PAIR OF LINEAR EQUATIONS IN TWO VARIABLES

https://www.entrancei.com/

https://www.entrancei.com/ncert-solutions-class-10-maths

https://www.entrancei.com/

PAIR OF LINEAR EQUATIONS IN TWO
VARIABLES

1. Linear Equation

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

2. Simultaneous pair of linear equations in two variables
A pair of linear equations in two variables is said to form a pair of linear equations.
a1x + b1y = c1 ;a2 x + b2y = c2 form a pair of linear equations in two variables.

The following cases occur: (x, y)

(i) if a1  b1 , it has a unique
a2 b2

solution. The graph of lines
intersect at one point. The
system is consistent

For example:

To find whether the given pair of linear equations are consistent

3x + 4y − 7 = 0 , 2x + 5y +1= 0

a1 = 3 , b1 = 4
a2 2 b2 5
As a1  b1
a2 b2
It has a unique solution. The system is consistent.

(ii) if a1 = b1 = c1 .It has infinite many
a2 b2 c2

solutions. Every solution of one
equation is a solution of other also. The
graph of both equations is coincident
lines. The system is dependent
consistent.

For Example:

To find the values of a and b so that the following system of linear equations has infinite
number of solutions.

2x − 3y = 7 and (a + b)x − (a + b − 3)y = 4a + b

Here a1 = 2 , b1 = 3 , c1 = 7
a2 a + b b2 a + b − 3 c2 4a + b

For infinite many solutions

a1 = b1 = c1 or 2 = 3 = 7
a2 b2 c2 a + b a + b − 3 4a + b

https://www.entrancei.com/ncert-solutions-class-10-maths

https://www.entrancei.com/

Taking first two terms
2= 3

a+b a+b−3
3a + 3b = 2a + 2b − 6
a + b = −6
a = −6 − b

Again, taking second two terms
3 =7

a + b − 3 4a + b
12a + 3b = 7a + 7b − 21
5a − 4b = −21
5(−6 − b) − 4b = −21
− 9b = 30 − 21 = 9
b = −1
 a = −6 + 1 = –5
a = −5 , b = – 1

(iii) If a1 = b1  c1 . It has no solution.
a2 b2 c2

The graph of both line are parallel to
each other. The system is inconsistent

For Example:

To find the value of k for which system has no solution 3x − 4y + 7 = 0 , kx + 3y − 5 = 0
.

a1 = 3 ; b1 = − 4 ; c1 = − 7
a2 k b2 3 c2 5

Condition for no solution is 3 =−4
k3

k =−9
4

https://www.entrancei.com/ncert-solutions-class-10-maths

https://www.entrancei.com/

Table for consistency/Inconsistency conditions

Pair of linear equations Algebraic Graphical Algebraic
a1x + b1y + c1 = 0 conditions
a2 x + b2 y + c2 = 0 representation interpretation
a1  b1
(i) a2 b2 Intersecting Exactly one
Consistent (Independent) lines solution
(unique
Coincident lines solution)

(ii) Consistent a1 = b1 = c1 Infinitely many
(Dependent) a2 b2 c2 solution
a1 = b1  c1
(iii) Inconsistent a2 b2 c2 Pair of parallel No solution
lines

A pair of value of x and y satisfying each one of the equations in x and y is called a
solution of the system.

Methods of solving simultaneous pair of linear equations in two variables:

(i) Graphical Method (ii) Algebraic Method

Let the system of linear equation in two variables be a1x + b1y = c1 and a2 x + b2y = c2

.

3. Graphical method:

• Method of solving simultaneous linear equations in two variables:

Step 1: Get three solutions of each of given linear equations.

Step 2: Plot these points on graph in order to draw the lines representing these
equations.

Step 3: Get the solution of equations.

4. Algebraic Method:

(i) Substitution method

Step 1: From one equation find the value of one variable, (say y) in terms of other
variable, i.e., x.

Step 2: Substitute the value of variable obtained in step 1, in the other equation to
get an equation in one variable.

Step 3: Solve the equation obtained in step 2 to get the value of one variable.

Step 4: Substitute the value of variable so obtained in any given equation to find the
value of other variable.

For Example:
Solve for x and y using Substitution method; ax + y = m and bx − y = n

ax + y = m …(i)
bx − y = n …(ii)

Step 1: From (i), we have
ax + y = m

y = m − ax …(iii)

https://www.entrancei.com/ncert-solutions-class-10-maths

https://www.entrancei.com/

Step 2: Substituting value of y in equation (ii)
bx − (m − ax) = n

bx − m + ax = n
bx + ax = n + m

x(a + b) = m + n

x = m+n
a+b

Step 3: Substituting value of x in equation (iii), we get
y = m − a m + n 
 a+b 

y = ma + mb − am − an
a+b

y = bm − an
a+b

Solution is x = m + n , y = bm − an
a+b a+b

(ii) Elimination method

Step 1: Obtain the two equations.

