
01. Real Numbers
9
Lecture1.1

Lecture1.2

Lecture1.3

Lecture1.4

Lecture1.5

Lecture1.6

Lecture1.7

Lecture1.8

Lecture1.9


02. Polynomials
11
Lecture2.1

Lecture2.2

Lecture2.3

Lecture2.4

Lecture2.5

Lecture2.6

Lecture2.7

Lecture2.8

Lecture2.9

Lecture2.10

Lecture2.11


03. Linear Equation
9
Lecture3.1

Lecture3.2

Lecture3.3

Lecture3.4

Lecture3.5

Lecture3.6

Lecture3.7

Lecture3.8

Lecture3.9


04. Quadratic Equation
8
Lecture4.1

Lecture4.2

Lecture4.3

Lecture4.4

Lecture4.5

Lecture4.6

Lecture4.7

Lecture4.8


05. Arithmetic Progressions
11
Lecture5.1

Lecture5.2

Lecture5.3

Lecture5.4

Lecture5.5

Lecture5.6

Lecture5.7

Lecture5.8

Lecture5.9

Lecture5.10

Lecture5.11


06. Some Applications of Trigonometry
7
Lecture6.1

Lecture6.2

Lecture6.3

Lecture6.4

Lecture6.5

Lecture6.6

Lecture6.7


07. Coordinate Geometry
17
Lecture7.1

Lecture7.2

Lecture7.3

Lecture7.4

Lecture7.5

Lecture7.6

Lecture7.7

Lecture7.8

Lecture7.9

Lecture7.10

Lecture7.11

Lecture7.12

Lecture7.13

Lecture7.14

Lecture7.15

Lecture7.16

Lecture7.17


08. Triangles
15
Lecture8.1

Lecture8.2

Lecture8.3

Lecture8.4

Lecture8.5

Lecture8.6

Lecture8.7

Lecture8.8

Lecture8.9

Lecture8.10

Lecture8.11

Lecture8.12

Lecture8.13

Lecture8.14

Lecture8.15


09. Circles
8
Lecture9.1

Lecture9.2

Lecture9.3

Lecture9.4

Lecture9.5

Lecture9.6

Lecture9.7

Lecture9.8


10. Areas Related to Circles
10
Lecture10.1

Lecture10.2

Lecture10.3

Lecture10.4

Lecture10.5

Lecture10.6

Lecture10.7

Lecture10.8

Lecture10.9

Lecture10.10


11. Introduction to Trigonometry
7
Lecture11.1

Lecture11.2

Lecture11.3

Lecture11.4

Lecture11.5

Lecture11.6

Lecture11.7


12. Surface Areas and Volumes
9
Lecture12.1

Lecture12.2

Lecture12.3

Lecture12.4

Lecture12.5

Lecture12.6

Lecture12.7

Lecture12.8

Lecture12.9


13. Statistics
12
Lecture13.1

Lecture13.2

Lecture13.3

Lecture13.4

Lecture13.5

Lecture13.6

Lecture13.7

Lecture13.8

Lecture13.9

Lecture13.10

Lecture13.11

Lecture13.12


14. Probability
9
Lecture14.1

Lecture14.2

Lecture14.3

Lecture14.4

Lecture14.5

Lecture14.6

Lecture14.7

Lecture14.8

Lecture14.9


15. Construction
7
Lecture15.1

Lecture15.2

Lecture15.3

Lecture15.4

Lecture15.5

Lecture15.6

Lecture15.7

NCERT Solutions – Linear Equation Exercise 3.1 – 3.7
Q.1 Aftab tells his daughter, “Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be”. (Isn’t this interesting)? Represent this situation algebraically and graphically.
Sol. Let’s denote the present age of daughter and her father Aftab as x years and y years respective. Then, algebraic representation is given by the following equations :
7(x – 7) = y – 7
⇒ 7x – 49 = y – 7
⇒ 7x – y = 42
and, 3(x + 3) = y + 3
⇒ 3x + 9 = y + 3
⇒ 3x – y = – 6
To obtain the equivalent graphical representation, we find two points on the line representing each equation. That is, we find two solutions of each equation. That is, we find two solutions of each equation.
These solutions are given below in the tables:
For 7x – y = 42
For 3x – y = – 6
To represent these equations graphically, we plot the points A(6, 0) and B(5, –7) to get the graph of (i) and the points C(0, 6) and D(–2, 0) give the graph of (ii).
Q.2 The coach of a cricket team buys 3 bats and 6 balls for Rs 3900. Later, she buys another bat and 2 more balls of the same kind for Rs 1300. Represent this situation algebraically and geometrically.
Sol. Let us denote the cost of bat be Rs. x and one ball be Rs. y. Then, the algebraic representation is given by the following equations :
3x + 6y = 3900 ⇒ x + 2y = 1300 …(1)
and, x + 3y = 1300 …(2)
To obtain the equivalent geometric representation, we find two points on the line representing each equation. That is, we find two solutions of each equation.
These solutions are given below in the table.
For x + 2y = 1300
For x + 3y = 1300
We plot the points A(0, 650), B(1300, 0) to obtain the geometric representation of x + 2y = 1300 and C(0,13003) and B (1300, 0) to obtain the geometric representation of x + 3y = 1300.
We observe that these lines intersect at B (1300, 0).
Q.3 The cost of 2 kg of apples and 1 kg of grapes on a day was found to be Rs. 160. After a months, the cost of 4 kg of apples and 2 kg of grapes is Rs. 300. Represent the situation algebraically and geometrically.
Sol. Let us denote the cost of 1 kg of apple by Rs. x and cost of 1 kg grapes by Rs.y.
Then the algebraic representation is given by the following equations :
2x + y = 160 …(1)
4x + 2y = 300 …(2)
⇒ 2x + y = 150
To find the equivalent geometric representation, we find two points on the line representing each equation. That is, we find two solutions of each equation.
2x + y = 160
2x + y = 150
Q.1 From the pair of linear equations in the following problems, and find their solution graphically.
(i) 10 students of Class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boy and girls who took part in the quiz.
(ii) 5 pencils and 7 pens together cost Rs. 50, whereas 7 pencils and 5 pens togeher cost Rs. 46. Find the cost of one pencil and that of one pen.
Sol. (i) Let us denote the number of boys by x and the number of girls by y. Then, the equations formed are
x + y = 10…(1)
and, y = x + 4
⇒ – x + y = 4…(2)
Let us draw the graphs of equations (1) and (2) by finding two solutions for each of these equations. The solutions of the equations are given in the table.
For x + y = 10
For – x + y = 4
Plotting the points and draw the lines passing through them to represent the equations as shown.
These two lines intersect at (3, 7). So, x = 3 and y = 7 is the required solution.
Hence, the number of boys and girls are 3 and 7 respectively.
Verification : Put x = 3 and y = 7 in (1) and (2), we find that both the equations are satisfied.
(ii) Let us denote the cost of one pencil by Rs. x and one pen by Rs. y. Then, the equations formed are
5x + 7y = 50 …(1)
and, 7x + 5y = 46 …(2)
Let us draw the graphs of equations (1) and (2) by finding two solutions for each of these equations.
The solutions of the equations are given in the table.
For 5x + 7y = 50
For 7x + 5y = 46
Plotting the points and draw the lines passing through them to represent the equations as shown.
These two lines intersect at (3, 5). So x = 3 and y = 5 is the required solution.
Hence, the cost of one pencil is Rs. 3 and that of one pen is Rs. 5.
Verification : Put x = 3 and y = 5 in (1) and (2), we find that both the equations are satisfied.
