Pair Of Linear Equations In Two Variables

We have,

5x – 4y + 8 = 0

7x + 6y – 9 = 0

Here, a₁ = 5, b₂ = -4, c₁ = 8

a₁ = 7, b₂ = 6, c₁ = -9

∴

∴ Two line are intersecting with each at a point.

We have,

9x + 3y + 12 = 0

18x + 6y – 24 = 0

Here, a₁ = 9, b₂ = 3, c₁ = 12

a₁ = 18, b₂ = 6, c₁ = -24

∴

∴ Both the lines will coincide.

We have,

6x – 3y + 10 = 0

2x - y + 9 = 0

Here, a₁ = 6, b₂ = -3, c₁ = 10

a₁ = 2, b₂ = -1, c₁ = 9

∴

∴ Two line are parallel to each other.

3x + 2y = 5 2x -3y = 7

Here, a₁ = 3, b₂ = 2, c₁ = 5

a₁ = 2, b₂ = -3, c₁ = 7

∴

∴ two equations are consistent.

2x - 3y = 8 4x -6y = 9

Here, a₁ = 2, b₂ = -3, c₁ = 8

a₁ = 4, b₂ = -6, c₁ = 9

∴

∴ both lines are inconsistent

9x -10y = 12

Here, a₁ = , b₂ = , c₁ = 7

a₁ = 9, b₂ = -10, c₁ = 12

∴

∴ both lines are consistent

5x - 3y = 11 -10x + 6y = -22

Here, a₁ = 5, b₂ = -3, c₁ = 11

a₁ = -10, b₂ = 6, c₁ = -22

∴

∴ two lines are consistent

2x + 3y = 12

Here, a₁ = , b₂ = 2, c₁ = 8

a₁ = 2, b₂ = 3, c₁ = 12

∴

∴ two equations are consistent

x + y = 5 2x + 2y = 10

Here, a₁ = 1, b₂ = 1, c₁ = 5

a₁ = 2, b₂ = 2, c₁ = 10

∴

∴ two lines are inconsistent

x - y = 8 3x + 3y = 16

Here, a₁ = 1, b₂ = -1, c₁ = 8

a₁ = 3, b₂ = 3, c₁ = 12

∴

∴ two lines are inconsistent

2x + y = 6 4x - 2y = 4

Here, a₁ = 2, b₂ = 1, c₁ = 6

a₁ = 4, b₂ = -2, c₁ = 4

∴

∴ two lines are consistent

2x - 2y = 2 4x - 4y = 5

Here, a₁ = 2, b₂ = 2, c₁ = 2

a₁ = 4, b₂ = 4, c₁ = 5

∴

∴ two lines are inconsistent.

Let the number of pants purchased by Neha be ‘x’ and no. of skirts purchased be ‘y’.

Twice the number of pants = 2x

From the given condition,

y = 2x – 2 (I)

i.e. number of skirts are 2 less than 2 twice the number of pants.

y = 4x – 4 (II)

i.e. number of skirts are 4 less than four times the number of pants.

Subtracting eq. I from II, we get

2x = 2

⇒ x = 1

So number of pants purchased = 1 and number of skirts = 0

Let the number of girls and boys in the class be x and y respectively

According to given condition, we have:

x + y = 10

x-y = 4

x + y = 10 ⇒ x = 10-y x-y = 4 ⇒ x = 4 + y

From the graph it can be observed that the two lines intersect each other at the point (7, 3)

So, x = 7 and y = 3

Number of Girls = 7;

Number of boys = 3

Let the cost of one pencil and one pen be ₹x and ₹y respectively.

According to given situation, we have :

5x + 7y = 50

7x + 5y = 46

5x + 7y = 50 7x + 5y = 46

From the graph. It can be observed that the two lines are intersecting each other at (3,5)

x = 3 and y = 5

Cost of pencil = ` 3;

Cost of pen = ` 5

Let l be the length and b be the breadth

l = b + 4

l-b = 4

Perimeter = 2(l + b)

× 2(l + b) = 36 (given)

l-b = 4 ……I

l + b = 36 ….II

adding eq. I and II

⇒ 2l = 40

⇒ l = = 20m

Substitute l = 20 in(1)

l-b = 4

20-b = 4

b = 20-4 = 16m

Length = 20 m

Width = 16 m

2x + 3y-8 = 0

3x + 2y-7 = 0(intersecting)

2x + 3y + 12 = 0 (parallel lines)

4x + 6y-16 = 0 (coincident lines)

Let the length be x and breadth be y

According to question,

xy-80 = (x-5)(x + 2)

⇒xy-80 = xy + 2x-5y-10

⇒-80 + 10 = 2x-5y

⇒ 2x-5y = -70 ….I

(x + 10)(y-5) = xy + 50

⇒ xy-5x + 10y-50 = xy + 50

⇒ -5x + 10y = 100

⇒ -x + 2y = 20 ….II

Multiplying Eq. II by 2

2x-5y = -70

-2x + 4y = 40

-y = -30

y = 30

substituting y = 30 in eq. II

-x + 2× 30 = 20

-x = 20-60

x = 40

length = 40units and breadth = 30 units

Let the number no. of students be x

Let the number no. of benches be y

According to given condition, we have

x = 3y + 1 ⇒ x-3y = 1 …..I

x = 4(y-1) ⇒ x = 4y-4 ⇒ x-4y = -4 ….II

Equating eq. I and II

x-3y = 1

-x + 4y = 4

y = 5

Substituting y = 5 in eq. I

x-3×5 = 1

⇒ x-15 + 1 ⇒x = 16

Number of students = 16;

Number of benches = 5

Let the monthly incomes be represented by x and their expenditure be represented by y.

According to the given income ratios,

Income of the first person is 9x and that of the second persons is 7x.

According to the given expenditure ratios;

Expenditure of the first person is 4y and the second person’s is 3y.

Both of them manage to save ₹2000 per month.

i.e. income – expenditure = savings

Then,

9x – 4y = 2000 …I

7x - 3y = 2000 …II

Multiplying I by 3 and II by 4, we get

27x – 12y = 6000 …III

28x – 12y = 8000 …IV

Subtracting III from IV, we get

x = 2000

Putting x = 2000 in I we get,

18000 – 4y = 2000

⇒ 16000 = 4y

⇒ y = 4000

Now,

Income of the first person = 9x = 9× 2000 = 18000

Income of the second person = 7x = 7× 2000 = 14000

Let the digit in the unit’s place be x and the digit in the tens place be y.

Then, the number = 10y + x

The number obtained by reversing the order of the digits = 10x + y

According to given conditions,

(10y + x) + (10x + y) = 66

⇒ 11(x + y) = 66

⇒ (x + y) = 6

According to second situation, digits differ by 2

So, either x – y = 2 or y – x = 2

Thus , we have the following sets of simuntaneous equations

x + y = 6 …I

x – y = 2 …II

or,

x + y = 6 …III

x – y = 2 …IV

solving equation I and II, we get x = 2 and y = 4

solving equation III and IV , we get x = 4 and y = 2

When x = 4 and y = 2,

Two digit number = (10y + x) = 10(4) + 2 = 42

When x = 2 and y = 4,

Two digit number = (10y + x) = 10(2) + 4 = 24

Hence, the required number is either 24 or 42.

Let the larger of the two supplementary angles be ‘y’ and the smaller angle be ‘x’

Therefore, x + y = 180 …I

(∵ sum of supplementary angles is 180° )

Also, y = x + 18 (∵ Given) …..II

Substituting II in I, we get,

x + x + 18 = 180

⇒ 2x + 18 = 180

⇒ 2x = 180 – 18

⇒ 2x = 162

⇒ x = 81°

Substitute x = 81 in II

y = 81 + 18 = 99°

Therefore, measure of larger angle is 99° and the measure of the smaller angle is 81°

i. Let the fixed charge be x.

Let the charges per km be y.

Total charges to be paid will be the sum of the fixed charges and the charges per km.

For 10kms, the amount paid is ₹220.

x + 10y = 220 …I

For 15kms, the amount paid is ₹ 310

x + 15y = 310 …II

Solving I and II we get

x + 10y = 220

x + 15y = 310

- - -

-5y = -90

y = 18

Substituting the value of y in Eq. I we get

x + 10(18) = 220

x = 220 – 180

x = 40

The Fixed charge are ₹ 40 and the charges per km are ₹18

ii. For 25 km ,

total = 40 + 25(18) = 40 + 450 = 490

hence, for 25 km, the taxi fare is 490

Let the numerator be x and denominator be y.

Hence, the fraction =

Then, according to first condition, we have

and

⇒ 5x + 5 = 4y + 4

⇒ 5x – 4y = -1 ….I

According to second condition,

⇒ 2x – 10 = y – 5 …II

Using the elimination method, multiplying eq. I by 2 and Eq. II by 5

10x – 8y = -2

10x – 5y = 25

- + - _

-3y = -27

y = 9

substitute the value of y = 9 in Eq. I

5x – 4(9) + 1 = 0

5x = -1 + 36

x = 7

hence, the given fraction is .

Let the speed of faster car = x km/h

Speed of slower car = y km/h

Relative speed when both car travels in same direction = x-y

If the cars travel in the same direction, they meet in 5 hours.

