NCERT Solutions for Class 10 Maths Chapter 3: Pair of Linear Equations in Two Variables

Let the present age of Aftab and his daughter be x and y respectively.
Seven years ago, Age of Aftab = x – 7 and Age of his daughter = y – 7
According to the given condition,

Thus, the given conditions can be algebraically represented as:
x – 7y = –42

And x – 3y = 6

The graphical representation is as follows:

 

 

Let the cost of a bat and a ball be Rs x and Rs y respectively.
The given conditions can be algebraically represented as:
3x + 6y = 3900
x + 2y = 1300

Three solutions of this equation can be written in a table as follows:

Three solutions of this equation can be written in a table as follows:
 

The graphical representation is as follows:

 

Let cost of 1 kg of apples = Rs x and let cost of 1 kg of grapes= Rs y

According to given conditions, we have

2x + y = 160… (1)

4x + 2y = 300

⇒ 2x + y = 150… (2)

So, we have equations (1) and (2), 2x + y = 160 and 2x + y = 150 which represent given situation algebraically.

For equation 2x + y = 160, we have following points which lie on the line

We plot the points for both of the equations and it is the graphical representation of the given situation.

 

(i) Let number of boys = x

Let number of girls = y

According to given conditions, we have

x + y = 10

And, x = 10 - y

putting y=0,5,10,we get,
X=10-0=10
X=10-5=5,
X=10-10=0

 

Number of girls is 4 more than number of boys ........Given,
so,
Y=x+4
putting x=-4,0,4 we get,
Y=-4+4=0
Y=0+4
Y=4+4=8

We plot the points for both of the equations to find the solution.

 

 

(ii)

Let the cost of one pencil=Rs.X
and Let the cost of one pen=Rs.Y
According to the given conditions, we have:
=5x + 7y = 50
=5x=50-7y
=x=10-7/5y

Three solutions of this equation can be written in a table as follows:

Three solutions of this equation can be written in a table as follows:

The graphical representation is as follows:

 

(i) 5x − 4y + 8 = 0, 7x + 6y – 9 = 0

Comparing equation 5x − 4y + 8 = 0 with a1x + b1y + c1 = 0 and 7x + 6y – 9 = 0 with a2x + b2y + c2 = 0,

We get,

Hence,

we find that,

(ii) 9x + 3y + 12 = 0, 18x + 6y + 24 = 0

Comparing equation 9x + 3y + 12 = 0 with a1x + b1y + c1 = 0 and 7x + 6y – 9 = 0 with a2x + b2y + c2 = 0,

We get,

Hence

We find that,

Hence, lines are coincident.

(iii) 6x − 3y + 10 = 0, 2x – y + 9 = 0

Comparing equation 6x − 3y + 10 = 0 with a1x + b1y + c1 = 0 and 7x + 6y – 9 = 0 with a2x + b2y + c2 = 0,

We get,

Hence

We find that,

Hence,
lines are parallel to each other.

 

 

(i) 3x + 2y = 5, 2x − 3y = 7

Comparing equation 3x + 2y = 5 with a1x + b1y + c1 = 0 and 7x + 6y – 9 = 0 with a2x + b2y + c2 = 0,

We get,

Hence,

Therefore,these linear equations will intersect at one point only and have only one possible solution.
And,pair of linear equations is consistent

(ii) 2x − 3y = 8, 4x − 6y = 9

Comparing equation 2x − 3y = 8 with a1x + b1y + c1 = 0 and 7x + 6y – 9 = 0 with a2x + b2y + c2 = 0,

We get,

Hence,

Therefore,these linear equations are parallel to each other and have no possible solution.in
And,pair of linear equations is inconsistent

(iii) 9x − 10y = 14

We get,

Hence,

Therefore, these linear equations will intersect each other at one point and have only one possible solution.
And,pair of linear equations is consistent

(iv) 5x − 3y = 11, −10x + 6y = −22

Comparing equation 5x − 3y = 11 with a1x + b1y + c1 = 0 and 7x + 6y – 9 = 0 with a2x + b2y + c2 = 0,

We get,

Hence,

Therefore these pair of lines have infinite number of solutions
And,pair of linear equations is consistent

 

(i) x + y = 5, 2x + 2y = 10

We get,
Hence,

(ii) x – y = 8, 3x − 3y = 16

 

We get,

Hence,

Therefore, these linear equations are intersecting each other at one point and thus have only one possible solution.
Hence,the pair of linear equations is inconsistent.