Q.2 On comparing the ratios a1a2,b1b2 and c1c2 find out whether the representing the following pairs of linear equations intersect at a point, parallel or coincide.
(i) 5x – 4y + 8 = 0; 7x + 6y – 9 = 0
(ii) 9x + 3y + 12 = 0; 18x + 6y + 24 = 0
(iii) 6x – 3y + 10 = 0; 2x – y + 9 = 0
Sol. (i) The given pair of linear equations are
5x – 4y + 8 = 0
and, 7x + 6y – 9 = 0
Here, 57≠−46
⇒ (1) and (2) are intersecting lines.
(ii) The given pair of linear equations are
9x + 3y +12 = 0
and, 18x + 6y + 24 = 0
Here, 918=36=1224 [Since, Each = 12]
Therefore, (1) and (2) are coincident lines.
(iii) The given pair of linear equations are
6x – 3y + 10 = 0
and, 2x – y + 9 = 0
Here, 62=−3−1≠109
Therefore, (1) and (2) are parallel lines
Q.3 On comparing the ratios a1a2,b1b2 and c1c2, find out whether the following pairs of linear equations are consistent, or inconsistent.
(i) 3x + 2y = 5; 2x – 3y = 7
(ii) 2x – 3y = 8; 4x – 6y = 9
(iii) 32x+53y=7; 9x – 10y = 14
(iv) 5x – 3y = 11; – 10x + 6y = – 22
(v) 43x+2y=8; 2x + 3y = 12.
Sol. (i) 3x + 2y = 5; 2x – 3y = 7
3x + 2y – 5 = 0
a1=3,b1=2,c1=−5,a2=2,b2=−3,c2=−7 and 2x – 3y – 7 = 0
a1a2=32,b1b2=2−3
Since, 32≠23,a1a2≠b1b2
Therefore, The pair of linear equations is consistent.
(ii) 2x – 3y = 8; 4x – 6y = 9; 2x – 3y – 8 = 0 and 4x – 6y – 3 = 0
a1=2,b1=−3,c1=−8; a2=4,b2=−6,c2=−9
a1a2=24=12,b1b2=−3−6=12,c1c2=−8−9=89
Since, 12=12≠89,i.e.,a1a2=b1b2≠c1c2
Therefore, The pair of linear equations is inconsistent.
(iii) 32x+53y=7; 9x – 10y = 14
a1=32,b1=53,c1=−7;a2=9,b1=−10,c2=−14
a1a2=329=16,b1b2=53−10=−16
Since, 16≠−16, a1a2≠b1b2
Therefore, The pair of linear equations is consistent.
(iv) 5x – 3y = 11; – 10x + 6y = – 22
a1=5,b1=−3,c1=−11;a2=−10,b2=6,c2=22
a1a2=510=−12,b1b2=−36=−12,c1c2=−1122=−12
Since,−12=−12=−12,i.e.,a1a2=b1b2=c1c2
Therefore, The pair of linear equation is consistent.
(v) 43x+2y=8;2x+3y=12
a1=43,b1=2,c1=−8;a2=2,b2=3,c2=−12
a1a2=432=23,b1b2=23,c1c2=−8−12=23
Since, 23=23=23,i.e.,a1a2=b1b2=c1c2
Therefore, The pair of linear equation is consistent.
Q.4 Which of the following pairs of linear equations are consistent ? Obtain the solution in such cases graphically.
(i) x + y = 5, 2x + 2y = 10
(ii) x – y = 8, 3x – 3y = 16
(iii) 2x + y – 6 = 0, 4x – 2y – 4 = 0
(iv) 2x – 2y – 2 = 0, 4x – 4y – 5 = 0
Sol. (i) Graph of x + y = 5 :
We have, x + y = 5
⇒ y = 5 – x
When x = 0, y = 5; when x = 5, y = 0
Thus, we have the following table :
Plotting the points A(0, 5) and (5, 0) on the graph paper. Join A and B and extend it on both sides to obtain he graph of x + y = 5.
Graph of 2x +2y = 10 :
We have 2x + 2y = 10
⇒ 2y = 10 – 2x
⇒ y = 5 – x
When x = 1, y = 5 – 1 = 4; when x = 2, y = 5 – 2 = 3
Thus, we have the following table :
Plotting the points C(1, 4) and D(2, 3) on the graph paper and drawing a line passing through these points on the same graph paper, we obtain the graph of 2x + 2y = 10
We find that C and D both lie on the graph of x + y = 5. Thus, the graphs of the two equations are coincident . Consequently, every solution of one equation is a solution the other.
Hence, the system of equations has infinitely many solutions, i.e., consistent.
(ii) Graph of x – y = 8 :
We have, x = 0, y = – 8; when x = 8, y = 0
Thus, we have the following table :
Plotting the points A(0, – 8) and B(8, 0) on a graph paper. Join A and B and extend it on both sides to obtain the graph of x – y = 8 as shown.
graph of 3x – 3y = 16:
We have, 3x – 3y = 16
⇒ 3y = 3x – 16
⇒y=3x−163
When x = 0, y = −163=−513;
When x=163=513, y = 0
Thus, we have the following table :
Plot the points C(0,−163) and D(163,0) on the same graph paper. Join C and D and extend it on both sides to obtain the graph of 3x – 3y = 16 as shown.
We find the graphs of x – y = 8 and 3x – 3y = 16 are parallel. So, the two lines have no common point. Hence, the given equations has no solution, i.e., inconsistent.
(iii) Graph of 2x + y – 6 = 0 :
We have, 2x + y – 6 = 0
⇒ y = 6 – 2x
When x = 0, y = 6 – 0 = 6; when x = 3, y = 6 – 6 = 0
Thus, we have the following table :
Plot the points A(0, 6) and B(3, 0) on a graph paper. Join A and B and extend it on both sides to obtain graph of 2x + y – 6 = 0 as shown.
Graph of 4x – 2y – 4 = 0 :
We have,
4x – 2y – 4 = 0
⇒ 2y = 4x – 4
⇒ y = 2x – 2
When x = 0 , y = – 2; when x = 1, y = 0
Thus, we have the following table :
Plotting the points C(0, –2) and D(2, 0) on the same graph and drawing a line joining them as shown.
Clearly, the two lines intersect at point P(2, 2).
Hence, x = 2, y = 2 is the solution of the given equations, i.e., consistent.
Verification : Putting x = 2, y = 2 in the given equations, we find that both the equations are satisfied.
(iv) Graph of 2x – 2y – 2 = 0;
We have,
2x – 2y – 2 = 0
⇒ 2y = 2x – 2
⇒ y = x – 2
When x = 2, y = 0; when x = 0, y = – 2
Thus, we have the following table :
Plot the points A(2, 0) and B(0, –2) on a graph paper. Join A and B and extend it on both sides to obtain the graph of 2x –2y – 2 = 0.
Graph of 4x – 4y – 5 = 0 :
We have,
4x – 4y – 5 = 0
⇒ 4y = 5 – 4x
⇒ y = 5−4x4
When x = 0, y = −54; when x = 54, y = 0
Thus, we have the following table :
Plot the points C(0,−54) and D(54,0) on the same
graph paper. Join C and D and extend it on both sides to obtain the graph of 4x – 4y – 5 = 0 as shown.
We find the graphs of these equations are parallel lines. So, the two lines have no common point. Hence, the given system of equations has no solution, i.e., inconsistent.
Q.5 Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the graden.
Sol. Let the length of the garden be x m and its width be y m.
Then, perimeter = 2(Length + Width) = 2(x + y)
But perimeter = 36 [Given]
Therefore, 2(x + y) = 36
⇒x+y=362=18
Also, x = y + 4
Therefore, y + 4 + y = 18
⇒ 2y = 18 – 4 = 14
⇒y=142=7
Therefore, x = y + 4 = 7 + 4 = 11
Hence, Length = 11m and width = 7m.
Q.6 Given the linear equation 2x + 3y – 8 = 0, write other linear equation in two variables such that the geometrical representation of the pair so formed
(i) intersecting lines
(ii) parallel lines
(iii) coincident lines
Sol. Given linear equation is 2x + 3y – 8 = 0 …(1)
(i) For intersecting lines, we know that
a1a2≠b1b2
Any line intersecting line may be taken as
5x + 2y – 9 = 0
(ii) For parallel lines, a1a2=b1b2≠c1c2
Therefore, Any line parallel to (1) may be taken as
4x + 6y –3 = 0
(iii) For coincident lines, a1a2=b1b2=c1c2
Any line coincident to (1) may be taken as
6x + 9x – 24 = 0