Step 2: Multiply the equations so as to make the coefficients of one of the variables
equal, to be eliminated.

Step 3: (a) If the coefficients of the variable to be eliminated are having same sign,
then subtract the equation obtained in step 2.

(b) If the coefficients of the variable to be eliminated are having opposite sign,
then add the equation obtained in step 2.

Step 4: Solve the equation obtained in step 3 to obtain the value of one variable.

Step 5: Substitute the value of variable so obtained in any of the given equation to
find the value of other variable.

For Example:

Solve for x and y using elimination method: 3x + 4y = 10 and 2x − 2y = 2

3x + 4y = 10 …(i)

2x − 2y = 2 …(ii)

Step 1: Multiplying (i) by 2 and (ii) by 3 to make the coefficients of x equal.

6x + 8y = 20 …(iii)

6x − 6y = 6 …(iv)

Step 2: Subtract equation (iv) from (iii) to eliminate x, because the coefficients of x
are the same.
6x + 8y = 20

6x − 6y = 6

−+ −

14y = 14
y = 14 = 1

14

https://www.entrancei.com/ncert-solutions-class-10-maths

https://www.entrancei.com/

Step 3: Substituting the value of y in (i), we get
3x + 4 1 = 10
3x = 10 − 4 = 6
x = 6=2
3

Solution is x = 2, y = 1

(iii) Comparison Method
Step 1: From each equation find the value of one variable in terms of other.
Step 2: Equate them to get an equation in one variable and solve.

For Example:

Solve for x and y by Comparison Method: 2x + 3y = 8 ; x − 5y = −9

2x + 3y = 8 …(i)

x − 5y = −9 …(ii)

For equation (i),
2x = 8 − 3y

or x = 8 − 3y …(iii)
2 …(iv)

From equation (ii),
x = −9 + 5y

From equation (iii) and (iv), we have
8 − 3y = − 9 + 5y
2
8 − 3y = −18 + 10y

8 + 18 = 10y + 3y

26 = 13 y

26 = y
13

2=y

Putting value of y in equation (iv), we have
x = 8−32
2
x =1

Solution x = 1, y = 2

(iv) Cross Multiplication Method

For system of linear equations a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 when,

a1  b1 the unique solution is given by x = b1c2 − b2c1 and y = a2c1 − c2a1 .
a2 b2 a1b2 − a2b1 a1b2 − a2b1

For Example:

Solve the following system of equations in x and y by Cross Multiplication Method

https://www.entrancei.com/ncert-solutions-class-10-maths

https://www.entrancei.com/

ax + by − a + b = 0 , bx − ay − a − b = 0 +1
By cross multiplication, we have

xy

b –a+b a b

–a –a–b b –a

x = y =1
(b)(−a − b) − (−a)(−a + b) (−a + b)(b) − (−a − b)(a) (a)(−a) − b(b)

or x = y = 1
(−ab − b2 ) − (a2 − ab) (−ab + b2 ) − (−a2 − ab) − a2 − b2

or x = y = 1
− a2 − b2 a2 + b2 − a2 − b2

or x = − a2 − b2 and y = a2 + b2
− a2 − b2 − (a2 + b2 )

or x = 1 and y = –1

5. Equations reducible to system of linear equations
(i) ax + by = cxy; dx + ey = fxy

We have two types of solutions:

(a) Trivial solution, when x = 0 and y = 0

(b) Non-trivial solutions, when x  0 and y  0

Steps for non-trivial solution:

Step 1: Divide both sides by xy to obtain a + b = c, d + e = f
yx yx

Step 2: Substitute 1 = u, 1 = v to get av + bu = c, dv + eu = f
xy

Step 3: Solve these equations to get values of u and v .

Step 4: Use 1 = u and  = v to get values of x and y .
xy

For Example:

Solve for u and v : 2u + v = 7 uv ; u + 3v = 11uv
33

2u + v = 7 uv …(i)
3

u + 3v = 11uv …(ii)
3

Dividing equation (i) and (ii) by uv, we have
2+1=7
vu 3
1 + 3 = 11
vu 3

https://www.entrancei.com/ncert-solutions-class-10-maths

https://www.entrancei.com/

Putting 1 = x and 1 = y ; we have
vu

2x + y = 7 …(iii)
3

x + 3y = 11 …(iv)
3

Multiplying equation (iv) by 2 and subtracting (iv) from (iii) to eliminate x, we get

2x + y = 7 / 3

2x + 6y = 22 / 3

–– –

–5y = − 15
3

y = 15 = 1
35

Putting y = 1 in equation (iii), we get
2x = 7 −1= 4
33
x=2
3

Now, 1 = x or 1=2  v=3
v v3 2

1 = 1 or u=1
u

Solution is u = 0, v = 0 and u = 1, v = 3
2

(ii) p + q = e, r + s = f
ax + by cx + dy ax + by cx + dy

Step 1: Substitute 1 = u and 1 = v
ax + by cx + dy

Step 2: Solve the equations obtained, for u and v .