(iii) 2x + y = 6, 4x − 2y = 4

We get,

Hence,

Therefore, these linear equations are intersecting each other at one point and thus have only one possible solution.
Hence,the pair of linear equations is consistent

 

(iv) 2x − 2y – 2 = 0, 4x − 4y – 5 = 0

We get,

Hence,

Therefore, these linear equations are parallel to each other and have no possible solution,
Hence,the pair of linear equations is inconsistent.

 

Let width of rectangular garden = x metre
and length=y

So,

Hence, the graphic representation is as follows.

 

(i) Let the second line be equal to a2x + b2y + c2 = 0,

Intersecting Lines:For this Condition,

The Second line such that it is intersecting the given line is
2x+4y-6=0

As,

(ii) Let the second line be equal to a2x + b2y + c2 = 0,

parallel Lines:
For this Condition,

Hence,the second line can be 4x+6y-8=0

As,

(iii) Let the second line be equal to a2x + b2y + c2 = 0,

Coincident lines: For coincident lines,

Hence,the second line can be 6x+9y-24=0

As,

 

For equation x – y + 1 = 0, we have following points which lie on the line

For equation 3x + 2y – 12 = 0, we have following points which lie on the line

 

(i) x + y = 14 …(1)

x – y = 4 … (2)

x = 4 + y from equation (2)

Putting this in equation (1), we get

4 + y + y = 14

⇒ 2y = 10⇒ y = 5

Putting value of y in equation (1), we get

x + 5 = 14

⇒ x = 14 – 5 = 9

Therefore, x = 9 and y = 5

(ii) s – t = 3 … (1)

…(2)

Using equation (1), we can say that s = 3 + t

Putting this in equation (2), we get

⇒ 2t + 6 + 3t =36

⇒ 5t + 6 = 36

⇒ 5t = 30⇒ t = 6

Putting value of t in equation (1), we get

s – 6 = 3⇒ s = 3 + 6 = 9

Therefore, t = 6 and s = 9

(iii) 3x – y = 3 … (i)

9x − 3y = 9 … (ii)

From equation (i),we get,
y =3x − 3 … (iii)

putting value of y from equation(iii) to equation(ii)
9x-3(3x-3)=9
=9x-9x+9=9
=9=9

This is always true,and pair of these equations have infinite possible solutions.
Therefore one possible solutions is x=1 and y=0

(iv) 0.2x + 0.3y = 1.3 … (1)

0.4x + 0.5y = 2.3 … (2)

Using equation (1), we can say that

0.2x = 1.3 − 0.3y

⇒ x =

Putting this in equation (2), we get

0.4x(6.5-1.5y)+ 0.5y = 2.3

⇒ 2.6 − 0.6y + 0.5y = 2.3

⇒ −0.1y = −0.3 ⇒ y = 3

Putting value of y in (1), we get

0.2x + 0.3 (3) = 1.3

⇒ 0.2x + 0.9 = 1.3

⇒ 0.2x = 0.4 ⇒ x = 2

Therefore, x = 2 and y = 3

(v) ……….(1)

……….(2)

Using equation (1), we can say that

x =

Putting this in equation (2), we get

⇒

⇒ ⇒ y = 0

Putting value of y in (1), we get x = 0

Therefore, x = 0 and y = 0

(vi)… (1)

… (2)

Using equation (2), we can say that

⇒ x =

Putting this in equation (1), we get

⇒

⇒

⇒

⇒⇒ y = 3

Putting value of y in equation (2), we get

⇒

⇒

⇒ x = 2

Therefore, x = 2 and y = 3

2x + 3y = 11 … (1)

2x − 4y = −24 … (2)

Using equation (2), we can say that

2x = −24 + 4y

⇒ x = −12 + 2y

Putting this in equation (1), we get

2 (−12 + 2y) + 3y = 11

⇒ −24 + 4y + 3y = 11

⇒ 7y = 35 ⇒ y = 5

Putting value of y in equation (1), we get

2x + 3 (5) = 11

⇒ 2x + 15 = 11

⇒ 2x = 11 – 15 = −4⇒ x = −2

Therefore, x = −2 and y = 5

Putting values of x and y in y = mx + 3, we get

5 = m (−2) + 3

⇒ 5 = −2m + 3

⇒ −2m = 2 ⇒ m = −1

Form a pair of linear equations for the following problems and find their solution by substitution method.

 

(i) Let first number be x and second number be y.