Total distance = 100 km

x-y = ⇒ x-y = 20 …1

Relative speed when both car travels in different direction = x + y

If they travel towards each other, they meet in 1 hour.

x + y = 100 …2

Adding Eq. 1 and 2

x-y = 20

x + y = 100

2x = 120

x =

substituting x = 60 in eq. 2

60 + y = 100

y = 100-60

y = 40 km

Speed of the faster car = 60km/h

Speed of the slower car = 40km/h

Let the larger of the two supplementary angles be ‘y’ and the smaller angle be ‘x’

Therefore, y = 2x – 3 …I

Given two angles are complementary.

x + y = 180°

⇒ x + (2x – 3) = 180

⇒ 3x - 3 = 180

⇒ 3x = 180 + 3

⇒ 3x = 183

⇒ x = 61°

Substitute x = 61 in I

y = 61× 2 – 3 = 122 – 3 = 119°

Therefore, measure of larger angle is 119° and the measure of the smaller angle is 61°

Let the number of pages in the first part be x and in the second part be y.

Total pages number of pages is 1382,

Hence, x + y = 1380 …I

The second part of the book has 64 pages more than the first part

y = x + 64 …II

substituting value of y in equation I

x + x + 64 = 1382

2x = 1382 – 64

2x = 1318

x = 659

substituting the value of x in equation (2),

y = 659 + 64

y = 723

the pages in the first book are 659 and the pages in the second part are 723.

Let the proportion of the first solution be represented by x and the second solution be represented by y.

First solution contains 50% solution =

Second solution contains 80% solution =

Given that together they can give 100ml of 68% solution.

5x + 8y = 680 …I

Total percentage of solution concentration is 100%

x + y = 100 …II

Multiplying Eq. II by 5 and solving with Eq. I

5x + 8y = 680

5x + 5y = 500

-__- -____

3y = 180

y = 60

substitute in Eq. II

x + 60 = 100

x = 40

Hence, the solutions are 40 and 20ml

Let the amount be invested at 10% be x and amount to be invested at 15% be y.

Total amount to be invested is ₹12000

Therefore, x + y = 12000 …I

Simple interest on ₹x at 10% for 1 year =

Simple interest on ₹y at 15% for 1 year =

Now, total amounts is to be invested at 12%

=

2x + 3y = 28800 …II

Multiplying Eq. I by 2 and solving with Eq. II

2x + 3y = 28800

2x + 2y = 12000

_- - - _

y = 4800

substitute y = 4800 in Eq. I

x + 4800 = 1200

x = 7200

Hence the amount to be invested at 10% is ₹7200 and the amount to be invested at 15% is ₹4800.

Let be p andbe q,

We get the following pair of linear equations:

5p + q = 2 …III

6p – 3q = 1 …IV

Using the elimination method:

Multiplying eq. III by 3 and adding to IV, we get

15p + 3q = 6

6p – 3q = 1

21p = 7

⇒ p =

Substitute the value of p in eq. III

⇒

⇒

But, ⇒ 3 = x – 1 ⇒ x = 4

⇒ 3 = y – 2 ⇒ y = 5

So, x = 4 and y = 5

⇒ …I

⇒

Substituting = p and = q

We get the pair of linear equation,

p + q = 2 …III

p – q = 6 ….IV

using the elimination method,

p + q = 2

p – q = 6

2p = 8

p = 4

substitute the value of p in Eq. III

4 + q = 2

q = -2

but, = p = 4 ⇒ x =

= q = -2 ⇒ x =

and

Let be p and be q

2p + 3q = 2 …III

4p – 9q = -1 …IV

Multiplying eq. III by 3 and subtracting with eq. IV, we get

6p + 9q = 6

4p – 9q = -1

10p = 5

p =

substitute value of p in Eq. III

+ 3q = 2

⇒ 3q = 2 – 1

⇒ q =

But, ⇒ ⇒ ⇒ x = 4

⇒ ⇒ ⇒ y = 9

6x + 3y = 6xy

2x + 4y = 5xy

⇒ ….I

⇒ ….II

Let = p and = q

6p + 3q = 6

2p + 4q = 5

Multiplying eq. IV by 3 and subtracting from Eq. III

6p + 3q = 6

6p + 12q = 15

-9q = -9

q = 1

substitute the value of q in III

6p + 3(1) = 6

p =

But, ⇒ x = 2

⇒ y = -1

and

Let = p and = q

5p – 2q = -1 ….I

15p + 7q = 10 ….II

Multiply Eq. I by 3 and subtract eq. II

15p – 6q = -3

-15p - 7q = -10

-13q = -13

q = 1

substitute the value of q in Eq. I

5p – 2(1) = -1

⇒ 5p = - 1 + 2

⇒ 5p = 1

⇒ p =

But, ⇒ 5 = x + y …III

⇒ 1 = x – y ⇒ y = x – 1

Substitute value of y in eq. III

x + x – 1 = 5

⇒ 2x = 6

⇒ x = 3

∴ y = 3 – 1 = 2

x = 3 and y = 2

and

Let = p and = q

2p + 3q = 13 …I

5p – 4q = -2 …II

Multiply Eq. I by 4 and Eq. II by 3 and solve

8p + 12q = 52

15p – 12q = -6

23p = 46

P = 2

Substitute the value of p in Eq. I

2(2) + 3q = 13

⇒ 4 + 3q = 13

⇒ 3q = 9

⇒ q = 3

But, = p = 2 ⇒ x =

= q = 3 ⇒ y =

and

Let = p and = q

10p + 2q = 4 …I

15p – 5q = -2 …II

Multiply Eq. I by 5 and Eq. II by 2

50p + 10q = 20

30p – 10q = -4

80p = 16

p =

substitute value of p in eq. I

10 + 2q = 4

⇒ 2 + 2q = 4

⇒ 2q = 2

⇒ q = 1

But, ⇒ 5 = x + y …III

⇒ 1 = x – y ⇒ y = x – 1

Substitute value of y in eq. III

x + x – 1 = 5

⇒ 2x = 6

⇒ x = 3

∴ y = 3 – 1 = 2

x = 3 and y = 2

and

Let = m and = n

m + n⇒ 4(m + n) = 3 ⇒4m + 4n = 3…(I)

⇒ 8(m-n) = -1 × 2 ⇒8m-8n = -2…(II)

Multiply Eq. I by 2

8m + 8n = 6

8m - 8n = -2

16m = 4

Substituting in Eq. II

– 8n = -2

⇒ 2 – 8n = -2

⇒ -8n = -4

⇒ n =

∴ ⇒ ⇒ 4 = 3x + y ⇒ 4 = 3x + y

∴ ⇒ ⇒ 2 = 3x-y ⇒ 2 = 3x-y ….IV

Equating Eq. III and IV

3x + y = 4

3x-y = 2

6x = 6

x = 1

Substituting x = 1

3×1 + y = 4

Hence, x = 1 and y = 1

Let the speed of the boat in still water be x km/h and speed of the stream be y km/h

Speed of boat upstream = (x – y) km/h

Speed of boat downstream = (x + y) km/h

But, total time of journey is 10hrs.

= 10

= 13

Let be a and be b

30a + 44b = 10 …I

40a + 55b = 13 …II

Multiply Eq. I by 4 and Eq. II by 3

120a + 176b = 40

120a + 165b = 39

_- - - _

11b = 1

b =

substituting value of b in Eq. I

30a + 44× = 10

⇒ 30a + 4 = 10

⇒ 30a = 6

⇒ a =

∴ ⇒ ⇒ 5 = x + y ⇒ 5 = x + y

∴ ⇒ ⇒ 11 = x - y ⇒ 11 = x - y ….IV

x + y = 5

x – y = 11

2x = 16

x = 8

substitute value of x in Eq. III

8 + y = 5

Y = 3

Hence, Speed of boat = 8 km/h;

Speed of stream = 3 km/h

Let the speed of the train be x km/h and that of the car be y km/h

Also, we know that time =

In situation 1, time spent travelling by train = h

And time spent travelling by car = h

So, total time taken in 8hour journey is

=

⇒ …I

Again, when he travels 200km by train and rest by car

Time taken by him travel 200km by train = h

And time spent travelling by car = h

Total time taken =

But given time of journey is 8hrs 20 min i.e = h

So, …II

Let = a and = b

30a + 120b = 2 …III

200a + 400b = …IV

Multiply Eq. III by 20 and Eq. IV by 3

600a + 2400b = 40

600a + 1200b = 25

_- - - _

1200b = 15

b =

substitute value of b in Eq. III

30a + 120 × = 2

a =

but, ⇒ x = 60

⇒ y = 80

Speed of train = 60 km/h

Speed of car = 80 km/h

Let the time taken by one woman to finish the work = x days

Work done by one women in one day = days

Let the time taken by one man to finish the work = y days

Work done by one men in one day = days

Now, 2 women and 5 men can finish the work in 4 days

So, ⇒ …I

Also, 3 women and 6 men can finish the work in 3 days

So, ⇒ …II

Let be a and be b

8a + 20b = 1 …III

9a + 18b = 1 …IV

Multiplying Eq. III by 9 and Eq. IV by 8 and solving

72a + 180b = 9

72a + 144b = 8

_- - - _

36b = 1

b =

substituting value of b in Eq. III

8a + 20 × = 1

288a + 20 = 36

288a = 16

a =

⇒ x = 18

⇒ y = 36

Number of days by man = 18;

Number of days by woman = 36