Q.7 Draw the graphs of the equations x – y + 1 = 0 and 3x + 2y – 12 = 0. Determine the coordinates of the vertices of the triangle formed by these lines and the xaxis, and shade the triangular region.
Sol. For the graph of x – y + 1 = 0.
We have,
x – y + 1 = 0
⇒ y = x + 1
When x = 0, y = 1; when x = – 1, y = 0
Thus, we have the following table :
Plot the points A(0, 1) and B(–1, 0) on a graph paper. Join A and B and extend it on both sides to obtain the graph of x – y + 1 = 0.
For the graph of 3x + 2y – 12 = 0 :
We have,
3x + 2y – 12 = 0
⇒ 2y = 12 – 3x
⇒ y=12−3x2
When x = 4, y=12−122=0
When x = 0, y=122=6
Thus, we have the following table :
Plot the points C(4, 0) and D(0, 6) on the same graph paper and draw a line passing through these two points to obtain the graph of equation 3x + 2y – 12 = 0.
Clearly, we obtain a ΔPBC formed by the given lines and the x – axis. The coordinates of the vertices are P(2, 3), B(–1, 0) and C(4, 0).
Q.1 Solve the following pair of linear equations by the substitution method :
(i) x + y = 14 x – y = 4
(ii) s – t = 3 s3+t2=6
(iii) 3x + y = 3 9x – 3y = 9
(iv) 0.2x + 0.3y = 1.3 0.4x + 0.5y = 2.3
(v) 2–√x+3–√y=0 3–√x−8–√y=0
(vi) 3x2−5y3=−2 x3+y2=136
Sol. (i) The given system of equations is
x + y = 14 …(1)
and, x – y = 4 …(2)
From (1), y = 14 – x
Substituting y = 14 – x in (2), we get
x – (14 – x) = 14
⇒ x – 14 + x = 4
⇒ 2x = 4 + 14
⇒ 2x = 18
⇒ x = 9
Putting x = 9 in (1), we get
9 + y = 14
⇒ y = 14 – 9
⇒ y = 5
Hence, the solution of the given system of equations is
x = 9, y = 5
(ii) The given system of equations is
s – t = 3 …(1)
s3+t2=6 …(2)
From (1), s = 3 + t
Substituting s = 3 + t in (2), we get
3+t3+t2=6
⇒2(3+t)+3t=36
⇒ 6 + 2t + 3t = 36
⇒ 5t = 30
⇒ t = 6
Putting t = 6 in (1), we get
s – 6 = 3
⇒ s = 3 + 6 = 9
Hence, the solution of the given system of equations is
s = 9, t = 6
(iii) The given system of equations is
3x – y = 3 …(1)
and, 9x – 3y = 9 …(2)
From (1), y = 3x – 3
Substituting y = 3x – 3 in (2), we get
9x – 3(3x – 3) = 9
⇒ 9x – 9x + 9 = 9
⇒ 9 = 9
This statement is true for all values of x. However, we do not get a specific value of x as a solution. So, we cannot obtain a specific value of y . This situation has arisen because both the given equations are the same.
Therefore, Equations (1) and (2) have infinitely many solutions.
(iv) The given system of equations is
0.2x + 0.3y = 1.3
⇒ 2x + 3y = 13…(1)
and, 0.4x + 0.5y = 2.3
⇒ 4x + 5y = 23 …(2)
From (2), 5y = 23 – 4x
⇒ y=23−4x5
Substituting y=23−4x5 in (1), we get
2x+3(23−4x5)=13
⇒ 10x + 69 – 12x = 65
⇒ – 2x = – 4
⇒ x = 2
Putting x = 2 in (1), we get
2 × 2 + 3y = 13
⇒ 3y = 13 – 4 = 9
⇒ y=93=3
Hence, the solution of the given system of equations is
x = 2, y = 3
(v) The given system of equations is
2x−−√+3y−−√=0 …(1)
and, 3x−−√−8y−−√=0 …(2)
From (2), 8y−−√=3x−−√
⇒ y=3x√8√
Substituting y=3x√8√ in (1), we get
2x−−√+3–√(3x√8√)=0
⇒ 2x−−√+3x8√=0
⇒ 16x−−−√+3x=0
⇒ 4x + 3x = 0
⇒ 7x = 0
⇒ x = 0
Putting x = 0 in (1), we get
0 + 3–√y = 0
⇒ y = 0
Hence, the solution of the given system of equations is
x = 0, y = 0
(vi) The given system of equations is
3x2−5y3=−2
⇒ 9x – 10y = – 12 …(1)
and, x3+y2=136
⇒ 2x + 3y = 13 …(2)
From (1), 10y = 9x + 12
⇒ y=9x+1210
Substituting y=9x+1210 in (2), we get
2x+3(9x+1210)=13
⇒ 20x + 27x + 36 = 130
⇒ 47x = 130 – 36
⇒ 47x = 94
⇒ x=9447=2
Putting x = 2 in (1), we get
9 × 2 – 10y = – 12
⇒ –10y = –12 – 18
⇒ – 10y = – 30
⇒ y=−30−10=3
Hence, the solution of the given system of equations is
x = 2, y = 3
Q.2 Solve 2x + 3y = 11 and 2x – 4y = – 24 and hence find the value of ‘m’ for which y = mx + 3.
Sol. The given system of equations is
2x + 3y = 11
and, 2x – 4y = – 24
From(1), 3y = 11 – 2x
⇒y=11−2x3
Substituting y=11−2x3 in (2), we get
2x−4(11−2x3)=−24
⇒ 6x – 44 + 8x = – 72
⇒ 14x = –72 + 44
⇒ 14x = –28
⇒ x=−2814=−2
Putting x = –2 in (1), we get
2(– 2) + 3y = 11
⇒ – 4 + 3y = 11
⇒ 3y = 11 + 4
⇒ 3y = 15
⇒ y=153=5
Putting x = – 2, y = 5 in y = mx + 3, we get
5 = m(–2) + 3
⇒ –2m = 5 – 3
⇒ – 2m = 2
⇒ m = – 1
Q.3 Form the pair of linear equations for the following problems and find their solution by substitution method.
(i) The difference between two numbers is 26 and one number is three times the other. Find them.
(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. find them.
(iii) The coach of a cricket team buys 7 bats and 6 balls for Rs 3800. Later, she buys 3 bats and 5 balls for Rs 1750. Find the cost of each bat and each ball.
(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is Rs. 105 and for a journey of 15 km, the charge paid Rs. 155. What are the fixed charges and the charge per km ? How much does a person have to pay for travelling a distance of 25 km ?
(v) A fraction becomes 911, if 2 is added to both the numerator and the denominator. If 3 is added to both the numerator and the denominator, it become 56. Find the fraction.
(vi) Five years hence, the age of Jacob will three times that of his son. Five years ago, Jacob age was seven times that of his son. What are the present ages?
Sol. (i) Let the numbers be x and y (x > y). Then,
x – y = 26
and, x = 3y
From (2), y=x3 in (1), we get
x−x3=26
⇒3x−x=78
⇒ 2x = 78
⇒ x = 39
Putting x = 39 in (2), we get
⇒ 39 = 3y
⇒ y=393=13
Hence, the required numbers are 39 and 13.
(ii) Let the angles be x° and y°(x > y). Then,
x + y = 180
and, x = y + 18
Substituting x = y + 18 in (1), we get
y + 18 + y = 180
⇒ 2y = 180 – 18
⇒ 2y = 162
⇒ y=1622=81
Putting y = 81 in (2), we get
x = 81 + 18 = 99
Thus, the angles are 99° and 81°.
(iii) Let the cost of one bat and one ball be Rs. x and Rs. y respectively. Then,
7x + 6y = 3800 …..(1)
and, 3x + 5y = 1750 …(2)
From (2), 5y = 1750 – 3x
⇒ y=1750−3x5
Substituting y=1750−3x5 in (1), we get
7x+6(1750−3x5)=3800
⇒ 35x + 10500 – 18x = 19000
⇒ 17x = 19000 – 10500
⇒ 17x = 8500
⇒ x=850017=500
Putting x = 500 in (2), we get
3(500) + 5y = 1750
⇒ 5y = 1750 – 1500
⇒ 5y = 250
⇒ y=2505=50
Hence, the cost of one bat is Rs. 500 and the cost of one ball is Rs 50.
(iv) Let the fixed charges of taxi be Rs. x per km and the running charges be Rs y km/hr.
According to the given condition, we have
x + 10y = 105 …(1)
x + 15y = 155 …(2)
From (1), x = 105 – 10y
Substituting x = 105 – 10y in (2), we get
105 – 10y + 15y = 155
⇒ 105 + 5y = 155
⇒ 5y = 155 – 105
⇒ 5y = 50
⇒ y = 10
Putting y = 10 in (1), we get
x + 10 × 10 = 105
⇒ x = 105 – 100
⇒ x = 5
Total charges for travelling a distance of 25 km
= x + 25y = Rs (5 + 25 × 10)