According to given conditions, we have

x – y = 26 (assuming x > y) … (1)

x = 3y(x > y)… (2)

Putting equation (2) in (1), we get

3y – y = 26

⇒ 2y = 26

⇒ y = 13

Putting value of y in equation (2), we get

x = 3y =

Therefore, two numbers are 13 and 39.

(ii) Let smaller angle =x and let larger angle =y

According to given conditions, we have

y = x + 18 … (1)

Also, (Sum of supplementary angles) … (2)

Putting (1) in equation (2), we get

x + x + 18 = 180

⇒ 2x = 180 – 18 = 162

⇒ x=162/2
x=81

Putting value of x in equation (1), we get

y = x + 18 = 81 + 18 =

Therefore, two angles are .

(iii) Let cost of each bat = Rs x and let cost of each ball = Rs y

According to given conditions, we have

7x + 6y = 3800 … (1)

And,3x + 5y = 1750 … (2)

Using equation (1), we can say that 7x+6y=3800
= 6y = 3800-7x

And,

Putting value of y from equation(1) to equation (2)

Therefore, cost of each bat = Rs 500 and cost of each ball = Rs 50

(iv) Let fixed charge = Rs x and let charge for every km = Rs y

According to given conditions, we have

x + 10y = 105… (1)

x + 15y = 155… (2)

Using equation (1), we can say that

x = 105 − 10y

Putting this in equation (2), we get

105 − 10y + 15y = 155

⇒ 5y = 50 ⇒ y = 10

Putting value of y in equation (1), we get

x + 10 (10) = 105

⇒ x = 105 – 100 = 5

Therefore, fixed charge = Rs 5 and charge per km = Rs 10

To travel distance of 25 Km, person will have to pay = Rs (x + 25y)

= Rs (5 + 25 × 10)

= Rs (5 + 250) = Rs 255

(v) Let numerator = x and let denominator = y

According to given conditions,
Fraction becomes 9/11 , if 2 is added in both numerator and denominator

=11x+22 =9y+18 (by cross multiplication)
=11x =9y - 4

And, if 3 is added in both numerator and denominator

=6x+18 =5y+15 .....(ii) (by cross multiplication)

putting value of x from equation (i) to equation (ii)

Putting value of y in equation (i), we get

(vi) Let present age of Jacob = x years

 

Let present age of Jacob’s son = y years

According to given conditions, we have

(x + 5) = 3 (y + 5) … (1)

And, (x − 5) = 7 (y − 5) … (2)

From equation (1), we can say that

x + 5 = 3y + 15

⇒ x = 10 + 3y

Putting value of x in equation (2) we get

10 + 3y – 5 = 7y − 35

⇒ −4y = −40

⇒ y = 10 years

Putting value of y in equation (1), we get

x + 5 = 3 (10 + 5)

⇒ x = 45 – 5 = 40 years

Therefore, present age of Jacob = 40 years and, present age of Jacob’s son = 10 years

 

(i) x + y = 5 … (1)

2x – 3y = 4 … (2)

Elimination method:

Multiplying equation (1) by 2, we get equation (3)

2x + 2y = 10 … (3)

2x − 3y = 4 … (2)

Subtracting equation (2) from (3), we get

5y = 6⇒ y = 6/5

Putting value of y in (1), we get

x + 6/5= 5

⇒ x = 5 − 6/5= 19/5

Therefore, x = 19/5 and y = 6/5

Substitution method:

x + y = 5 … (1)

2x − 3y = 4 … (2)

From equation (1), we get,

x = 5 − y

Putting this in equation (2), we get

2 (5 − y) − 3y = 4

⇒ 10 − 2y − 3y = 4

⇒ 5y = 6 ⇒ y = 6/5

Putting value of y in (1), we get

x = 5 − 6/5= 19/5

Therefore, x = 19/5 and y = 6/5

(ii) 3x + 4y = 10… (1)

2x – 2y = 2… (2)

Elimination method:

Multiplying equation (2) by 2, we get (3)

4x − 4y = 4 … (3)

3x + 4y = 10 … (1)

Adding (3) and (1), we get

7x = 14⇒ x = 2

Putting value of x in (1), we get

3 (2) + 4y = 10

⇒ 4y = 10 – 6 = 4

⇒ y = 1

Therefore, x = 2 and y = 1

Substitution method:

3x + 4y = 10… (1)

2x − 2y = 2… (2)

From equation (2), we get

2x = 2 + 2y

⇒ x = 1 + y … (3)

Putting this in equation (1), we get

3 (1 + y) + 4y = 10

⇒ 3 + 3y + 4y = 10

⇒ 7y = 7⇒ y = 1

Putting value of y in (3), we get x = 1 + 1 = 2

Therefore, x = 2 and y = 1

(iii) 3x − 5y – 4 = 0 … (1)

9x = 2y + 7… (2)

Elimination method:

Multiplying (1) by 3, we get (3)

9x − 15y – 12 = 0… (3)

9x − 2y – 7 = 0… (2)

Subtracting (2) from (3), we get

−13y – 5 = 0

⇒ −13y = 5

⇒ y = −5/13

Putting value of y in (1), we get

3x – 5 (−5/13)− 4 = 0

⇒ 3x = 4 −

⇒ x =

Therefore, x = and y =

Substitution Method:

3x − 5y – 4 = 0 … (1)

9x = 2y + 7… (2)

From equation (1), we can say that

3x = 4 + 5y⇒ x =

Putting this in equation (2), we get

9 − 2y = 7

⇒ 12 + 15y − 2y = 7

⇒ 13y = −5 ⇒ y =

Putting value of y in (1), we get

3x – 5 = 4

⇒ 3x = 4 −

⇒ x =

Therefore, x = and y =

(iv) … (1)

… (2)

Elimination method:

Multiplying equation (1) by 2, we get

… (3)

substracting equation (2)from (3) we get

Putting value of x in (2), we get

=x-4=-2
=x=2

Substitution method:

… (1)

… (2)

From equation (2), we can say that

Putting this in equation (1), we get

Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method:

 

(i) Let numerator =x and let denominator =y

According to given condition, we have

⇒ x + 1 = y – 1 and 2x = y + 1

⇒ x – y = −2 … (1) and 2x – y = 1… (2)

So, we have equations (1) and (2), multiplying equation (1) by 2 we get (3)

2x − 2y = −4… (3)

2x – y = 1… (2)

Subtracting equation (2) from (3), we get

−y = −5⇒ y = 5

Putting value of y in (1), we get

x – 5 = −2⇒ x = −2 + 5 = 3

Therefore, fraction =

(ii) Let present age of Nuri = x years and let present age of Sonu = y years

5 years ago, age of Nuri = (x – 5) years

5 years ago, age of Sonu = (y – 5) years

According to given condition, we have

(x − 5) = 3 (y − 5)

⇒ x – 5 = 3y – 15

⇒ x − 3y = −10… (1)

10 years later from present, age of Nuri = (x + 10) years

10 years later from present, age of Sonu = (y + 10) years

According to given condition, we have

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

⇒ x + 10 = 2y + 20

⇒ x − 2y = 10 … (2)

Subtracting equation (1) from (2), we get

y = 10 − (−10) = 20 years

Putting value of y in (1), we get

x – 3 (20) = −10

⇒ x – 60 = −10

⇒ x = 50 years

Therefore, present age of Nuri = 50 years and present age of Sonu = 20 years

(iii) Let digit at ten’s place = x and Let digit at one’s place = y

According to given condition, we have

x + y = 9 … (1)

And 9 (10x + y) = 2 (10y + x)

⇒ 90x + 9y = 20y + 2x

⇒ 88x = 11y

⇒ 8x = y

⇒ 8x – y = 0 … (2)

Adding (1) and (2), we get

9x = 9⇒ x = 1

Putting value of x in (1), we get

1 + y = 9

⇒ y = 9 – 1 = 8

Therefore, number = 10x + y = 10 (1) + 8 = 10 + 8 = 18

(iv) Let number of Rs 100 notes = x and let number of Rs 50 notes = y

According to given conditions, we have

x + y = 25 … (1)

and 100x + 50y = 2000

⇒ 2x + y = 40 … (2)

Subtracting (2) from (1), we get

−x = −15⇒ x = 15

Putting value of x in (1), we get

15 + y = 25

⇒ y = 25 – 15 = 10

Therefore, number of Rs 100 notes = 15 and number of Rs 50 notes = 10

(v) Let fixed charge for 3 days = Rs x

Let additional charge for each day thereafter = Rs y

According to given condition, we have

x + 4y = 27 … (1)

x + 2y = 21 … (2)

Subtracting (2) from (1), we get

2y = 6⇒ y = 3

Putting value of y in (1), we get

x + 4 (3) = 27

⇒ x = 27 – 12 = 15

Therefore, fixed charge for 3 days = Rs 15 and additional charge for each day after 3 days = Rs 3

 