Step 3: Solve for x and y by substituting the values of u and v in Step 1.

For Example:

Solve for x and y : 10 + 2 = 4 ; 15 − 5 = −2 ; x  y, x  – y
x+y x−y x+y x−y

10 + 2 = 4 …(i)
x+y x−y

15 − 5 = −2 …(ii)
x+y x−y

Put 1 = u and 1 = v
x+y x−y

10u + 2v = 4 …(iii)

15u − 5v = −2 …(iv)

Multiplying equation (iii) by 5 and equation (iv) by 2 to eliminate v

50u + 10v = 20

https://www.entrancei.com/ncert-solutions-class-10-maths

https://www.entrancei.com/

30u − 10v = −4

80u = 16

u = 16 = 1
80 5

Putting value of u in equation (iii), we get

10  1 + 2v = 4
5

2 + 2v = 4
2v = 2
v=1

 1 = 1; 1 =1
x+y 5 x−y

or x + y = 5 …(v)

x−y =1 …(vi)

Adding equation (v) and (vi), we get

2x = 6  x = 3

Putting value of x in equation (v), we get
3+y =5

y=2

Solution is x = 3; y = 2
(iii) ax + by = c, bx + ay = d

Step 1: Add both the equations and reduce it to similar form by dividing by a + b .

Step 2: Subtract one equation from other and reduce it to simpler form by dividing
throughout by a − b.

Step 3: Solve equations obtained by step 1 and 2.

For Example:
Solve for x and y : 217x + 131y = 913 and 113x + 217y = 827

We have …(i)
217x + 131y = 913 …(ii)
131x + 217y = 827

Adding equation (i) and (ii), we have …(iii)
348x + 348y = 1740

or x + y = 5

Subtracting equation (ii) from (i), we have …(iv)

86x − 86y = 86
or x − y = 1
From equation (iii) and (iv), we get

x+y =5
x−y =1

2x = 6
x =3

https://www.entrancei.com/ncert-solutions-class-10-maths

https://www.entrancei.com/

Putting x = 3 in equation (iii), we have
3+y =5
y =2

Solution is x = 3, y = 2

6. To solve the Word Problems

(i) Read the problem carefully and identify the unknown quantities. Give these
quantities a variable name like x, y, u, v etc.

(ii) Identify the variables to be determined.
(iii) Read the problem carefully and formulate the equations in terms of the variables

to be determined.
(iv) Solve the equations obtained in step (iii) using any one of the methods used

earlier.

For Example:
A train covered a certain distance at a uniform speed. If the train would have been
6 km/hr faster it would have taken 8 hrs less than the scheduled time. And if the train
were slower by 6 km/hr, it would have taken 12 hrs more than the scheduled time. Find
the distance of the journey.

OR
Let the uniform speed of train = x km/hr
and scheduled time of train = y hrs
 Distance (length) of journey = xy km [ Distance = speed × time]

Case 1:

Increased speed = (x + 6) km/hr

and changed time = (y – 8)hrs

According to given condition (x + 6)(y – 8) = xy

or xy – 8x + 6y – 48 = xy

or – 8x + 6y = 48 ..... (i)

Case 2:
Changed speed = (x – 6) km/hr

Changed time = (y + 12) hrs

 According to given condition (x – 6)(y + 12) = xy

xy + 12x – 6y – 72 = xy

12x – 6y = 72 ..... (ii)

From (i) and (ii), we have
– 8x + 6y = 48
12x – 6y = 72

Adding (i) and (ii), we have

4x = 120

or x = 30 km/hr

https://www.entrancei.com/ncert-solutions-class-10-maths

https://www.entrancei.com/

Putting value of x in (i), we have
– 8(30) + 6y = 48
6y = 48 + 240
y = 288 = 48 hrs
6

 Distance of the journey = xy
= 30 × 48
= 1440 km

https://www.entrancei.com/ncert-solutions-class-10-maths

https://www.entrancei.com/

https://www.entrancei.com/ncert-solutions-class-10-maths