= Rs. 255
Hence, the fixed charge is Rs 5, the charge per km is Rs. 10 and the total charge for travelling a distance of 25 km is Rs 255.
(v) Let the fraction be xy.
Then, according to the given conditions, we have
x+2y+2=911 and x+3y+3=56
⇒ 11x + 22 = 9y + 18 and 6x + 18 = 5y + 15
⇒ 11x – 9y = – 4 …(1)
and, 6x – 5y = – 3….(2)
From (2), 5y = 6x + 3
⇒ y=6x+35
Substituting y=6x+35 in (1), we get
11x−9(6x+35)=−4
⇒ 55x – 54x – 27 = –20
⇒ x = – 20 + 27
⇒ x = 7
Putting x = 7 in (1), we get
11(7) – 9y = – 4
⇒ – 9y = – 4 – 77
⇒ –9y = – 81
⇒ y=−81−9=9
Hence, the given fraction is 79.
(vi) Let the present age of Jacob be x years and the present age of his son be y years.
Five years hence, Jacob’s age = (x + 5) years
Son’s age = (y + 5) years
Five years ago, Jacob’s age = (y – 5) years
Son’s age = (y – 5) years
As per question, we get
(x + 5) = 3(y + 5)
⇒ x + 5 = 3y + 15
⇒ x – 3y =10
⇒ x – 3y = 15 – 5
and, (x – 5) = 7(y – 5)
⇒ x – 5 = 7y – 35
⇒ x – 7y = – 30
⇒ x – 7y = – 35 + 5
From (1), x = 3y + 10
Substituting x = 3y +10 in (2), we get
3y + 10 – 7y = – 30
⇒ – 4y = – 30 – 10
⇒ – 4y = – 40
⇒ y = 10
Putting y = 10 in (1),we get
x – 3 × 10 = 10
⇒ x = 10 + 30 = 40
Hence, present age of Jacob is 40 years and that his son is 10 years.
Q.1 Solve the following pair of linear equation by the elimination method and the substituti method :
(i) x + y = 5 and 2x – 3y = 4
(ii) 3x + 4y = 10 and 2x – 2y = 2
(iii) 3x – 5y – 4 = 0 and 9x = 2y + 7
(iv) x2+2y3=−1 and x−y3=3
Sol. (i) By elimination method :
The given system of equation is
x + y = 5
and, 2x – 3y = 4
Multiplying (1) by 3, we get
3x + 3y = 15
Adding (2) and (3), we get
5x = 19
⇒ x=195
Putting x=195 in (1), we get
195 + y = 5
⇒ y = 5 – 195=25−195
Hence, x=195, y=65
By substitution method :
The given system of equations is
x + y = 5 …(1)
and, 2x – 3y = 4 …(2)
From (1), y = 5 – x
Substituting y = 5 – x in (2), we get
2x – 3 (5 – x) = 4
⇒ 2x – 15 + 3x = 4
⇒ 5x = 4 + 15
⇒ 5x = 19
⇒ x=195
Putting x=195 in (1), we get
195+y=5
⇒ y = 5 – 195=25−195=65
Hence, x=195,y=65
By geometrical method :
Graph of x + y = 5 :
We have, x + y = 5
⇒ y = 5 – x
When x = 0 , y = 5; When x = 5, y = 0
Thus, we have the following table :
Plotting the points A(0, 5) and B(5, 0) and drawing a line joining them, we get the graph of the equation x + y = 5 as shown.
Graph of 2x – 3y = 4
We have, 2x – 3y = 4
⇒ 3x = 2x – 4
⇒ y=2x−43
When x = 2, y=4−43=0; when x = – 1, y=−2−43=−62=−2
Thus, we have the following table :
Plotting the points C(2, 0) and D(–1,–2) on the same graph paper and drawing a line joining them, we obtain the graph of the equation.
Clearly, two lines intersect at the point P(195,65).
Hence, x=195,y=65 is the solution of the given system.
The method of elimination is the most efficient in this case.
(ii) By elimination method :
The given system of equation is
3x + 4y = 10 …(1)
and, 2x – 2y = 2 …(2)
Multiplying (2) by 2 and adding to (1), we get
3x + 4y + 4x – 4y = 10 +4
⇒ 7x = 14
⇒ x = 2
Putting x = 2 in (1), we get
3(2) + 4y = 10
⇒ 4y = 10 – 6 = 4
⇒ y = 1
Hence, x = 2 y = 1
By substitution method :
The given system of equaitons is
3x + 4y = 10 …(1)
and, 2x – 2y = 2
⇒ x – y = 1 …(2)
From (2), y = x – 1
Substituting y = x – 1 in (1), we get
3x + 4(x – 1) = 10
⇒ 3x + 4x – 4 = 10
⇒ 7x = 14
⇒x = 2
Putting x = 2 in (1), we get
3(2) + 4y = 10
⇒ 4y = 10 – 6 = 4
⇒ y = 1
Hence, x = 2, y = 1
By geometrical method :
The given system of equation is
3x + 4y = 10 …(1)
and, 2x – 2y = 2
⇒ x – y = 1 …(2)
For the graph of 3x + 4y = 10
We have, 3x + 4y = 10
⇒ 4y = 10 – 3x
⇒ y=10−3x4
When x = 2, y=10−64=44=1; when x = – 2, y=10+64=164=4
Thus, we have the following table :
Plotting the points A(2, 1) and B(–2, 4) and drawing a line joining them, we get the graph of the equation 3x + 4y = 10.
For the graph of x – y = 1 :
We have, x – y = 1
⇒ y = x – 1
When x = 3, y = 2; When x = 0, y = – 1
Thus, we have the following table :
Plotting the points C(3, 2) and D(0, – 1) on the same graph paper and drawing a line joining them, we obtain the graph of the equation.
Clearly, two lines intersect at the point A(2, 1).
Hence, x = 2, y = 1
In this case, almost all the methods are equally efficient, however, geometrical method takes more time.
(iii) By elimination method :
The given system of equations is
3x – 5y – 4 = 0
⇒ 3x – 5y = 4 ….(1)
and 9x = 2y + 7
⇒ 9x – 2y = 7 …(2)
Multiplying (1) by 3, we get
9x – 15y = 12 …(3)
Sibtracting (2) from (3), we get
– 13y = 5
⇒ y=−513
Putting y=−513 in (1), we get
3x –5(−513) = 4
⇒ 3x+2513=4
3x = 4 – 2513
⇒ 3x=52−2513
⇒3x=2713
⇒ x=913
Hence, x=913, y=−513
By, substitution method :
The given system of equations is
3x – 5y – 4 = 0
⇒ 3x – 5y = 4 …(1)
and, 9x = 2y + 7
⇒ 9x – 2y = 7….(2)
From (2), 2y = 9x – 7
⇒ y = 9x−72
Substituting y = 9x−72 in (1), we get
3x−5(9x−72)=4
⇒ 6x – 45x + 35 = 8
⇒ – 39x = 8 – 35
⇒ – 39x = – 27
⇒ x = −27−39=913
Putting x = 913 in (2), we get
3×913−5y=4
⇒5y=2713−4
⇒5y=27−5213=−2513
⇒y=−513
Hence, x=913 y=−513
By geometrical method :
The given system of equations is
3x – 5y – 4 = 0
⇒ 3x – 5y = 4 …(1)
and, 9x = 2y + 7
⇒ 9x – 2y = 7…(2)
For the graph of 3x – 5y = 4 :
We have, 3x – 5y = 4
⇒ 5y = 3x – 4
⇒ y=3x−45]
When x = 3 y = 9−45=55 = 1;
When x = – 2, y = −6−45=−105=−2
Thus, we have the following table :
Plotting the points A(3, 1) and B(–2,–2) and drawing a line joining them, we get the graph of the equation 3x – 5y = 4.
For the graph of 9x – 2y = 7 :
We have, 9x – 2y = 7
⇒ 2y = 9x – 7
⇒ y=9x−72
When x = 1, y=9−72=22=1 ;
When x = 3, y=27−72=202=10 ;
Thus, we have the following table :
Plotting the points C(1, 1) and D(3, 10) on the same graph paper and drawing a line joining them, we obtain the graph of equation.
Clearly, two lines intersect at the point P(913,−513)
Hence, x=913,y=−513
In this case, the method of elimination is the most efficient.
(iv) By elimination method ;
The given system of equations is
x2+2y3=−1
⇒ 3x + 4y = – 6 …(1)
and, x−y3=3
⇒ 3x – y = 9…(2)
Multiplying (2) by 4 and adding to (1), we get
15x = 30
⇒ x = 2
Putting x = 2 in (2), we get
3(2) – y = 9
⇒ –y = 9 – 6 = 3
⇒ y = – 3
Hence, x = 2, y = – 3
By substitution method :
x2+2y3=−1
⇒ 3x + 4y = – 6 …(1)
and, x−y3=3
⇒ 3x – y = 9 …(2)
From (2), y = 3x – 9
Putting y = 3x – 9 in (1), we get
3x + 4(3x – 9) = – 6
⇒ 3x + 12x – 36 = – 6
⇒ 15x = 30
⇒ x = 2
Putting x = 2 in (2), we get
3(2) – y = 9
⇒ – y = 9 – 6 = 3
⇒ y = – 3
Hence, x = 2, y = – 3
By geometrical method :
