., it is inconsistent.

Step 3 : Solve the equation in one variable (x or y) so obtained to get its value.

Step 4 : Substitute this value of x (or y) in either of the original equations to get the value of the other variable.

Consistent & Unique Solution

Let's solve following pair of linear equations by elimination method
x + y = 5 ...(i)
2x + 3y = 4 ...(ii)

↬Multiply eq (i) by 3 to make the coefficients of y equal. Then we get the equations as :
3x + 3y = 15 ...(iii)

↬Subtracting eq (ii) from Equation (iii),
3x + 3y = 15
2x + 3y = 4
−   −        −   
 x + 0y = 11
 x = 11

↬Substituting x = 11 in (i),
y = 5 – 11
⇒ y = −6

∴ The Solution of the given pair of equations is (11,−6)

Verification
Substituting the values of x & y in both equations
Eq(i), LHS = 11+(−6) = 5 = RHS
Eq(ii), LHS =2×11+3(−6) = 22−18 = 4 = RHS
∵ Both equations are satisfied, the solution is correct.

Inconsistent Solution

Let's solve following pair of linear equations by elimination method
2x + 3y = 8 ...(i)
4x + 6y = 7 ...(ii)

↬Multiply eq (i) by 2 to make the coefficients of x equal. Then we get the equations as :,
4x + 6y = 16 ...(iii)

↬Subtracting eq (ii) from Equation (iii),
4x + 6y = 16
4x + 6y =  7
−   −        −  
0x + 0y = 9

which is a false statement.
∴ The pair of equations has no solution.

Cross Multiplication

For any pair of linear equations in two variables of the form
a₁x + b₁y + c₁ = 0 ...(i)
a₂x + b₂y + c₂ = 0 ...(ii)
To obtain the values of x and y as shown above, we follow the following steps:

Step 1 : Multiply equation (i) by b₂ and equation (ii) by b₁, to get
b₂a₁x + b₂b₁y + b₂c₁ = 0 ...(iii)
b₁a₂x + b₁b₂y + b₁c₂ = 0 ...(iv)

Step 2 : Subtracting equation (iii) from (iv) to eliminate y, we get:
(b₂a₁ – b₁a₂) x + (b₂b₁ – b₁b₂) y + (b₂c₁– b₁c₂) = 0
⇒ (b₂a₁ – b₁a₂) x = b₁c₂ – b₂c₁
⇒ x = (b₁c₂ – b₂c₁)/(a₁b₂ – a₂b₁)   ...(v)
provided a₁b₂ – a₂b₁ ≠ 0

Step 3 : Substituting this value of x in (i) or (ii), we get
y = (c₁a₂−c₂a₁)/(a₁b₂−a₂b₁)   ...(vi)

Note that we can write the solution given by equations (v) and (vI) in the following form :

x	=	y	=	1	...(vii)
b1c2−b2c1		c1a2−c2a1		a1b2−a2b1

The above result can be remembered using the following diagram :

The arrows between the two numbers indicate that they are to be multiplied and the second product is to be subtracted from the first.

Condition for solution to be consistent or inconsistent

(i) When	a1	≠	b1
	a2		b2

then, there is a unique solution.

(ii) When	a1	=	b1	≠	c1
	a2		b2		c2

then, there are infinitely many solutions.

(iii) When	a1	=	b1	=	c1
	a2		b2		c2

then, there is no solution.

Graphical Method of Solution of a Pair of Linear Equations
	Quadratic Equations ➤