The document NCERT Exemplar: Pair of Linear Equations in Two Variables Notes | EduRev is a part of the Class 10 Course Mathematics (Maths) Class 10.

All you need of Class 10 at this link: Class 10

**Exercise 3.1**

**Choose the correct answer from the given four options: 1. Graphically, the pair of equations 6x – 3y + 10 = 0 2x – y + 9 = 0 represents two lines which are (a) Intersecting at exactly one point. (b) Intersecting at exactly two points. (c) Coincident (d) parallel.Ans:** (d)

The given equations are,

6x - 3y + 10 = 0

dividing by 3

⇒ 2x - y + 10/3 = 0… (i)

And 2x - y + 9 = 0…(ii)

Table for 2x - y + 10/3 = 0,

x | 0 | -5/3 |

y | 10/3 | 0 |

Table for 2x - y + 9 = 0

x | 0 | -9/2 |

y | 9 | 0 |

Hence, the pair of equations represents two parallel lines.**2. The pair of equations x + 2y + 5 = 0 and –3x – 6y + 1 = 0 have(a) a unique solution(b) exactly two solutions(c) infinitely many solutions(d) no solutionAns:** (d)

The equations are:

–3

a

a

a

b

c

Here,

a

Therefore, the pair of equation has no solution.

Explanation:

Condition for a pair of linear equations to be consistent are:

Intersecting lines having unique solution,

a

Coincident or dependent

a

(a) one solution

(b) two solutions

(c) infinitely many solutions

(d) no solution

Explanation:

The given pair of equations are y = 0 and y = – 7.

Graphically, both lines are parallel and have no solution**5. The pair of equations x = a and y = b graphically represents lines which are(a) parallel(b) intersecting at (b, a)(c) coincident(d) intersecting at (a, b)Ans:** (d)

Graphically in every condition,

a, b>>0

a, b< 0

a>0, b< 0

a<0, b>0 but a = b≠ 0.

The pair of equations x = a and y = b graphically represents lines which are intersecting

at (a, b).

Hence, the cases two lines intersect at (a, b).

**Exercise 3.2**

**1. Do the following pair of linear equations have no solution? Justify your answer.(i) 2x + 4y = 312y + 6x = 6(ii) x = 2yy = 2x(iii) 3x + y – 3 = 02x + 2/3y = 2Solution:**

The Condition for no solution = a

(i) Yes.

Given pair of equations are,

2x + 4y – 3 = 0 and 6x + 12y – 6 = 0

Comparing the equations with ax+ by +c = 0;

We get,

a

a

a

b

c

Here, a

Hence, the given pair of linear equations has no solution.

(ii) No.

Given pair of equations,

x = 2y or x – 2y = 0

y = 2x or 2x – y = 0;

Comparing the equations with ax+ by +c = 0;

We get,

a

a

a

b

Here, a

Hence, the given pair of linear equations has unique solution.

(iii) No.

Given pair of equations,

3x + y – 3 = 0

2x + 2/3 y = 2

Comparing the equations with ax+ by +c = 0;

We get,

a

a

a

b

c

Here, a

(i) 3x + 1/7y = 3

7x + 3y = 7

(ii) –2x – 3y = 1

6y + 4x = – 2

(iii) x/2 + y + 2/5 = 0

4x + 8y + 5/16 = 0

Solution:

Condition for coincident lines,

a

(i) No.

Given pair of linear equations are:

3x + 1/7y = 3

7x + 3y = 7

Comparing the above equations with ax + by + c = 0;

Here, a

And a

a

b

c

Here, a

Hence, the given pair of linear equations has unique solution.

(ii) Yes,

Given pair of linear equations.

– 2x – 3y – 1 = 0 and 4x + 6y + 2 = 0;

Comparing the above equations with ax + by + c = 0;

Here, a

And a

a

b

c

Here, a

Hence, the given pair of linear equations is coincident.