The given system of equations is
x2+2y3=−1
⇒ 3x + 4y = – 6 …(1)
and, x−y3=3
⇒ 3x – y = 9 …(2)
For the graph of 3x + 4y = – 6 :
We have,
3x + 4y = – 6
⇒ 4y = – 6 – 3x
⇒ y = −6−3x4
When x = 2, y = −6−64=−124=−3;
when x = – 2, y = −6+64=04=0
Thus, we have the following table :
Plotting the points A(2,–3) and B(–2, 0) and drawing a line joining them, we get the graph of the equation 3x +4y = – 6.
For the graph of 3x – y = 9 :
We have, 3x – y = 9
⇒ y = 3x – 9
When x = 3, y = 6
⇒ y = 3x – 9
When x = 3, y = 9 – 9 = 0;
When x = 4, y = 12 – 9 = 3
Thus, we have the following table :
Plotting the points C(3, 0) and D(4, 3) on the same graph paper and drawing a line joining them, we obtain the graph of the equation.
Clearly, the two lines intersect at A(2, –3).
Hence, x = 2, y = – 3
In this case, almost all the methods are equally efficient, however, geometrical method takes more time.
Q.2 Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method :
(i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes 12 if we only add 1 to the denominator. What is the fraction ?
(ii) Five year ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu ?
(iii) The sum of the digits of a two – digit number 9. Also, nine times this number is twice the number obtained by reversingthe order of the digits. Find the number.
(iv) Meena went to a bank to withdraw Rs. 2000. She asked the cashier to give her Rs. 50 and Rs.100 notes only. Meena got 25 notes in all. Find how man notes of Rs. 50 and RS 100 she received.
(v) A lending library has a fixed charge for first three days and an additional charge for each day thereafter. Saritha paid Rs. 27 for a book she kept for seven days, while Susy paid Rs. 21 for book she kept for five days. Find the fixed charge and the charge for each extra day.
Sol. (i) Let x be the numerator and y be the denominator of the fraction. So, the fraction is xy.
By given conditions : x+1y−1=1
⇒x+1=y−1
⇒ x – y = – 2…(1)
and, xy+1=12
⇒ 2x = y + 1 …(2)
⇒ 2x – y = 1
Subtracting (2) from (1), we get
(x – y) – (2x – y) = – 2 – 1
⇒ x – y – 2x + y = – 3
⇒ – x = – 3
⇒ x = 3
Substituting x = 3 in (1), we get
3 – y = –2
⇒ y = 5
Hence, the required fraction is 35.
(ii) Let the present age of Nuri = x years
and, the present age of Sonu = y years
Five years ago, Nuri’s age = (x – 5) years
Sonu’s age = (y – 5) years
As per conditions, x – 5 = 3(y – 5)
⇒ x – 5 = 3y – 15
⇒ x – 3y = – 15 + 5
⇒ x – 3y = – 10 …(1)
Ten years later, Nuri’s age = (x + 10) years
Sonu’s age = (y + 10)
As per condition, x + 10 = 2(y + 10)
⇒ x + 10 = 2y + 20
⇒ x – 2y = 20 – 10
⇒ x – 2y = 10 …(2)
Subtracting (2) from (1), we get
– y = – 20
⇒ y = 20
Putting y = 20 in (2), we get
x – 2(20) = 10
⇒ x = 10 + 40 = 50
Therefore, Nuri’s present age = 50 years
and Sonu’s present age = 20 years
(iii) Let the digits in the units’s place and ten’s place be x and y respectively.
Therefore, Number = 10y + x
If the digits age reversed, the new number = 10x + y
As per conditions : x + y = 9 ….(1)
and, 9(10y + x) = 2(10x + y)
⇒ 90y + 9x = 20x +2y
⇒ 20x – 9x + 2y – 90y = 0
⇒ 11x – 88 y = 0 …(2)
From (1), y = 9 – x
Putting y = 9 – x in (2), we get
11x –88(9 – x) = 0
⇒ 11x – 88 × 9 + 88x = 0
⇒ 99x = 88 × 9
⇒ x = 88×999=8
Putting x = 8 in (1), we get
8 + y = 9
⇒ y = 1
Hence, the number = 10y + x = 10 × 1 + 8 = 18
(iv) Let the number of Rs. 50 notes be x and number of Rs. 100 notes be y. Then,
x + y = 25
and, 50x + 100y = 2000
⇒ x + 2y = 40
On subtracting (1), from (2), we get
y = 15
Putting y = 15 in (1), we get
x + 15 = 25
⇒ x = 10
Hence, number of Rs. 50 notes = 10
and, number of Rs. 100 notes = 15
(v) Let the fixed charge for 3 days be Rs. x and charge per gday be Rs. y.
Therefore, By the given conditions,
x + 4y = 27
and, x + 2y = 21
Subtracting (2) form (1), we get
2y = 6
⇒ y = 3
Put y = 3 in (1), we get
x + 4 × 3 = 27
⇒ x = 27 – 12 = 15
Fixed charges = Rs. 15
and, charges per day = Rs. 3
Q.1 Which of the following pairs of linear equations has unique solution, no solution, or finitely many solutions ? In case there is a unique solution, find it by using cross multiplication method.
(i) x – 3y – 3 = 0 3x – 9y – 2 = 0
(ii) 2x + y = 5 3x + 2y = 8
(iii) 3x – 5y = 20 6x – 10y = 40
(iv) x – 3y – 7 = 0 3x – 3y – 15 = 0
Sol. (i) The given system of equations is
x – 3y – 3 = 0
and, 3x – 9y – 2 = 0
These are of the form a1x+b1y+c1=0
and a2x+b2y+c2=0
where a1=1,b1=−3,c1=−3 and a2=3,b2=−9,c2=−2
We have, a1a2=13,b1b2=−3−9=13andc1c2=−3−2=32
Clearly, a1a2=b1b2≠c1c2
So, the given system of equations has no solution i.e., it is inconsistent.
(ii) The given system of equations may be written as
2x + y – 5 = 0 and 3x + 2y – 8 = 0
These are of the form a1x+b1y+c1=0anda2x+b2y+c2=0
where a1=2,b1=1,c1=−5anda2=3,b2=2andc2=−8
We have, a1a2=23,b1b2=12
Clearly, a1a2≠b1b2
So, the given system of equations has a unique solution.
To find the solution, we use the cross – multiplication method. By cross – multiplication, we have
x1×−8−2×(−5)=y−5×3−(−8)×2=12×2−3×1
⇒x−8+10=y−15+16=14−3
⇒x2=y1=11
⇒ x = 2, y = 1
Hence, the given system of equations has a unique solution given by x = 2, y = 1.
(iii) The system of equations may be written as
3x – 5y – 20 = 0 and 6x – 10y – 40 = 0
The given equations are of the form
a1x+b1y+c1=0anda2x+b2y+c2=0
where, a1=3,b1=−5,c1=−20anda2=6,b2=−10,c2=−40
We have, a1a2=36=12,b1b2=−5−10=12andc1c2=−20−40=12
Clearly, a1a2=b1b2=c1c2
So, the given system of equations has infinitely many solutions.
(iv) The given system of equations is
x – 3y – 7 = 0 and 3x – 3y – 15 = 0
The given equations are of the form
a1x+b1y+c1=0anda2x+b2y+c2=0
where a1=1,b1=−3,c1=−7anda2=3,b2=−3,c2=−15
we have, a1a2=13,b1b2=−3−3=1 Clearly, a1a2≠b1b2
So, the given system of equations has a unique solution.To find the solution, we use crossmultiplication method. By crossmultiplication, we have
x45−21=y−21+15=1−3+9
x24=y−6=16
⇒ x=246=4,y=−66=−1
Hence, the given system of equations has a unique solution given by
x = 4, y = – 1
Q.2 (i) For which values of a and b does the following pair of linear equations have an infinite number of solutions ? 2x + 3y = 7 (a – b)x + (a + b)y = 3a + b – 2
(ii) For which value of k will the following pair of linear equations have no solution ?
3x + y = 1 (2k – 1)x + (k – 1)y = 2k + 1
Sol. (i) We know that the system of equations
a1x+b1y=c1anda2x+b2y=c2
has infinite number of solutions if
a1a2=b1b2=c1c2
Therefore, The given system of equations will have infinite number of solutions if