(i) Comparing equation 2x + 3y – 7 = 0 with a1x +b1y + c1 = 0 and (a − b) x + (a + b)

y − 3a – b + 2 = 0 with a2x +b2y + c2 = 0

We get andand

Linear equations have infinite many solutions if

⇒

⇒

⇒ 2a + 2b = 3a − 3b and 6 − 3b − 9a = −7a − 7b

⇒ a = 5b… (1) and −2a = −4b – 6… (2)

Putting (1) in (2), we get

−2 (5b) = −4b – 6

⇒ −10b + 4b = −6

⇒ −6b = –6 ⇒ b = 1

Putting value of b in (1), we get

a = 5b = 5 (1) = 5

Therefore, a = 5 and b = 1

(ii) Comparing (3x + y – 1 = 0) with a1x +b1y + c1 = 0 and (2k − 1)x + (k − 1)y −2k – 1 = 0) with a2x +b2y + c2 = 0,

We get andand c2 = −2k − 1

Linear equations have no solution if

⇒

⇒

⇒ 3 (k − 1) = 2k – 1

⇒ 3k – 3 = 2k − 1

⇒ k = 2

Substitution Method:

8x + 5y = 9 … (i)

3x + 2y = 4 … (ii)

From equation (ii),we get

x = … (iii)

Substituting this value in equation (i), we obtain

32-16y+15y =27
-y =-5

Y = 5 … (iv)

Substituting this value in equation (ii), we obtain

 

3x+10=4
x=-2
Hence, x=-2,y=5

Cross multiplication method

8x + 5y = 9 … (1)

3x + 2y = 4 … (2)

⇒

⇒

⇒ x = −2 and y = 5

 

Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method:

 

 

(i) x − 3y – 3 = 0

3x − 9y – 2 = 0

Comparing equation x − 3y – 3 = 0 with a1x +b1y + c1 = 0 and 3x − 9y – 2 = 0 with a2x +b2y + c2 = 0,

We get

Here

Therefore, the given sets of lines are parallel to each other. Therefore, they will not intersect each other and thus, there will not be any solution for these equations.

(ii) 2x + y = 5

3x + 2y = 8

Comparing equation 2x + y -5= 0 with a1x +b1y + c1 = 0 and 3x + 2y -8= 0 with a2x +b2y + c2 = 0,

We get

Herethis means that there is unique solution for the given equations.

By cross-multiplication method,

⇒

⇒ x = 2 and y = 1

(iii) 3x − 5y = 20

6x − 10y = 40

Comparing equation 3x − 5y = 20 with a1x +b1y + c1 = 0 and 6x − 10y = 40 with a2x +b2y + c2 = 0 ,

We get

Here

It means lines coincide with each other.

Hence, there are infinite many solutions.

(iv) x − 3y – 7 = 0

3x − 3y – 15 = 0

Comparing equation x − 3y – 7 = 0 with a1x +b1y + c1 = 0 and 3x − 3y – 15 = 0 with a2x +b2y + c2 = 0,

We get

Here this means that we have unique solution for these equations.

By cross-multiplication,

⇒

⇒

⇒ x = 4 and y = –1

(i)Let fixed monthly charge = Rs x and let charge of food for one day = Rs y

According to given conditions,

x + 20y = 1000 … (1),

and x + 26y = 1180 … (2)

Subtracting equation (1) from equation (2), we get

6y = 180

⇒ y = 30

Putting value of y in (1), we get

x + 20 (30) = 1000

⇒ x = 1000 – 600 = 400

Therefore, fixed monthly charges = Rs 400 and, charges of food for one day = Rs 30

 

(ii) Let numerator = x and let denominator = y

According to given conditions,

⇒ 3x – 3 = y … (1) 4x = y + 8 … (1)

⇒ 3x – y = 3 … (1) 4x – y = 8 … (2)

Subtracting equation (1) from (2), we get

4x – y − (3x − y) = 8 – 3

⇒ x = 5

Putting value of x in (1), we get

3 (5) – y = 3

⇒ 15 – y = 3

⇒ y = 12

Therefore, numerator = 5 and, denominator = 12

 

(iii) Let number of correct answers = x and let number of wrong answers = y

According to given conditions,

3x – y = 40 … (1)

And, 4x − 2y = 50 … (2)

From equation (1), y = 3x − 40

Putting this in (2), we get

4x – 2 (3x − 40) = 50

⇒ 4x − 6x + 80 = 50

⇒ −2x = −30

⇒ x = 15

Putting value of x in (1), we get

3 (15) – y = 40

⇒ 45 – y = 40

⇒ y = 45 – 40 = 5

Therefore, number of correct answers = x = 15and number of wrong answers = y = 5

Total questions = x + y = 15 + 5 = 20

 

(iv)Let speed of car which starts from part A = x km/hr

Let speed of car which starts from part B = y km/hr

According to given conditions,5(x-y) =100(Assuming x > y)

(Assuming x > y)

 

⇒ 5x − 5y = 100

⇒ x – y = 20 … (1)

And,1(x+y) =100

⇒ x + y = 100 … (2)

Adding (1) and (2), we get

2x = 120

⇒ x = 60 km/hr

Putting value of x in (1), we get

60 – y = 20

⇒ y = 60 – 20 = 40 km/hr

Therefore, speed of car starting from point A = 60 km/hr

And, Speed of car starting from point B = 40 km/hr

 

(v) Let length of rectangle = x units and Let breadth of rectangle = y units

Area =xy square units. According to given conditions,

xy – 9 = (x − 5) (y + 3)

⇒ xy – 9 = xy + 3x − 5y – 15

⇒ 3x − 5y = 6 … (1)

And, xy + 67 = (x + 3) (y + 2)

⇒ xy + 67 = xy + 2x + 3y + 6

⇒ 2x + 3y = 61 … (2)

By cross-multiplication method,we obtain,

Therefore, length = 17 units and, breadth = 9 units

 

(i) … (1)

… (2)

Let =p and =q

then the equations becomes.

Using cross-multiplication method,we obtain,

⇒p = 2 andq = 3

(ii)

Let =p and =q

Putting this in (1) and (2), we get

2p + 3q = 2 … (i)

4p − 9q = −1 … (ii)

Multiplying (i) by 3, we get

6p + 9q =6 … (iii)

Adding equation (ii) and (iii),we obtain

10p =5

p =

Putting value of p in (i), we get

Hence,x = 4 and y = 9.

(iii) + 3y = 14

− 4y = 23 and Let =p

Putting value of p in equation, we get

4p + 3y = 14 … (i)

3p − 4y = 23 … (ii)

By cross-multiplication,we get

Also,

(iv)

putting= p and=q,we get,

=5p+q =2 … (i)
=6p-3q =1 … (ii)

Now, multiplying equation (i) by 3,we get

15p +3q = 6 … (iii)

Adding equation (ii) and (iii)

21 p = 7

p =

putting value of p in equation (iii),we get,

= -3q = -1

q =

we know that,
p =
= 3 = x - 1
= x = 4

q =

=y - 2 = 3
y = 5

Hence,x=4 and y =5

(v) 7x − 2y = 5xy … (1)

8x + 7y = 15xy … (2)

Dividing both the equations by xy, we get

… (i)

… (ii)

putting =p and =q

we get,

7q − 2p = 5 … (iii)

8q + 7p = 15 … (iv)

multiplying equation (iii) by 7 and equation (iv) by 2,we get,

49q - 14p =5 … (v)

16q + 14p = 30 … (vi)

After adding equation (v) and (vi),we get,
65q = 65
=q = 1

putting value of q in equation (iv), we get,

8 + 7p = 15
= 7p = 15 -8 = 7
= p =1

Now,
p =

q =

Hence,x = 1 and y = 1

(vi) 6x + 3y − 6xy = 0 … (1)

2x + 4y − 5xy = 0 … (2)

Dividing both the equations by xy, we get

Let =p and =q

Putting these in (3) and (4), we get

6q + 3p – 6 = 0 … (5)

2q + 4p – 5 = 0 … (6)

From (5),

3p = 6 − 6q

⇒p = 2 − 2q

Putting this in (6), we get

2q + 4 (2 − 2q) – 5 = 0

⇒ 2q + 8 − 8q – 5 = 0

⇒ −6q = −3⇒q = ½

Putting value of q in (p = 2 – 2q), we get

p = 2 – 2 (½) = 2 – 1 = 1

Putting values of p and q in (=p and =q), we getx = 1 andy = 2

(vii) … (1)

…(2)

Let

Putting this in (1) and (2), we get

10p + 2q = 4 … (3)

15p − 5q = −2 … (4)

From equation (3),

2q = 4 − 10p

⇒q = 2 − 5p … (5)

Putting this in (4), we get

15p – 5 (2 − 5p) = −2

⇒ 15p – 10 + 25p = −2

⇒ 40p = 8⇒p =

Putting value of p in (5), we get

q = 2 – 5 () = 2 – 1 = 1

Putting values of p and q in (), we get

⇒x +y = 5 … (6) andx –y = 1 … (7)

Adding (6) and (7), we get

2x = 6 ⇒x = 3

Puttingx = 3 in (7), we get

3 –y = 1

⇒y = 3 – 1 = 2

Therefore,x = 3 andy = 2

(viii)… (1)

… (2)

Let

Putting this in (1) and (2), we get

p +q = and

⇒ 4p + 4q = 3 … (3) and 4p − 4q = −1 … (4)

Adding (3) and (4), we get

8p = 2 ⇒p = ¼

Putting value of p in (3), we get

4 (¼) + 4q = 3

⇒ 1 + 4q = 3

⇒ 4q = 3 – 1 = 2

⇒q = ½

Putting value of p and q,we get,

⇒ 3x +y = 4 … (5) and 3x –y = 2 … (6)

Adding (5) and (6), we get

6x = 6 ⇒x = 1

Puttingx = 1 in (5) , we get

3 (1) +y = 4

⇒y = 4 – 3 = 1

Therefore,x = 1 and y = 1

(i) Let speed of rowing in still water =x km/h

Let speed of current =y km/h

So, speed of rowing downstream = (x +y) km/h

And, speed of rowing upstream = (x −y) km/h

According to given conditions,

2(x+y) = 20 and 2(x-y) = 4

⇒ 2x + 2y = 20 and 2x − 2y = 4

⇒x +y = 10 … (1) andx –y = 2 … (2)

Adding (1) and (2), we get

2x = 12⇒x = 6

Puttingx = 6 in (1), we get

6 +y = 10

⇒y = 10 – 6 = 4

Therefore, speed of rowing in still water = 6 km/h

Speed of current = 4 km/h

(ii) Let time taken by 1 woman alone to finish the work =x days

Let time taken by 1 man alone to finish the work =y days

Therefore,work done by a woman in 1 day =

work done by a man in 1 day =

According to the question,

putting = p and = q, in these equations
we obtain,

2p + 5q =
8p + 20q = 1

3p + 6q =
= 9p + 18q = 1

By cross-multiplication,we obtain

x = 18, y = 36
Hence, number of days taken by a woman = 18 Number of days taken by a man = 36

(iii) Let speed of train =x km/h and let speed of bus =y km/h

According to given conditions,

putting in these equations,we obtain

Putting this in the above equations, we get

60p + 240q = 4 … (3)

And 100p + 200q =

600p + 1200q = 25 … (4)

Multiplying (3) by 10, we get

600p +2400q = 40 … (5)

substracting equation (4) from (5),we obtain
1200q =15
q =… (6)

Substituting in equation(3),we obtain
60p + 3 = 4
60p = 1

u = 60km/h and v = 80km/h
Hence,speed of train = 60 km/h
speed of bus = 80 km/h

 

 

C = 3B = 2(A + B)

Taking 3B = 2(A + B)

B = 2A

2 -B = 0 ….(i)

We know that the sum of the measures of all angles of a triangle is 180°.

A + B + C =

A + B + 3B =

A + 4B = …….(2)

Multiplying equation (1) by 4, we obtain:

8A – 4B = 0 …….(3)

Adding equations (2) and (3), we get

9A =

A =

From eq. (2), we get,

+ 4B =

4B =

B =

C = 3B

3* =

Hence the measures of A, B and C are respectively.

 

The difference between the ages of Biju and Ani is 3 years. Either Biju is 3 years older than Ani or Ani is 3 years older than Biju. However, it is obvious that in both cases, Ani's father's age will be 30 years more than that of Cathy's age.

Let the age of Ani and Biju be x and y years respectively.

Therefore, age of Ani's father, Dharam = 2 × x = 2x years

Biju's sister Cathy And age of = years

By using the information given in the question,

Case (I) When Ani is older than Biju by 3 years, x - y = 3 (i)

2x -= 30

4x - y = 60 (ii)

Subtracting (i) from (ii), we obtain 3x = 60 - 3 = 57

x -= 19

Therefore, age of Ani = 19 years

And age of Biju = 19 - 3 = 16 years

Case (II) When Biju is older than Ani, y - x = 3 (i)

2x -= 30

4x - y = 60 (ii)

Adding (i) and (ii), we obtain 3x = 63

x = 21

Therefore, age of Ani = 21 years

And age of Biju = 21 + 3 = 24 years

 

Let the money with the first person and second person be Rs x and Rs y respectively.

According to the question,
x + 100 = 2(y - 100)
x + 100 = 2y - 200
x - 2y = - 300    ... (1)

6(x - 10) = (y + 10)
6x - 60 = y + 10
6x - y = 70        ... (2)

Multiplying equation (2) by 2, we obtain:
12x - 2y = 140    ... (3)
Subtracting equation (1) from equation (3), we obtain:
11x = 140 + 300
11x = 440
x = 40
Putting the value of x in equation (1), we obtain:
40 - 2y = -300
40 + 300 = 2y
2y = 340
y = 170

Thus, the two friends had Rs 40 and Rs 170 with them.

Let the speed of the train be x km/h and the time taken by train to travel the given distance be t hours and the distance to travel be d km.

We know that, Speed =

x =

Or, d = xt (i)

Using the information given in the question, we obtain

(x + 10) =

(x+10) (t-2) = d

Xt+10t-2x-20 = d

By using equation (i),

we obtain - 2x + 10t = 20 (ii)

(x-10) =

(x-10) (t+3) = d

Xt-10t+3x-30 = d

By using equation (i), we

obtain 3x - 10t = 30 (iii)

Adding equations (ii) and (iii),

we obtain x = 50

 

Using equation (ii), we

obtain ( - 2) × (50) + 10t = 20

- 100 + 10t =

10t = 120

t = 12 hours

From equation (i), we obtain

Distance to travel = d = xt

= 50 × 12

= 600 km

Hence, the distance covered by the train is 600 km.

 

Let the number of rows be x and the number of students in a row be y.
Total number of students in the class = Number of rows x Number of students in a row
= xy


According to the question,
Total number of students = (x - 1) (y + 3)
xy = (x - 1) (y + 3)

= xy - y + 3x - 3
3x - y - 3 = 0
3x - y = 3 ... (1)

Total number of students = (x + 2) (y - 3)
xy = xy + 2y - 3x - 6
3x - 2y = -6 ... (2)

Subtracting equation (2) from (1), we obtain:
y = 9

Substituting the value of y in equation (1), we obtain:
3x - 9 = 3
3x = 9 + 3 = 12
x = 4

Number of rows = x = 4
Number of students in a row = y = 9

Total number of students in a class = xy = 4 x 9 = 36

5x - y =5

= y = 5x - 5

Three solutions of this equation can be written in a table as follows:

3x - y = 3 or,
y = 3x – 3
The solution table will be as follows.

The graphical representation of these lines will be as follows.

It can be observed that the required triangle is ΔABC formed by these lines and y-axis. The coordinates of vertices are A (1, 0), B (0, - 3), C (0, - 5).

(i) px + qy = p - q … (1)
qx - py = p + q … (2)

Multiplying equation (1) by p and equation (2) by q,

we obtain p2x + pqy = p2 - pq … (3)
q2x - pqy = pq + q2 … (4)
Adding equations (3) and (4),
we obtain p2x + q2 x = p2 + q2
(p2 + q2) x = p2 + q2

x == 1

From equation (1),

we obtain p (1) + qy = p - q
qy = - q,so, y = - 1

(ii) ax + by = c … (1) bx + ay = 1 + c … (2)
Multiplying equation (1) by a and equation (2) by b, we obtain:
… (3)
… (4)
Subtracting equation (4) from equation (3),

From equation (1), we obtain ax + by = c

(iii)

……..(1)

……..(2)

Multiplying equation (1) and (2) by b and a respectively, we obtain:

……..(3)

……..(4)

Adding equations (3) and (4), we obtain:

By using (1), we obtain b (a) - ay = 0
ab - ay = 0 ay = ab
y = b

(iv)… (1)

……..(2)
Subtracting equation (2) from (1), we obtain:

Using equation (1) ,we obtain,

(v)152x – 378y = –74… (1)
–378x + 152y = –604 … (2)
Adding the equations (1) and (2), we obtain:
–226x – 226y = –678

Subtracting the equation (2) from equation (1), we obtain:
530x – 530y = 530

Adding equations (3) and (4), we obtain:
2x = 4

x = 2
Substituting the value of x in equation (3), we obtain:
y = 1

 

We know that the sum of the measures of opposite angles in a cyclic quadrilateral is.

Therefore,

4y + 20 - 4x = 180 - 4x + 4y = 160

x - y = - 40 (i)

Also

3y - 5 - 7x + 5 = 180 - 7x + 3y = 180 (ii)

Multiplying equation (1) by 3, we obtain:
3x - 3y = - 120 (iii)

Adding equations (2) and (3), we obtain:
- 7x + 3x = 180 - 120
- 4x = 60
x = -15

By using equation (i), we obtain x - y = - 40
-15 - y = - 40
y = -15 + 40 = 25