(iii) No,

Given pair of linear equations are

x/2 + y + 2/5 = 0

4x + 8y + 5/16 = 0

Comparing the above equations with ax + by + c = 0;

Here, a

And a

a

b

c

Here, a

Hence, the given pair of linear equations has no solution.

(i) –3x– 4y = 12

4y + 3x = 12

(ii) (3/5)x – y = ½

(1/5)x – 3y= 1/6

(iii) 2ax + by = a

1. ax + 2by – 2a = 0; a, b ≠ 0

(iv) x + 3y = 11

2 (2x + 6y) = 22

Solution:

Conditions for pair of linear equations to be consistent are:

a

a

(i) No.

The given pair of linear equations

– 3x – 4y – 12 = 0 and 4y + 3x – 12 = 0

Comparing the above equations with ax + by + c = 0;

We get,

a

a

a

b

c

Here, a

Hence, the pair of linear equations has no solution, i.e., inconsistent.

(ii) Yes.

The given pair of linear equations

(3/5)x – y = ½

(1/5)x – 3y= 1/6

Comparing the above equations with ax + by + c = 0;

We get,

a

a

a

b

c

Here, a

Hence, the given pair of linear equations has unique solution, i.e., consistent.

(iii) Yes.

The given pair of linear equations –

2ax + by –a = 0 and 4ax + 2by – 2a = 0

Comparing the above equations with ax + by + c = 0;

We get,

a

a

a

b

c

Here, a

Hence, the given pair of linear equations has infinitely many solutions, i.e., consistent

(iv) No.

The given pair of linear equations

x + 3y = 11 and 2x + 6y = 11

Comparing the above equations with ax + by + c = 0;

We get,

a_{1} = 1, b_{1} = 3, c_{1} = 11

a_{2} = 2, b_{2} = 6, c_{2} = 11

a_{1} /a_{2} = ½

b_{1} /b_{2} = ½

c_{1} /c_{2} = 1

Here, a_{1}/a_{2} = b_{1}/b_{2 }≠ c_{1}/c_{2}.

Hence, the given pair of linear equations has no solution.

**Exercise 3.3**

**1. For which value(s) of λ, do the pair of linear equationsλx + y = λ ^{2} and x + λy = 1 have(i) no solution?(ii) infinitely many solutions?(iii) a unique solution?Solution:**

The given pair of linear equations is

λx + y = λ

a

a

The given equations are;

λ x + y – λ

x + λ y – 1 = 0

Comparing the above equations with ax + by + c = 0;

We get,

a

a

a

b

c

(i) For no solution,

a

i.e. λ = 1/ λ ≠ λ

so, λ

and λ

Here, we take only λ = – 1,

Since the system of linear equations has no solution.

(ii) For infinitely many solutions,

a

i.e. λ = 1/ λ = λ

so λ = 1/ λ gives λ = + 1;

λ = λ

Hence satisfying both the equations

λ = 1 is the answer.

(iii) For a unique solution,

a

so λ ≠1/ λ

hence, λ

λ ≠ + 1;

So, all real values of λ except +1

The given pair of linear equations is

kx + 3y = k – 3 …(i)

12x + ky = k …(ii)

On comparing the equations (i) and (ii) with ax + by = c = 0,

We get,

a

a

Then,

a

b

c

For no solution of the pair of linear equations,

a

k/12 = 3/k ≠ (k-3)/k

Taking first two parts, we get

k/12 = 3/k

k

k = + 6

Taking last two parts, we get

3/k ≠ (k-3)/k

3k ≠ k(k – 3)

k

so, k ≠ 0, 6

Therefore, value of k for which the given pair of linear equations has no solution is k = – 6.

The given pair of linear equations are:

x + 2y = 1 …(i)

(a - b)x + (a + b)y = a + b – 2 …(ii)

On comparing with ax + by = c = 0 we get

a

a

a

b

c

For infinitely many solutions of the, pair of linear equations,

a

so, 1/(a - b) = 2/ (a + b) = 1/(a + b - 2)

Taking first two parts,

1/(a-b) = 2/ (a+b)

a + b = 2(a – b)

a = 3b …(iii)

Taking last two parts,

2/ (a + b) = 1/(a + b - 2)

2(a + b – 2) = (a + b)

a + b = 4 …(iv)

Now, put the value of a from Eq. (iii) in Eq. (iv), we get

3b + b = 4

4b = 4

b = 1

Put the value of b in Eq. (iii), we get

a = 3

So, the values (a, b) = (3, 1) satisfies all the parts. Hence, required values of a and b are 3 and 1 respectively for which the given pair of linear equations has infinitely many solutions.

Given pair of linear equations is

3x – y – 5 = 0 …(i)

6x – 2y – p = 0 …(ii)

On comparing with ax + by + c = 0 we get

We get,

a

a

a

b

c

Since, the lines represented by these equations are parallel, then

a

Taking last two parts, we get ½ ≠ 5/p

So, p ≠ 10

Hence, the given pair of linear equations are parallel for all real values of p except 10.

**(ii) – x + py = 1 and px – y = 1, if the pair of equations has no solution.**

Given pair of linear equations is

– x + py = 1 …(i)

px – y – 1 = 0 …(ii)

On comparing with ax + by + c = 0,

We get,

a

a

a

b

c

Since, the lines equations has no solution i.e., both lines are parallel to each other.

a

-1/p = – p ≠ 1

Taking last two parts, we get

p ≠ -1

Taking first two parts, we get

p

p = + 1

Hence, the given pair of linear equations has no solution for p = 1.

**(iii) – 3 x + 5y = 7 and 2px – 3y = 1, if the lines represented by these equations are intersecting at a unique point.**

Given, pair of linear equations is

– 3x + 5y = 7

2px – 3y = 1

On comparing with ax + by + c = 0, we get

Here, a

And a

a

b

c

Since, the lines are intersecting at a unique point i.e., it has a unique solution

a

-3/2p ≠ -5/3

p ≠ 9/10

Hence, the lines represented by these equations are intersecting at a unique point for all real values of p except 9/10

**(iv) 2 x + 3y – 5 = 0 and px – 6y – 8 = 0, if the pair of equations has a unique solution.**

Given, pair of linear equations is

2x + 3y – 5 = 0

px – 6y – 8 = 0

On comparing with ax + by + c = 0 we get

Here, a

And a

a

b

c

Since, the pair of linear equations has a unique solution.

a

so 2/p ≠ – ½

p ≠ – 4

Hence, the pair of linear equations has a unique solution for all values of p except – 4.

**(v) 2 x + 3y = 7 and 2px + py = 28 – qy, if the pair of equations have infinitely many solutions.**

Given pair of linear equations is

2x + 3y = 7

2px + py = 28 – qy

or 2px + (p + q)y – 28 = 0

On comparing with ax + by + c = 0,

We get,

Here, a

And a

a

b

c

Since, the pair of equations has infinitely many solutions i.e., both lines are coincident.

a

1/p = 3/(p+q) = ¼

Taking first and third parts, we get

p = 4

Again, taking last two parts, we get

3/(p+q) = ¼

p + q = 12

Since p = 4

So, q = 8

Here, we see that the values of p = 4 and q = 8 satisfies all three parts.

Hence, the pair of equations has infinitely many solutions for all values of p = 4 and q = 8.

Given linear equations are

x – 3y – 2 = 0 …(i)

-2x + 6y – 5 = 0 …(ii)

On comparing with ax + by c = 0,

We get

a

a

a

b

c

i.e., a

Hence, two straight paths represented by the given equations never cross each other, because they are parallel to each other.

Condition for the pair of system to have unique solution

a

Let the equations be,

a

a

Since, x = – 1 and y = 3 is the unique solution of these two equations, then

It must satisfy the equations –

a

– a

and a

– a

Since for the different values of a

Hence, infinitely many pairs of linear equations are possible.**7. If 2 x + y = 23 and 4x – y = 19, find the values of 5y – 2x and y/x – 2.**

Given equations are

2x + y = 23 …(i)

4x – y = 19 …(ii)

On adding both equations, we get

6x = 42

So, x = 7

Put the value of x in Eq. (i), we get

2(7) + y = 23

y = 23 – 14

so, y = 9

Hence 5y – 2x = 5(9) – 2(7) = 45 – 14 = 31

y/x – 2 = 9/7 -2 = -5/7

Hence, the values of (5y – 2x) and y/x – 2 are 31 and -5/7 respectively.

**Solution:**

Using property of rectangle,

We know that,

Lengths are equal,

i.e., CD = AB

Hence, x + 3y = 13 …(i)

Breadth are equal,

i.e., AD = BC

Hence, 3x + y = 7 …(ii)

On multiplying Eq. (ii) by 3 and then subtracting Eq. (i),

We get,

8x = 8

So, x = 1

On substituting x = 1 in Eq. (i),

We get,

y = 4

Therefore, the required values of x and y are 1 and 4, respectively.

**Exercise 3.4**

**1. Graphically, solve the following pair of equations:2x + y = 62x – y + 2 = 0Find the ratio of the areas of the two triangles formed by the lines representing these equations with the x-axis and the lines with the y-axis.Solution: **

Given equations are 2x + y = 6 and 2x – y + 2 = 0

Table for equation 2x + y – 6 = 0, for x = 0, y = 6, for y = 0, x = 3.

x | 0 | 3 |

y | 6 | 0 |

Table for equation 2x – y + 2 = 0, for x = 0, y = 2, for y = 0,x = – 1

x | 0 | -1 |

y | 2 | 0 |

Let A_{1} and A_{2} represent the areas of triangles ACE and BDE respectively.

Let, Area of triangle formed with x -axis = T_{1}

T_{1} = Area of △ACE = ½ × AC × PE

T_{1} = ½ × 4 × 4 = 8

And Area of triangle formed with y – axis = T_{2}

T_{1} = Area of △BDE = ½ × BD × QE

T_{1} = ½ × 4 × 1 = 2

T_{1}:T_{2} = 8:2 = 4:1

Hence, the pair of equations intersect graphically at point E(1, 4)

i.e., x = 1 and y = 4.**2. Determine, graphically, the vertices of the triangle formed by the lines y = x, 3y = x, x + y = 8 Solution: **Given linear equations are

y = x …(i)

3y = x …(ii)

and x + y = 8 …(iii)

Table for line y = x,

x | 0 | 1 | 2 |

y | 0 | 1 | 2 |

Table for line x = 3y,

x | 0 | 3 | 6 |

y | 0 | 1 | 2 |

Table for line x + y = 8

x | 0 | 4 | 8 |

y | 8 | 4 | 0 |

Plotting the points A (1, 1), B(2, 2), C (3, 1), D (6, 2), we get the straight lines AB and CD.

Similarly, plotting the point P (0, 8), Q(4, 4) and R(8, 0), we get the straight line PQR.

AB and CD intersects the line PR on Q and D, respectively.

So, △OQD is formed by these lines. Hence, the vertices of the △OQD formed by the given lines are O(0, 0), Q(4, 4) and D(6, 2).**3. Draw the graphs of the equations x = 3, x = 5 and 2x – y – 4 = 0. Also find the area of the quadrilateral formed by the lines and the x–axis.****Solution: **

Given equation of lines x = 3, x = 5 and 2x - y - 4 = 0.

Table for line 2x – y – 4 = 0

x | 0 | 2 |

y | -4 | 0 |

Plotting the graph, we get,

From the graph, we get,

AB = OB-OA = 5-3 = 2

AD = 2

BC = 6

Thus, quadrilateral ABCD is a trapezium, then,

Area of Quadrileral ABCD = ½ × (distance between parallel lines) = ½ × (AB) × (AD + BC)

= 8 sq units**4. The cost of 4 pens and 4 pencil boxes is Rs 100. Three times the cost of a pen is Rs 15 more than the cost of a pencil box. Form the pair of linear equations for the above situation. Find the cost of a pen and a pencil box.****Solution:**

Let the cost of a pen and a pencil box be Rs x and Rs y respectively.

According to the question,

4x + 4y = 100

Or x + y = 25 …(i)

3x = y + 15

Or 3x - y = 15 …(ii)

On adding Equation (i) and (ii), we get,

4x = 40

So, x = 10

Substituting x = 10, in Eq. (i) we get

y = 25 - 10 = 15

Hence, the cost of a pen = Rs. 10

The cost of a pencil box = Rs. 15**5. Determine, algebraically, the vertices of the triangle formed by the lines****3 x – y = 3**

3x – y = 2 …(i)

2x -3y = 2 …(ii)

x + 2y = 8 …(iii)

Let the equations of the line (i), (ii) and (iii) represent the side of a ∆ABC.

On solving (i) and (ii),

We get,

[First, multiply Eq. (i) by 3 in Eq. (i) and then subtract]

(9x - 3y)-(2x - 3y) = 9 - 2

7x = 7

x = 1

Substituting x=1 in Eq. (i), we get

3 × 1 - y = 3

y = 0

So, the coordinate of point B is (1, 0)

On solving lines (ii) and (iii),

We get,

[First, multiply Eq. (iii) by 2 and then subtract]

(2x + 4y)-(2x-3y) = 16-2

7y = 14

y = 2

Substituting y=2 in Eq. (iii), we get

x + 2 × 2 = 8

x + 4 = 8

x = 4

Hence, the coordinate of point C is (4, 2).

On solving lines (iii) and (i),

We get,

[First, multiply in Eq. (i) by 2 and then add]

(6x - 2y) + (x + 2y) = 6 + 8

7x = 14

x = 2

Substituting x=2 in Eq. (i), we get

3 × 2 – y = 3

y = 3

So, the coordinate of point A is (2, 3).

Hence, the vertices of the ∆ABC formed by the given lines are as follows,

A (2, 3), B (1, 0) and C (4, 2).

Let the speed of the rickshaw and the bus are x and y km/h, respectively.

Now, she has taken time to travel 2 km by rickshaw, t

Speed = distance/ time

she has taken time to travel remaining distance i.e., (14 – 2) = 12km

By bus t

By first condition,

t

Now, she has taken time to travel 4 km by rickshaw, t

and she has taken time to travel remaining distance i.e., (14 – 4) = 10km, by bus = t

By second condition,

t

(4/x) + (10/y) = (13/20) …(ii)

Let (1/x) = u and (1/y) = v

Then Equations. (i) and (ii) becomes

2u + 12v = ½ …(iii)

4u + 10v = 13/20…(iv)

[First, multiply Eq. (iii) by 2 and then subtract]

(4u + 24v) – (4u + 10v) = 1–13/20

14v = 7/20

v = 1/40

Substituting the value of v in Eq. (iii),

2u + 12(1/40) = ½

2u = 2/10

u = 1/10

x = 1/u = 10km/hr

y = 1/v = 40km/hr

Hence, the speed of rickshaw = 10 km/h

And the speed of bus = 40 km/h.

Offer running on EduRev: __Apply code STAYHOME200__ to get INR 200 off on our premium plan EduRev Infinity!

51 videos|346 docs|103 tests

- Important Questions: Pair of Linear Equations in Two Variables
- Facts That Matter: Pair of Linear Equations in Two Variables
- RS Aggarwal Test: Pair of Linear Equations in Two Variables
- Additional Question - Answers: Pair of Linear Equations in Two Variable
- HOTS Questions: Pair of Linear Equations in Two Variables
- Value-Based Questions: Pair of Linear Equations in Two Variables