2a−b=3a+b=73a+b−2
⇒ 2a−b=3a+band2a−b=73a+b−2
⇒ 2a + 2b = 3a – 3b and 6a + 2b – 4 = 7a – 7b
⇒ a = 5b and a – 9b = – 4
Putting a = 5b in a – 9b = – 4, we get
5b – 9b = – 4
⇒ – 4b = – 4
⇒ b = 1
Putting b = 1 in a – 9b = – 4, we get
a = 5(1) = 5
Hence, the given system of equations will have infinitely many solutions if
a = 5 and b = 1
(ii) We know that the system of equations
a1x+b1y=c1anda2x+b2y=c2
has no solution if a1a2=b1b2≠c1c2
So, the given system of equations will have no solution if
32k−1=1k−1≠12k+1
⇒ 32k−1=1k−1and1k−1≠12k+1
Now, 32k−1=1k−1
⇒ 3k – 3 = 2k – 1
⇒ k = 2
Clearly, for k = 2, we have
1k−1≠12k+1
Hence, the given system of equations will have no solution if k = 2.
Q.3 Solve the following pair of linear equations by the substitution and cross – multiplication methods :
8x + 5y = 9
3x + 2y = 4
Sol. The given system of equations is
8x + 5y = 9…(1)
and, 3x + 2y = 4…(2)
By graphical method :
For the graph of 8x + 5y = 9
We have, 8x + 5y = 9
⇒ 5y = 9 – 8x
⇒y=9−8x5
⇒ When x = 3, y = 9−245=−155=−3 ; when x = – 2, y = 9+165=255=5
Thus, we have the following table :
Plot the points A(3, – 3) and B(–2, 5) on a graph paper. Join A and B and extend it on both sides to obtain the graph of 8x + 5y = 9 as shown.
For the graph of 3x + 2y = 4
We have,
3x + 2y = 4
⇒ 2y = 4 – 3x
⇒ y = 4−3x2
When x = 0, y = 4−02=42=2;
When x = 2, y = 4−62=−22=−1
Thus, we have the following table :
Plot the points C(0, 2) and D(2, – 1) on the same graph paper. Join C and D and extend it on both sides to obtain the graph of 3x + 2y = 4.
Clearly, the two lines intersect at B(–2, 5).
Hence, x = – 2, y = 5 is the solution of the given system of equations.
By substitution method :
Substituting y = 9−8x5 in (2), we get
3x+2(9−8x5)=4
⇒ 15x + 18 – 16x = 20
⇒ – x = 2
⇒ x = – 2
Putting x = – 2 in (1), we get
8(–2) +5y = 9
⇒ 5y = 9 + 16 = 25
⇒ y=255=5
Hence, x = – 2, y = 5 is the solution of the given system of equations.
By cross – multiplication method :
x−20+18=y−27+32=116−15
⇒ x−2=y5=11
⇒ x=−21=−2andy=51=5
Hence, x = – 2, y = 5 is the solution of the given system of equations.
The method of cross – multiplication is more efficient.
Q.4 Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method :
(i) A part of monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 20 days she has to pay Rs. 1000 as hostel charges whereas a student B, who takes food for 26 days, pays Rs. 1180 as hostel charges. Find the fixed charges and the cost of food per day.
(ii) A fraction becomes 13 when 1 is subtracted from the numerator and it becomes 14 when 8 is added to its denominator. find the fraction.
(iii) Yash scored 40 marks in a test, getting 3 marks for each right answer and losing 1 mark for each wrong answer. Had 4 marks been awarded for each correct answer and 2 marks been deducted for each incorrect answer, then Yash would have scored 50 marks. How many questions were there in the test ?
(iv) Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars ?
(v) The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the rectangle.
Sol. (i) Let the fixed charges be Rs. x and charge per day be Rs. y.
Therefore, By the given conditions,
x + 20y = 1000 …(1)
and, x + 26y = 1180 …(2)
Subtracting (1) from (2), we get
6y = 180
⇒ y = 30
From (1), x + 20 × 30 = 1000
⇒ x = 1000 – 600 = 400
Therefore, Fixed charge = Rs. 400
and cost of food per day = Rs. 30
(ii) Let the fraction be xy.
According to the given conditions :
x−1y=13
⇒ 3x – 3 = y
⇒ 3x – y – 3 = 0…(1)
and, xy+8=14
⇒ 4x = y + 8
⇒ 4x – y – 8 = 0
Solving (1) and (2) by cross – multiplication method.
x8−3=y−12+24=1−3+4
⇒ x5=y12=11
⇒ x = 5, y = 12
Hence, the required fraction is 512.
(iii) Let Yash answer x questions correctly and y questions incorrectly. According to the given conditions :
3x – y = 40 …(1)
and, 4x – 2y = 50 …(2)
Multiplying (1) by 2 and subtracting (2) from the result so obtained, we get
2x = 30
⇒ x = 15
From (1), 3 × 15 – y = 40
⇒ – y = 40 – 45
⇒ – y = – 5
⇒ y = 5
Therefore, Total number of questions in the test are 15 + 5 = 20.
(iv) Let X and Y be two cars starting from points A and B respectively. Let the speed of car X be x km/hr and that of Y be y km/hr.
Case I : When two cars move in the same direction.
Let these cars meet at point Q. Then,
Distance travelled by car X = AQ
Distance travelled by car Y = BQ
It is given that two cars meet in 5 hours.
Therefore, Distance travelled by car A in 5 hours = 5x km
⇒ AQ = 5x
Distance travelled by’ car Y in 5 hours = 5y km
⇒ BQ = 5y
Clearly, AQ – BQ = AB
⇒ 5x – 5y = 100 [Since, AB = 100 km]
⇒ x – y = 20
Case II : When two cars move in opposite directions.
Let these cars meet at point P. Then,
Distance traveled be car X = AP
Distance traveled be car Y = BP
In this case, two cars meet in 1 hour.
Therefore, Distance traveled by car X in 1 hour = x km
⇒ AP = x
Distance travelled by car Y in 1 hour = y km
⇒ BP = y
Clearly, AP + PB = AB
⇒ x + y = 100 …(2) [Since, AB = 100 km]
Solving (1) and (2), we have
x = 60, y = 40
Hence, speed of car X is 60 km/hr and speed of car Y 40 km/hr.
(v) Let the length and breadth of the rectangle be x and y units respectively. Then,
Area = xy sq. units
If length is reduced by 5 units and breadth is increased by 3 units, then area is reduced by 9 sq. units.
Therefore, xy – 9 = (x – 5) (y + 3)
⇒ xy – 9 = xy + 3x – 5y – 15
⇒ 3x – 5y – 6 = 0 …(1)
When length is increased by 3 units and breadth by units, the area is increased by 67 sq. units.
Therefore, xy + 67 = (x + 3) (y + 2)
⇒ xy + 67 = xy + 2x + 3y + 6
⇒ 2x + 3y – 61 = 0 …(2)
Solving (1) and (2), we get
x305+18=−y−183+12=19+10
⇒ x=32319=17, y=17119=9
Hence, the length and breadth of the rectangle are 17 units and 9 units respectively.
Q.1 Solve the following pairs of equations by reducing them to a pair of linear equations :
(i) 12x+13y=2
13x+12y=136
(ii) 2x√+3y√=2
4x√−9y√=−1
(iii) 4x+3y=14
3x−4y=23
(iv) 5x−1+1y−2=2
6x−1−3y−2=1
(v) 7x−2yxy=5
8x+7yxy=15
(vi) 6x + 3y = 6xy
2x + 4y = 5xy
(vii) 10x+y+2x−y=4
15x+y−5x−y=−2
(viii) 13x+y+13x−y=34
12(3x+y)−12(3x−y)=−18
Sol. (i) Taking 1x=u and 1y=v. The given system of equations become
12u+13v=2
⇒ 3u + 2v = 12 …(1)
13u+12v=136
⇒ 2u + 3v = 13…(2)
Multiplying (1) by 3 and (2), we have
9u + 6v = 36 …(3)
and, 4u + 6v = 26 …(4)
Subtracting (4) from (3), we get
5u = 10
⇒ u = 2
Putting u = 2 in (3), we get
18 + 6v = 36
⇒ 6v = 18
⇒ v = 3
Now, u = 2
⇒ 1x=2
⇒x=12
and, v = 3
⇒ 1y=3
⇒ y=13
Hence, the solution is x = 12, y=13.
(ii) The given system of equations is
2x√+3y√=2 and 4x√−9y√=−1
Putting u=1x√ and v=1y√ .Then , the given equations become
2u + 3v = 2 …(1)
and, 4u – 9v = – 1 …(2)
Multiplying (1) by 3, we get
6u + 9v = 6…(3)
Adding (2) and (3), we get
10u = 5
⇒ u=510=12
Putting u=12 in (1), we get
2×12+3v=2
⇒ 3v = 1
⇒ v=13
Now, u=12
⇒ 1x√=12
⇒ x−−√ = 2
⇒ x = 4
and, v=13
⇒ 1y√=13
⇒ y√ = 3
⇒ y = 9
Hence, the solution is x = 4, y = 9.
(iii) The given system of equations is
4x+3y=14…(1)
and, 3x−4y=23…(2)
Multiplying (1) by 4 and (2) by 3, we get
16x+12y=56…(3)
and, 9x−12y=69…(4)
Adding (3) and (4), we get
25x=125
⇒ x=25125=15
Putting x=15 in (1), we get
4×5+3y=14
⇒ 3y = 14 – 20
⇒ 3y = – 6
⇒ y = – 2
Hence, the solution is x=15, y = – 2.
(iv) Let u=1x−1,v=1y−2. Then, the given system of equations becomes
5u + v = 2 …(1)
and, 6u – 3v = 1 …(2)
Multiplying (1) by 3, we get
15u + 3v = 6 …(3)
Adding (2) and (3), we get
21u = 7
⇒u=13
Putting u=13 in (1), we get
53+v=2
⇒ v=2−53=6−53=13
Now, u=13
⇒ 1x−1=13
⇒ x – 1 = 3
⇒ x = 4
v=13
⇒ 1y−2=13
⇒ y – 2 = 3
⇒ y = 5
Hence, the solution is x = 4, y = 5.
(v) The given system of equations is
7x−2yxy=5
⇒ 7y−2x=5
and, 8x+7yxy=15
⇒ 8y+7x=15
Let u = 1x, v = 1y. Then, the above equations become
7 – 2u = 5 …(1)
and, 8v + 7u = 15 …(2)
Multiplying (1) by 7 and (2), we get
49v – 14u = 35 …(3)
and, 16v + 14u = 30
Adding (3) and (4), we get
65v = 65
⇒ v = 1
Putting v = 1 in (1), we get
7 – 2u = 5
⇒ – 2u = – 2
⇒ u = 1
Now, u = 1
⇒ 1x=1
⇒ x = 1
and, v = 1
⇒ 1y=1
⇒ y = 1
Hence, the solution is x = 1, y = 1.
(vi) The given system of equations is 6x + 3y = 6xy and 2x + 4y = 5xy, where x and y are non zero.
Since x≠0,y≠0, , we have xy ≠ 0.
On dividing each one of the given equations by xy we get
3x+6y=6 and 4x+2y=5
Taking 1x=u and 1y=v, the above equations become
3u + 6v = 6…(1)
and, 4u + 2v = 5…(2)
Multiplying (2) by 3, we get
12u + 6v = 15 …(3)
Subtracting (1) from (3), we get
9u = 15 – 6 = 9
⇒ u = 1
Putting u = 1 in (1), we get
3×1 + 6v = 6
⇒ 6v = 6 – 3 = 3
⇒ v=12
Now, u = 1
⇒ 1x=1
⇒ x = 1
and, v=12
⇒ 1y=12
⇒ y = 2
Hence, the given system of equations has one solution x = 1, y = 2.
(vii) The given system of equations is
10x+y+2x−y=4
15x+y−5x−y=−2
Putting u=1x+y and v=1x−y. Then, the given equations become
10u +2v = 4
⇒ 5u + v = 2 …(1)
and, 15u – 5v = – 2…(2)
Multiplying (1) by 5, we get
25u + 5v = 10 …(3)
Adding (2) and (3), we get
40u = 8
⇒ u=15
Putting u=15 in (1), we get
5(15)+v=2
⇒ v = 2 – 1 = 1
Now, u=15
⇒ x + y = 5 …(4)
v = 1
⇒ 1x−y=1
⇒ x – y = 1
Adding (4) and (5), we get
2x = 6
⇒ x = 3
When x = 3, then from (4), we get
3 + y = 5
⇒ y = 5 – 3 = 2
Hence, the given system of equations has one solution
x = 3, y = 2
(viii) Taking u=13x+y and v=13x−y. The give system of equations becomes
u + v =34…(1)
12u−12v=−18
⇒ u – v = −14…(2)
Adding (1) and (2), we get
2u = 34−14
⇒ 2u=24=12
⇒ u=14
Putting u=14 in (1), we get
14+v=34
⇒ v=34−14=24=12
Now, u=14
⇒ 13x+y=14
⇒ 3x + y = 4…(3)
and, v=12
⇒ 3x – y = 2…(4)
⇒13x−y=12
Adding (3) and (4), we get
Putting x = 1 in (3), we get
3x + y = 4
⇒ y = 4 – 3 = 1
Hence, the solution is x = 1, y = 1.
Q.2 Formulate the following problems as a pair if equations, and hence find their solutions :
(i) Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current.
(ii) 2 women and 5 men can together finish an embroidery work in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone to finish the work, and also that taken be 1 man alone.
(iii) Roohi travels 300 km to her home partly be train and partly be bus. She takes 4 hours if she travels 60 km by train and the remaining by bus. If she travels 100 km by train and the remaining by bus, she takes 10 minutes linger. find the speed of the train and the bus separately.
Sol. (i) Let the speed of the boat in still water be x km/hr. and the speed of the stream by y km/hr. Then,
Speed upstream = (x – y) km/hr
Speed downstream = (x + y) km/hr
Time taken to cover 20 km downstream = 2 hours
⇒ 20x+y=2
⇒ x +y = 10…(1)
Time taken to cover 4 km upstream = 2 hours
⇒ 4x−y=2
⇒ x – y = 2…(2)
Adding (1) and (2), we get
2x = 12
⇒ x = 6
Putting x = 6 in (1), we get
6 + y = 10
⇒ y = 10 – 6 = 4
Hence, the speed of boat in still water = 6km/hr
and, speed of stream = 4 km/hr
(ii) Let 1 woman can finish the embroidery in x days and 1 man can finish the embroidery in y days.
Then,
1 woman’s 1 day’s work = 1x
1 man’s 1 day’s work = 1y
Therefore, 2x+5y=14 and 3x+6y=13
Putting 1x=u and 1y=v, these equations become
2u + 5v = 14…(1)
and, 3u + 6v = 13…(2)
Multiplying (1) by 3 and (2) by 2 and subtracting, we get
3v=112
⇒ v=136
Putting v=136 in (1), we get
2u=(14−536)=9−536
=436=19
⇒u=118
Now, u=118
⇒1x=118
⇒ x = 18
and,
v=136
⇒ 1y=136
⇒ y = 36
Thus, I woman alone can finish the embroidery in 18 days and 1 man alone can finish it in 36 days.
(iii) Let the speed of train be c km/hr and speed of bus be y km/hr.
Then, 60x+240y=4 …(1)
and, 100x+200y=4+1060=256 …(2)
Let 1x=u and 1y=v, than eqns. (1) and (2) becomes
60u +240v = 4 …(3)
100u + 200v = 256 … (4)
Multiplying eqn. (3) by 5 and eqn. (4) by 6, we get
⇒ u=−5−300=160
From eqn. (3)
60 × 160 + 240 × v = 4
⇒ v=3240=180
Since, x = 1u=60 and y = 1v=80
Therefore, Speed of train is 60 km/h and speed of bus is 80 km/h.
Q.1 The ages of two friends Ani and Biju differ by 3 years. Ani’s father Dharam is twice as old as Ani and Biju is twice as old as his sister Cathy. The ages of Cathy and Dharam differ by 30 years. Find the ages of Ani and Biju.
Sol. Let the ages of Anil and Biju be x years and y years respectively. Then,
x – y = ± 3 [Given]
Dharam’s age = 2x, and Cathy’s age = y2
Clearly, Dharam is older than Cathy
2x−y2=30
⇒ 4x – y = 60
Thus, we have the following two systems of linear equations :
x – y = 3 …(1)
and, 4x – y = 60 …(2)
x – y = – 3…(3)
and, 4x – y = 80 ….(4)
Subtracting (1) from (2), we get
3x = 57
⇒ x = 19
Putting x = 19 in (1), we get
19 – y = 3
⇒ y = 19 – 3 = 16
Subtracting (4) from (3),we get
3x = 83
⇒ x=833=2723
Putting 833−y=−3
⇒ y=833+3=83+93
=923=3023
Hence, Ani’s age = 19 years
and Biju’s age = 16 years
or Ani’s age = 2723 years
and Biju’s age = 3023 years
Q.2 One says, “Give me a hundred, friend ! I shall then become twice as rich as you.” The other replies, ” If you give me ten, I shall be six times as rich as you.” Tell me what is the amount of their (respective) capital ? [From the Bijaganita of Bhaskara II]
Sol. Let the friends be named as A and B. Let A has Rs. x and B has Rs. y.
As per question, we have
x + 100 = 2(y – 100)
⇒ x + 100 = 2y – 200
⇒ x – 2y + 300 = 0 ….(1)
and, y + 10 = 6(x – 10)
⇒ y + 10 = 6x – 60
⇒ 6x – y – 70 = 0…(2)
Multiplying (2) by 2, we get
12x – 2y – 140 = 0 …(3)
Subtracting (3) from (1), we get
– 11x + 440 = 0
⇒ – 11x = – 440
⇒ x = 40
Putting x = 40 in (1), we get
40 – 2y + 300 = 0
⇒ – 2y = – 340
⇒ y = 170
Q.3 A train covered a certain distance at a uniform speed . If the train would have been 10 km/h faster, it would have taken 2 hours less than the scheduled time. And, if the train were slower by 10 km/m, it would have taken 3 hours more than the scheduled time. Find the distance covered by the train.
Sol. Let the original speed of the train be x kmph and let the time taken to complete the journey be y hours.
Then, the distance covered = xy km
Case I : When speed = (x + 10) kmph
and, time taken = (y – 2) hours
In this case, distance = (x + 10) (y – 2)
⇒ xy = (x + 10) (y – 2)
⇒ xy = xy – 2x + 10y – 20
⇒ 2x – 10y + 20 = 0…(1)
Case II : When speed = (x – 10) kmph
and, time taken = (y + 3)hours
In this case, distance = (x – 10) (y + 3)
⇒ xy = (x – 10) (y + 3)
⇒ xy = xy + 3x – 10y – 30
⇒ 3x – 10y – 30 = 0 …(2)
Subtracting (1) from (2), we get
100 – 10y + 20 = 0
⇒ x = 50
Putting x = 50 in (1), we get
100 – 10y + 20 = 0
⇒ – 10y = – 120
y = 12
Therefore, The original speed of the train = 50kmph
The time taken to complete the journey = 12 hours
Therefore, The length of the journey = Speed × Time
= (50 × 12) km = 600 km
Q.4 The students of a class are made to stand in rows. If 3 students are extra in a row, there would be 1 row less. If 3 students are less in a row, there would be 2 rows more. Find the number of students in the class.
Sol. Let originally there x students in each row. Let there be y rows in total. Therefore, total number of students is xy.
Clase I : When 3 students are taken extra in each row, then the number of rows in this case becomes (y –1).
Therefore, xy = (x + 3) (y – 1)
⇒ xy = xy – x + 3y – 3
⇒ x – 3y + 3 = 0
Case II : When 3 students are taken less in each row, then the number of rows becomes (y + 2).
Therefore, xy = (x – 3) (y + 2)
⇒ xy = xy + 2x – 3y – 6
⇒ 2x – 3y – 6 = 0…(2)
Solving (1) and (2), we get
x = 9, y = 4
Hence, the number of students = xy = 9 × 4 = 36.
Q.5 In a ΔABC, ∠C = 3∠B = 2(∠A + ∠B). Find the three angles.
Sol. Let ∠A = x°,∠B = y°.Then,
∠C = 3∠B
⇒∠C = 3y°
⇒ 3∠B = 2(∠A + ∠B)
⇒ 3y = 2(x + y)
3y = 2x +2y
⇒ y = 2x
⇒ 2x – y = 0 …(1)
Since ∠A, ∠B, ∠C are angles of a triangle.
Therefore, ∠A + ∠B + ∠C = 180°
⇒ x + y + 3y = 180°
⇒x + 4y = 180° …(2)
Putting y = 2x in (2), we get
x + 8x = 180°
⇒ 9x = 180°
⇒ x = 20°
Putting x = 20° in (1), we get
y = 2 × 20° = 40°
Hence,∠A = 20°, ∠B = 40° and ∠C = 3y° =3 × 40° = 120°.
Q.6 Draw the graphs of the equations 5x – y = 5 and 3x – y = 3. Determine the coordinate of the vertices of the triangle formed by these lines and the y axis.
Sol. The given system of equations is
5x – y = 5 …(1)
and, 3x – y = 3 …(2)
Since both the equations are linear. so their graphs must be a straight line.
Consider the equation (1),
5x – y = 5
⇒ y = 5x – 5
When x = 1, y = 5 – 5 = 0 ;
When x = 2, y = 10 – 5 = 5
Thus, we have the following table :
Plot the points A(1, 0) and (2, 5) on the graph paper. Join AB and extend it on both sides to obtain the graph of 5x – y = 5.
Now, consider the equation (2),
3x – y = 3
⇒ y = 3x – 3
When x = 0, y = 0 – 3 = 3;
When x = 2, y = 3(2) – 3 = 3
Thus, we have the following table :
Plot the points C(0, – 3) and D(2, 3) on the same graph paper. Join CD and extend it on both sides to obtain the graph of 3x – y = 3.
We know that when a line meets y axis, the value of x is zero. In the equation 5x – y = 5, if we put x = 0, we get y = – 5. Thus, the line 5x – y = 5 meets the yaxis at (0, – 5).
Similarly, the line 3x – y = 3 meets the y – axis at (0, –3)
The area of the Δ ACE = 12 × CE × OA
= (12×2×1) sq. units
= 1sq. units
Q.7 Solve the following pair of linear equations :
(i) px + qy = p – q qx – py = p + q
(ii) ax + by = c bx + ay = 1+ c
(iii) xa−yb=0 ax + by = a2+b2
(iv) Solve for x and y :
(a – b)x + (a + b)y = a2−2ab−b2
(a + b) (x + y) = a2+b2
(v) 152x – 378y = – 74 – 378x + 152y = – 604
Sol. (i) The given system of equations is
px + qy = p – q
⇒ px + qy – (p – q) = 0
qx – py = p + q
⇒ qx – py – (p + q) = 0
By cross – multiplication, we get
x−q(p+q)−p(p−q) =y−q(p−q)+p(p+q) =1−p2−q2
⇒ x−pq−q2−p2+pq =y−pq+q2+p2+pq =1−(p2+q2)
⇒x−(p2+q2)=yp2+q2=1−(p2+q2)
⇒x=−(p2+q2)−(p2+q2)=1
and y=p2+q2−(p2+q2)=−1
Hence, the required solution is x = 1, y = – 1.
(ii) The given system of equations may be written as
ax + by – c = 0
bx + ay – (1 + c) = 0
By cross – multiplication, we have
⇒x−b(1+c)+ac=y−bc+a(1+c)=1a2−b2
⇒x−b−bc+ac=y−bc+a+ac=1a2−b2
⇒xc(a−b)−b=yc(a−b)+a=1(a−b)(a+b)
⇒x=c(a−b)−b(a−b)(a+b)
=ca+b−b(a−b)(a+b)
and, y=c(a−b)+a(a−b)(a+b)
=ca+b+b(a−b)(a+b)
Hence, the required solution is
x=ca+b−b(a−b)(a+b)
y=ca+b+b(a−b)(a+b)
(iii) The given system of equations is
xa−yb=0
⇒ bx – ay = 0 …(1)
and, ax + by – (a2+b2) = 0 …(2)
By crossmultiplication, we have
xa(a2+b2)−0=y0+b(a2+b2)=1b2+a2
⇒xa(a2+b2)=xb(a2+b2)=1a2+b2
x=a(a2+b2)a2+b2=a
y=b(a2+b2)a2+b2=b
Hence, the required solution is x = a, y = b.
(iv) The given system of equations may be rewritten as
(a – b)x + (a + b)y – (a2−2ab−b2)=0
and, (a + b)x + (a + b)y – (a2+b2)=0
By cross – multiplication, we have
x−(a+b)(a2+b2)+(a+b)(a2−2ab−b2)
y−(a+b)(a2−2ab−b2)(a−b)(a2+b2)
=1(a−b)(a+b)−(a+b)2
=x(a+b)−a2−b2+a2−2ab−b2
=y−a3+2a2b+ab2−a2b+2ab2−b3+a3+a3+ab2−a2b−b3
=1a2−b2−a2−b2−2ab
⇒x(a+b)(−2ab−2b2)
⇒x(a+b)(−2ab−2b2)=y4ab2=1−2a2−2ab
⇒x(a+b)(−2b)(a+b)=y4ab2
=1−2b(a+b)
⇒ x = a + b
and y=−2aba+b
Hence, the required solution is x = a + b, y=−2aba+b
Hence, the required solution is x = a + b, y=−2aba+b
(v) We have, 152x – 378y = – 74 …(1)
and, – 378x + 152y = – 604 …(2)
Adding (1) and (2), we get
– 226x – 256y = – 678
⇒ x + y = 3 …(3)
Subtracting (1) from (2), we get
– 530x + 530y = – 530
⇒ x – y = 1 …(4)
Adding (3) and (4), we get 2x = 4
⇒ x = 2
Putting x = 2 in (1), we get 2 + y = 3
⇒ y = 1
Hence, the required solution is x = 1, y = 2.
Q.8 ABCD is a cyclic quadrilateral (see figure). Find the angles of the cyclic quadrilateral.
Sol. We know that the sum of the opposite angles of cyclic quadrilateral is 180°. In the cyclic quadrilateral ABCD, angles A and C, angles B and D form pais of opposite angles
Therefore, ