Pair Of Linear Equations In Two Variables

Let the present age of father = x

Let the present age of son = y

Now,

Given, Five years ago, father was seven times as old as son.

⇒ (x–5) = 7(y–5)

x – 5 = 7y – 35

= x – 7y + 30 = 0 …[1]

Also, after five years, father will be three times as old as son

⇒ (x + 5) = 3(y + 5)

x + 5 = 3y + 15

= x – 3y – 10 = 0 …[2]

eq[1] and eq[2] are required equations

Let the cost of 1 kg apple = x

Let the cost of 1 kg pine-apple = y

Given that, sum of the cost of 1 kg apple and 1 kg pine-apple is Rs. 150

⇒ x + y = 150 …[1]

Also, cost of 1 kg apple is twice the cost of 1kg pine-apple

⇒ x = 2y …[2]

eq[1] and eq[2] are the required equations.

let marks obtained by Ilesh be x and marks obtained by Nilesh by y.

Now,

given, Nilesh got twice the marks as obtained by Ilesh.

⇒ y = 2x …[1]

Also, the sum of the marks as obtained by them is 135.

x + y = 135 …[2]

These are the two required linear equations.

Let the length of the rectangle be x and breadth be y.

Now,

given, Length of a rectangle is four less than the thrice of its breadth

x = 3y – 4 …[1]

and

also, the perimeter is 110

2(x + y) = 110

⇒ x + y = 55 …[2]

eq[1] and eq[2] are the required equations.

let x and y be the weight of father and son respectively.

Now,

Given, The sum of the weights of a father and a son is 85 kg

x + y = 85 …[1]

and,

also, The weight of the son is of the weight of his father

…[2]

These are the required linear equations.

Let the score of Sachin and Sehwag be x and y respectively

given, Both of them together make a total score of 200 runs.

⇒ x + y = 200 …[1]

also, Sachin Tendulkar makes his score thrice the Sehwag's score

⇒ x = 3y …[2]

These are the two required equations.

Let the probabilities of getting heads and tails are x and y respectively.

Given, the probability of getting head on its face is twice to the probability of getting tail on its face

⇒ x = 2y …[1]

also, the sum of probabilities is 1

⇒ x + y = 1 …[2]

∴ these two equations are the required equations.

Generally, we substitute x = 0 or y = 0 in the given linear equations to get y and x. So we get two points on the straight line. To find more points on the line, take different values of x related to it, we get different values for y from the equation.

We get the following tables for the given linear equations.

For 2x + y = 8

y = 8 – 2x

For x + 6y = 15

x = 15 – 6y

Plotting these points on a graph and joining them, we get two straight lines.

From the graph, we can see that both lines intersect at (3, 2), hence the solution to this pair is (3, 2).

Generally, we substitute x = 0 or y = 0 in the given linear equations to get y and x. So we get two points on the straight line. To find more points on the line, take different values of x related to it, we get different values for y from the equation.

for x + y = 1

⇒ y = 1 – x

for, 3x + 3y = 2

Plotting these points on a graph and joining them, we get two straight lines.

From the graph, we can see that both lines are parallel and have no intersection points.

Generally, we substitute x = 0 or y = 0 in the given linear equations to get y and x. So we get two points on the straight line. To find more points on line, take different values of x related to it, we get different values for y from the equation.

We get the following tables for the given linear equations.

for

2x + 3y = 5

for

x + y = 2

Plot it

Plotting these points on a graph and joining them, we get two straight lines.

From the graph, we can see that both lines intersect at (1, 1), hence solution to this pair is (1, 1).

Generally, we substitute x = 0 or y = 0 in the given linear equations to get y and x. So we get two points on the straight line. To find more points on line, take different values of x related to it, we get different values for y from the equation.

We get the following tables for the given linear equations.

for

x—y = 6

for, 3x—3y = 18

Plotting these points on a graph and joining them, we get two straight lines.

From the graph, we can see that both lines coincide and hence there are infinite number of solutions.

Generally, we substitute x = 0 or y = 0 in the given linear equations to get y and x. So we get two points on the straight line. To find more points on the line, take different values of x related to it, we get different values for y from the equation.

We get the following tables for the given linear equations.

for

(x + 2) (y — 1) = xy

It is a straight line

for,

(x — 1) (y + 1) = xy

Plotting these points on a graph and joining them, we get two straight lines.

From the graph, we can see that both lines intersect at (4, 3), hence the solution to this pair is (4, 3).

Generally, we substitute x = 0 or y = 0 in the given linear equations to get y and x. So, we get two points on the straight line. To find more points on line, take different values of x related to it, we get different values for y from the equation.

We get the following tables for the given linear equations.

for

3x + 2y = 5

for, 2x–3y = –1

when y = 0, lines would intersect the x-axis

so the coordinates of triangle at x-axis would be

So the coordinates of the triangle will be

Let the number of girls be x and the number of boys be y.

Given, total no. of participants is 15.

⇒ x + y = 15

Also, boys participants is 5 less than the number of girls participants

⇒ x = y – 5

We have two equations we will plot them and their intersection will give the required result.

x + y = 15

x = y – 5

Plotting these points on a graph and joining them, we get two straight lines.

From the graph, we can see that both lines intersect at (5, 10), hence solution to this pair is (5, 10).

∴ no. of boys is 5 and no. of girls is 10

Generally, we substitute x = 0 or y = 0 in the given linear equations to get y and x. So we get two points on the straight line. To find more points on line, take different values of x related to it, we get different values for y from the equation.

We get the following tables for the given linear equations.

for

2x + 3y = 5

for

Plotting these points on a graph and joining them, we get two straight lines.

From the graph, we can see that both lines coincide.

⇒ They don’t have consistent solutions.

x + y = 7

⇒ x = 7 – y …[1]

By substituting eq[1] in 3x – y = 1

⇒ 3(7 – y) – y = 1

⇒ 21 – 3y – y = 1

⇒ 4y = 20

⇒ y = 5

If we put the above result in eq[1], we get

⇒ x = 7 – 5

⇒ x = 2

3x – y = 0

⇒ y = 3x …[1]

By substituting eq[1] in x – y + 6 = 0

x – (3x) + 6 = 0

⇒ x = 3

Now, by eq[1]

3x – y = 0

∴ 3(3) – y = 0

⇒ y = 9

⇒ x = 3, y = 9

2x + 3y = 5

⇒ 2x = 5–3y

⇒ …[1]

By substituting eq[1] in 2x + 3y = 7

⇒ 5–3y + 3y = 7

⇒ 5 = 7

it is not true

we got this independent of any variable and it is a false statement, no real solution exists to these linear pair of equations.

x – y = 3

⇒ x = y + 3 …[1]

By substituting eq[1] in 3x – 3y = 9

3(y + 3) – 3y = 9

9 = 9

The equation is independent of any variable and is true and therefore

the pair of linear equations are concurrent and satisfies all real values of x and y.

multiplying 6 to L.H.S and R.H.S to the first equation to remove fraction

we get,

9x – 10y = –12 …[1]

multiplying 6 to L.H.S and R.H.S to the second equation to remove fraction

we get, 2x + 3y = 13

⇒ …[2]

Substituting eq[2] in eq[1]

⇒

⇒ 27x – 130 + 20x = –36

⇒ x = 2

Using this result in eq[2]

⇒ y = 3

x – y = 28 eq[1] eq[2]

and x – 3y = 0

⇒ x = 3y

Substituting eq[2] in eq[1]

3y – y = 28

⇒ 2y = 28

⇒ y = 14

Now by, x = 3y

⇒ x = 3(14)

⇒ x = 42

∴ in y = mx + 5

putting values of x and y.

14 = 42m + 5

⇒ 14–5 = 42m

⇒

Let the numerator and denominator of the actual fraction be x and y.

Given, fraction becomes if 3 is added to both the numerator and the denominator

⇒ 5x = 4y – 3

⇒ …[1]

also, If 5 is added to the numerator and the denominator, it becomes .

⇒

⇒ 6x – 5y = –5 …[2]

substituting eq[1] in eq[2]

⇒ y = 7

by eq[1] and y = 7

⇒ x = 5

∴ original fraction =

Let the present ages of the father be x and the son be y.

given, The sum of present ages of a father and his son is 50 years.

⇒ x + y = 50

⇒ x = 50 – y …[1]

also, After 5 years, the age of the father becomes thrice the age of his son

⇒ x + 5 = 3(y + 5) …[2]

By substituting eq[1] in eq[2]

we get,

50–y + 5 = 3(y + 5)

⇒ y = 10

and by eq[1] and y = 10

x = 40

so the present age of father is 40 years and present ages of son is 10 years.

Let the cost of one ticket from Ahmedabad to Anand be x and the cost of one ticket form Ahmedabad to Vadodara be y.

given, 5 tickets from Ahmedabad to Anand and 10 tickets from Ahmedabad to Vadodara for Rs. 1100.

⇒ 5x + 10y = 1100 …[1]

Also, cost of one ticket from Ahmedabad to Anand and one ticket from Ahmedabad to Vadodara is Rs. 140

⇒ x + y = 140

⇒ y = 140 – x …[2]

By substituting eq[2] in eq[1] we get

5x + 10(140 – x) = 1100

⇒ 5x + 1400 – 10x = 1100

⇒ 5x = 300

x = 60

and by eq[2] and x = 60

⇒ y = 140 – 60

⇒ y = 80

multiply 15 to the first equation to remove fraction

we get, 3x – 5y = 4 …[1]

Now let’s, multiply 18 to the second equation to remove fraction

we get, 9x – 2y = 7 …[2]

Now, let’s find eq[2] – eq[1] × 3

⇒ 9x – 2y – 3(3x – 5y) = 7 – 4×3

⇒

Now if we put this result in eq[1]

⇒ 39x = 27

⇒

multiplying first equation by 13 and second equation by 4 to get same coefficient of x.

we get, 52x – 247 y = –169 …[1]

52x – 92y = –76 …[2]

eq[2] – eq[1]

155y = 93

⇒

Put this result in 4x — 19y + 13 = 0 to find the value of x

4x — 19() + 13 = 0

⇒ x =

Multiply the first equation by b

⇒ bx + by = ab + b² …[1]

⇒ ax — by = a² — b² …[2]

Now, eq[2] + eq[1]

x(a + b) = a² + ab

⇒ x(a + b) = a(a + b)

⇒ x = a

by x + y = a + b and x = a

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

⇒ y = b

Multiply the first equation by 2 and the second equation by 3.

⇒ 10ax + 12by = 56 …[1]

⇒ 9ax + 12by = 54 …[2]

on subtracting eq[2] from eq [1]

ax = 2

⇒ x =

by x = and 3ax + 4by = 18

3a() + 4by = 18

we get

Let the larger number be x and smaller number y.

given, sum of number is 35

⇒ x + y = 35 …[1]

Also, Four times the larger number is 5 more than 5 times the smaller number.

4x = 5y + 5 …[2]

Multiplying eq[1] by 5 and adding eq[2] to result, we get

5(x + y) + 4x = 175 + 5y + 5

⇒ 9x = 180

⇒ x = 20

From eq[1]

⇒ y = 15

∴ The larger number is 20 and the smaller number is 15.

Let the number of 50 paise coins be x, and the number of 25 paise coins be y.

Given, the total no. of coins is 140

⇒ x + y = 140 …[1]

also, the amount in the bag is Rs. 50

⇒ 2x + y = 200 …[2]

eq[2] – eq[1]

x = 60

and by eq[1]

⇒ y = 80

amount of 50 paise cons =

amount of 25 paise cons =

Let the digit at 10s place y and digit at one’s place be x.

∴ 10y + x = The number

Now if we interchange the digits we get a number that is 9 less than the original number.

interchanged number = 10x + y

⇒ 10x + y = 10y + x – 9

⇒ 9x – 9y + 9 = 0

⇒ x – y + 1 = 0 …[1]

and the sum of digits is 3

⇒ x + y = 3 eq[2]

adding eq[1] and …[2]

2x = 2

⇒ x = 1

From eq[1]

x – y + 1 = 0

⇒ y = 2

∴ the original number is 21

Let the length be

x and breadth by y.

Given, the length of a rectangle is twice its breadth

⇒ x = 2y

⇒ x – 2y = 0 …[1]

and also, perimeter of the rectangle is 120 cm

⇒ 2(x + y) = 120

⇒ 2x + 2y = 120 …[2]

eq[1] + eq[2]

3x = 120

⇒ x = 40

and x = 2y

⇒ y = 20

The area is product of length and breadth

⇒ area = 20×40

⇒ area = 800 cm²

Let the amount at 8% be x and the amount at 6% be y.

Where I is interest, P is the principal amount N is the time in years and R is the rate per annum.

Now,

I in case 1 of 8%

⇒ I = …[1]

in the second case of 6%

I = …[2]

The total interest is 500.

eq[1] + eq[2]

⇒

⇒ 4x + 3y = 25000 …[3]

Now if he interchanges the amount he gets 50 more

⇒

⇒ 4y + 3x = 27500 …[4]

Now, adding and subtracting equation eq[1] and eq[2].

7x + 7y = 52500

⇒ x + y = 7500 …[5]

x – y = –2500 …[6]

adding [5] and [6]

2x = 5000

⇒ x = 2500

and by [5]

x + y = 7500

⇒ 2500 + y = 7500

⇒ y = 5000

Given equations are

0.3x + 0.4y = 2.5 and 0.5x — 0.3y = 0.3

Change the given equations to the form,

a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 …(i)

⇒ 0.3x + 0.4y – 2.5 = 0 and

0.5x — 0.3y – 0.3 = 0

On comparing with (i) we get,

a₁ = 0.3, b₁ = 0.4, c₁ = – 2.5; a₂ = 0.5, b₂ = – 0.3, c₂ = – 0.3

Applying cross multiplication method which says,

Putting the given values in the above equation we get,

Similarly,

∴The solution of the pair of equations is (3, 4).

Given equations are

5x + 8y = 18, 2x — 3y = 1

Change the given equations to the form,

a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 …(i)

⇒ 5x + 8y – 18 = 0 and

2x — 3y – 1 = 0

On comparing with (i) we get,

a₁ = 5, b₁ = 8, c₁ = – 18; a₂ = 2, b₂ = – 3, c₂ = – 1

Applying cross multiplication method which says,

Putting the given values in the above equation we get,

Similarly,

∴The solution of the pair of equations is (2, 1).

Given equations are

, 7x— 15y = 21

Change the given equations to the form,

a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 …(i)

⇒

And7x — 15y – 21 = 0

⇒ 5x + 3y – 15 = 0 and 7x — 15y – 21 = 0

On comparing with (i) we get,

a₁ = 5, b₁ = 3, c₁ = – 15; a₂ = 7, b₂ = – 15, c₂ = – 21

Applying cross multiplication method which says,

Putting the given values in the above equation we get,

Similarly,

∴The solution of the pair of equations is (3, 0).

Given equations are

3x + y = 5, 5x + 3y = 3

Change the given equations to the form, a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 …(i)

⇒ 3x + y – 5 = 0 and

5x + 3y – 3 = 0

On comparing with (i) we get,

a_{1 =} 3, b₁ = 1, c₁ = – 5; a₂ = 5, b₂ = 3, c₂ = – 3

Applying cross multiplication method which says,

Putting the given values in the above equation we get,

Similarly,

∴The solution of the pair of equations is (3, – 4).

Let the required two digit number be (10x + y) where x is the digit at tens place and y is the digit at unit place.

On interchanging the digits the two digit number formed will be (10y + x).

According to the question,

y = 2x and (10y + x) = 36 + (10x + y)

⇒ 2x – y = 0 and 10x + y – 10 y – x + 36 = 0

⇒ 2x –y = 0 and 9x – 9y + 36 = 0

⇒ 2x – y = 0 and x – y + 4 = 0

On comparing with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0we get,

a_{1 =} 2, b₁ = – 1, c₁ = 0; a₂ = 1, b₂ = – 1, c₂ = 4

Applying cross multiplication method which says,

Putting the given values in the above equation we get,

Similarly,

∴The solution of the pair of equations is (4, 8).

So, the required two digit number will be (10x + y) = (10× 4 + 8) = 48

Let the two numbers be x and y.

According to the question,

x + y = 70 and x – y = 6

⇒ x + y – 70 = 0 and x –y – 6 = 0

On comparing with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0we get,

a_{1 =} 1, b₁ = 1, c₁ = – 70; a₂ = 1, b₂ = – 1, c₂ = – 6

Applying cross multiplication method which says,

Putting the given values in the above equation we get,

Similarly,

∴The solution of the pair of equations is (38, 32).

So, the required numbers will be 38 and 32.

Let the number of rows be x and number of students in each row = y.

⇒ Number of students arranged = x × y

According to the question,

If three rows are reduced, number of rows = (x – 3)

Then three more students have to be arranged in each row, number of students in each row = (y + 3)

Number of students arranged = (x – 3)(y + 3)

If three more rows are formed, number of rows = (x + 3)

Then two students have to be taken off from each row, number of students in each row = (y – 2)

Number of students arranged = (x + 3)(y – 2)

In above both the cases the number of students arranged will remain same.

⇒ (x – 3)(y + 3) = xy and (x + 3)(y – 2) = xy

⇒ xy + 3x – 3y – 9 = xy

And xy – 2x + 3y – 6 = xy

⇒ 3x – 3y – 9 = 0 and – 2x + 3y – 6 = 0

⇒ x – y – 3 = 0 and – 2x + 3y – 6 = 0

On comparing with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0we get,

a_{1 =} 1, b₁ = – 1, c₁ = – 3; a₂ = – 2, b₂ = 3, c₂ = – 6

Applying cross multiplication method which says,

Putting the given values in the above equation we get,

Similarly,

∴The solution of the pair of equations is (15, 12).

So, Number of students arranged = x×y = 15× 12 = 180.

Given ∠ B = 3 ∠ C …(i)

And …(ii)

Putting (i) in (ii),

⇒ ∠A = 2∠C

We know that the angle sum property of a triangle states that ∠ A + ∠ B + ∠ C = 180°

⇒ ∠ A + 3∠C + ∠ C = 180°

⇒ ∠ A + 4∠ C – 180° = 0

Consider ∠A – 2∠C = 0 and ∠ A + 4∠ C – 180 = 0

On comparing with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0we get,

a_{1 =} 1, b₁ = – 2, c₁ = 0; a₂ = 1, b₂ = 4, c₂ = – 180

Applying cross multiplication method which says,

Putting the given values in the above equation we get,

Similarly,

∴ ∠ B = 3∠C = 3× 30° = 90°

∵ One of the angles of this triangle is 90° i.e. ∠ B

∴ ∆ABC is a right angled triangle.

Putting these value in given equations, we get

So, the given equations transforms into linear equation in two variables i.e.

15a + 4b = 42 … (i)

3a + 2b = 12 …..(ii)

Multiply equation (ii) by 5, so

15a + 4b = 42 … (iii)

15a + 10b = 60 …..(ii)

Subtract (iii) from (ii),

15a + 10b – 15a – 4b = 60 – 42

⇒ 6b = 18

⇒ b = 3

Putting above value in (ii),

3a + 2× 3 = 12

⇒ 3a = 12 – 6

⇒ 3a = 6

⇒ a = 2

Given: 2x + 3y = 2xy, 6x + 12y = 7xy

One of the solutions for given pair of equations will be x = 0 and y = 0

But let us find out some other solution as well.

Divide both the equations with xy,

Putting these value in given equations, we get

2b + 3a = 2, 6b + 12a = 7

So, the given equations transforms into linear equation in two variables i.e.

2b + 3a = 2… (i)

6b + 12a = 7….. (ii)

Multiply equation (i) by 3, so

6b + 9a = 6 … (iii)

6b + 12a = 7….. (ii)

Subtract (iii) from (ii),

6b + 12a – 6b – 9a = 7 – 6

⇒ 3a = 1

Putting above value in (ii),

⇒ 2b = 2 – 1

Putting these value in given equations, we get

4a + 5b = 2, 8a + 15b = 3

So, the given equations transforms into linear equation in two variables i.e.

4a + 5b = 2 … (i)

8a + 15b = 3 …..(ii)

Multiply equation (i) by 2, so

8a + 10b = 4 … (iii)

8a + 15b = 3 …..(ii)

Subtract (iii) from (ii),

8a + 15b – 8a – 10b = 3 – 4

⇒ 5b = – 1

Putting above value in (i),

⇒ 4a = 2 + 1

⇒ – y + 1 = 5

⇒ y = – 4

Putting these value in given equations, we get

So, the given equations transforms into linear equation in two variables i.e.

4a + 4b = 3 … (i)

4a – 4b = 1 …..(ii)

Add (i) and (ii),

4a + 4b + 4a – 4b = 3 + 1

⇒ 8a = 4

Putting above value in (i),

⇒ 4b = 3 – 2

⇒ 4b = 1

… (iii)

…. (iv)

Add (iii) and (iv),

3x + y + 3x – y = 2 + 4

⇒ 6x = 6

⇒ x = 1

Putting the above value in (iii),

3(1) + y = 2

⇒ y = 3 – 2

⇒ y = 1

Putting these value in given equations, we get

So, the given equations transforms into linear equation in two variables i.e.

3a + 4b = 2 … (i)

60a + 84b = 41 …..(ii)

Multiply (i) by 20,

60a + 80b = 40 …. (iii)

60a + 84b = 41 …..(ii)

Subtract (iii) from (ii),

60a + 84b – 60a – 80b = 41 – 40

⇒ 4b = 1

Putting above value in (i),

⇒ 3a = 2 – 1

⇒ 3a = 1

Squaring both the sides,

x = 9

Similarly,

Squaring both the sides,

y = 16

Let W and M represent the work done by a woman and a man respectively.

It is given that 5 women and 2 men can do the work in 4 days.

So, work done by 5 women and 2 men in a day will be

⇒ 20W + 8M = 1 …. (i)

Also,

It is given that 6 women and 3 men can do the same work in 3 days.

So, work done by 6 women and 3 men in a day will be

⇒ 18W + 9M = 1 …. (ii)

From (i) and (ii) equation,

20W + 8M = 18W + 9M

⇒ 2W = M

This means that the work done by 2 women in a day is equal to the work done by a man in a day. Substituting the value of M in (i) equation, we get

20W + 8M = 1

⇒ 20W + 8× 2W = 1

⇒ (20 + 16) W = 1

So, a woman do work in a day.

This means a woman will finish the work in 36 days if working alone.

Now, put the value of W in (ii),

18W + 9M = 1

So, a man do work in a day.

This means a man will finish the work in 18 days if working alone.

Let the speed of the boat in still water be x km/hr and the speed of stream be y km/hr. It is necessary that x > y.

The speed of the boat in downstream = (speed of boat in still water + speed of stream) = (x + y) km/hr.

The speed of boat in upstream = (speed of boat in still water – speed of stream) = (x – y) km/hr.

Also, we know that

In the first case, it is given that boat goes 21 km upstream and 18 km downstream in 9 hours

⇒ Time taken by boat in upstream and downstream = 9 hr

In the second case, it is given that boat goes 30 km upstream and 27 km downstream in 13 hours

⇒ Time taken by boat in upstream and downstream = 13 hr

Putting these value in (i) and (ii), we get

21a + 18b = 9

30a + 27b = 13

So, the given equations transforms into linear equation in two variables

7a + 6b = 3 …. (iii)

30a + 27b = 13 … (iv)

Multiply (iii) by 9 and (iv) by 2,

63a + 54b = 27 …. (v)

60a + 54b = 26 … (vi)

Subtract (vi) from (v),

63a + 54b – 60a – 54b = 27 – 26

⇒ 3a = 1

Putting above value in (iii),

… (vii)

…. (viii)

Add above two equations,

x – y + x + y = 3 + 9

⇒ 2x = 12

⇒ x = 6

Putting x = 6 in (viii),

y = 9 – 6

⇒ y = 3

Putting these value in given equations, we get

4b + 7a = 16… (i), 10b + 3a = 11 ….. (ii)

So, the given equations transforms into linear equation in two variables.

Multiply equation (i) by 5 and (ii) by 2,

20b + 35a = 80 … (iii)

20b + 6a = 22….. (iv)

Subtract (iv) from (iii),

20b + 35a – 20b – 6a = 80 – 22

⇒ 29a = 58

⇒ a = 2

Putting above value in (ii),

10b + 3× 2 = 11

⇒ 10b = 11 – 6

Let the speed of the train and the bus be x km/hr and y km/hr respectively.

We know that

Total distance travelled by Mahesh = 250 km

In the first case it is given that he travels 50 km by train and 200 km by bus.

Given that total time taken by Mahesh to reach in this case = 6 hr

In the second case it is given that he travels 100 km by train and 150 km by bus.

Given that total time taken by Mahesh to reach in this case = 7 hr

Putting these value in above equations, we get

50a + 200b = 6 … (i), 100a + 150b = 7 ….. (ii)

So, the given equations transforms into linear equation in two variables.

Multiply equation (i) by 2,

100a + 400b = 12 … (iii)

100a + 150b = 7 ….. (ii)

Subtract (ii) from (iii),

100a + 400b – 100a – 150b = 12 – 7

⇒ 250b = 5

Putting above value in (ii),

Let the rate of tea per kg be Rs x and the rate of sugar per kg be Rs y.

According to the question,

x = 7y

⇒ x – 7y = 0 … (i)

Also, cost of 2 kg of tea = 2x and cost of 5 kg of sugar = 5y

⇒ 2x + 5y = 570 … (ii)

Hence, equation (i) and (ii) is the required pair of linear equation.

For drawing the given pair of linear equations we need few coordinates to plot of these equations.

In x + 3y = 6,

If x = 0 then 0 + 3y = 6

⇒ y = 2

B(0, 2)

If y = 0 then x + 0 = 6

⇒ x = 6

D(6, 0)

In 2x — y = 5,

If x = 0 then 0 – y = 5

⇒ y = – 5

C(0, – 5)

If y = 0 then 2x + 0 = 5

⇒ x = 2.5

E(2.5, 0)

Now plot B, C, D and E on graph,

We find that A(3, 1) is the intersecting point of the lines. Hence, {(3, 1)} is the solution set of the pair of linear equations.

Using elimination method:

Multiply (i) by 4 and (ii) by 5,

Subtract (iii) from (iv),

⇒ x = 2

Putting this value of x in (i),

Given equations are

(a + b)x + (a — b)y = a² + 2ab — b²

And (a — b) (x + y) = a² — b²

Change the given equations to the form,

a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 …(i)

⇒ (a + b)x + (a — b)y – a² – 2ab + b² = 0

And

(a – b)x + (a – b)y – a² + b² = 0

On comparing with (i) we get,

a_{1 =} (a + b),

b₁ = (a — b),

c_{1 =} – a² – 2ab + b²;

a₂ = (a – b),

b₂ = (a – b),

c₂ = – a² + b² = b² – a² = (b + a)(b – a) {Using a² – b² = (a + b)(a – b)}

Applying cross multiplication method which says,

Putting the given values in the above equation we get,

Similarly,

∴The solution of the pair of equations is (a, b).

Putting these value in given equations, we get

4a + 7b = 2, 2(10a + 14b) = 9

So, the given equations transforms into linear equation in two variables i.e.

4a + 7b = 2… (i)

20a + 28b = 9 …..(ii)

Multiply equation (i) by 4, so

16a + 28b = 8 … (iii)

20a + 28b = 9…..(ii)

Subtract (iii) from (ii),

20a + 28b – 16a – 28b = 9 – 8

⇒ 4a = 1

Putting above value in (i),

⇒ 7b = 2 – 1

⇒ y + 2 = 7

⇒ y = 5

Let the two natural numbers be x and y such that x is the larger number and y is the smaller number.

According to the question,

x – y = 6

and 2x + 10 = 3y – 2

⇒ x – y – 6 = 0 …(i)

And 2x – 3y + 12 = 0 ..(ii)

Multiply (i) by 2,

2x – 2y – 12 = 0 .. (iii)

2x – 3y + 12 = 0 ..(ii)

Subtract (iii) from (ii),

2x – 3y + 12 – 2x + 2y + 12 = 0

⇒ – y + 24 = 0

⇒ y = 24

Putting this value in (i),

x – 24 – 6 = 0

⇒ x = 30

So the numbers are 30 and 24.

Let the length be x units and breadth be y units of a rectangle.

⇒ Area of rectangle = x × y

According to the question,

If length is reduced by 3, length = (x – 3)

And breadth is increased by 5, breadth = (y + 5)

Area = (xy + 30)

⇒ (x – 3)(y + 5) = (xy + 30)

⇒ xy + 5x – 3y – 15 = xy + 30

⇒ 5x – 3y = 45 …(i)

If length is incresed by 5, length = (x + 5)

And breadth is reduced by 3, breadth = (y – 3)

Area = (xy – 10)

⇒ (x + 5)(y – 3) = (xy – 10)

⇒ xy – 3x + 5y – 15 = xy – 10

⇒ – 3x + 5y = 5 …(ii)

Multiply (i) by 3 and (ii) by 5,

15x – 9y = 135 .. (iii)

– 15x + 25y = 25 …(iv)

Add (iii) and (iv),

15x – 9y – 15x + 25y = 135 + 25

⇒ 16y = 160

⇒ y = 10

Putting this in (i),

5x – 3×10 = 45

⇒ 5x = 45 + 30

⇒ 5x = 75

⇒ x = 15

So, the length is 15 units and breadth is 10 units.

Let the fixed charge be Rs x and the cost of food per day be Rs y.

According to the question,

For Yash,

He took food for 25 days

⇒ x + 25 y = 2200..(i)

For Niyati,

She took food for 20 days

⇒ x + 20 y = 1800….(ii)

Subtract (ii) from (i),

x + 25y – x – 20 y = 2200 – 1800

⇒ 5y = 400

⇒ y = 80

Putting this value in (i),

x + 25× 80 = 2200

⇒ x = 2200 – 2000

⇒ x = 200

Hence, the fixed cost = Rs 200 and the cost for food per day = Rs 80

Let the required fraction be

According to the question,

⇒ 5x – 10 = 2y – 4

⇒ 5x – 2y = 6 …(i)

Also,

⇒ 4x + 20 = 3y + 15

⇒ 4x – 3y = – 5 …(ii)

Multiply (i) by 3 and (ii) by 2,

15x – 6y = 18 ..(iii)

8x – 6y = – 10 ….(iv)

Subtract (iv) from (iii),

15x – 6y – 8x + 6y = 18 + 10

⇒ 7x = 28

⇒ x = 4

Putting this in (i),

5× 4 – 2y = 6

⇒ – 2y = 6 – 20

⇒ – 2y = – 14

⇒ y = 7

Hence the required fraction is

Given: x — 3y = 1…(i)

3x + y = 3 …(ii)

Multiply (i) by 3,

3x – 9y = 3 ..(iii)

3x + y = 3…(ii)

Subtract (ii) from (iii),

3x – 9y – 3x – y = 3 – 3

⇒ – 10y = 0

⇒ y = 0

Putting this in (i),

x – 0 = 1

⇒ x = 1

So, the solution set will be {1, 0}.

Given: 2x + y = 6 and 4x + 2y = 5

Change the given equations to the form,

a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 …(i)

⇒ 2x + y – 6 = 0

And

4x + 2y – 5 = 0

On comparing with (i) we get,

a_{1 =} 2,

b₁ = 1,

c₁ = – 6;

a₂ = 4,

b₂ = 2,

c₂ = – 5;

This means that the equations are inconsistent and have no solution.

Given: 3x + y = 7 ..(i)and —x + 2y = 2 …(ii)

Using elimination method,

Multiply (ii) by 3,

3x + y = 7 …(i)

– 3x + 6y = 6 …(iii)

Add (i) and (iii),

3x + y – 3x + 6y = 7 + 6

⇒ 7y = 13

Hence, x is eliminated by multiplying second equation by 3.

Given: 2x + 3y = 7 …(i)

3x + 2y = 3 …(ii)

Subtract (i) from (ii),

3x + 2y – 2x – 3y = 3 – 7

⇒ x – y = – 4

Given: ax + 2y = 7 and 2x + 3y = 8

Change the given equations to the form,

a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 …(i)

⇒ ax + 2y – 7 = 0

And

2x + 3y – 8 = 0

On comparing with (i) we get,

a_{1 =} a,

b₁ = 2,

c₁ = – 7;

a₂ = 2,

b₂ = 3,

c₂ = – 8;

For unique solution,

Given: 2x + y — 3 = 0 and 6x + 3y = 9

Change the given equations to the form,

a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 …(i)

⇒ 2x + y – 3 = 0

And

6x + 3y – 9 = 0

On comparing with (i) we get,

a_{1 =} 2,

b₁ = 1,

c₁ = – 3;

a₂ = 6,

b₂ = 3,

c₂ = – 9;

This means that the equations are consistent and have infinitely many solutions.

A two digit number is of the form 10y + x where x is the digit at unit place and y is the digit at tens place.

According to the question,

y = 5

Then number will be 10 × 5 + x = 50 + x

Let the number be 10x + y where x is the digit at tens place and y is the digit at unit place.

According to the question,

x = 7

Sum of the digits = 7 + y

⇒ 7 + y = 8y

⇒ 7y = 7

⇒ y = 1

So, the number = 71

Let the two numbers be x and y.

According to the question,

x + y = 10 ..(i)

x – y = 2 ..(ii)

Adding (i) and (ii),

x + y + x –y = 10 + 2

⇒ 2x = 12

⇒ x = 6

Putting x = 6 in (i),

6 + y = 10

⇒ y = 4

Then the greater number is 6.

Let the present ages of father and son be x years and y years respectively.

Age of father three years ago = x – 3

Age of son three years ago = y – 3

According to the question,

x – 3 + y – 3 = 40

⇒ x + y = 40 + 6

⇒ x + y = 46 .. (i)

After 2 years,

Age of father = x + 2

Age of son = y + 2

Sum of their ages after 2 years = x + 2 + y + 2 = (x + y) + 4 = 46 + 4 = 50 {Using (i)}

Given: of 2x + 4y = 8 …(i) and

x + 2y = 4 …(ii)

Using elimination method,

Multiply (ii) by 2,

2x + 4y = 8 …(i)

2x + 4y = 8 …(iii)

Since both the equations are same therefore they have infinite solution.

Standard form of linear equation in two variable: ax + by + c = 0 where a, b and c are the constants.

Taking LCM (2, 3) = 6 in the denominator,

⇒ 3x – 2y = 6

⇒ 3x – 2y – 6 = 0

This is the standard form of the given equation.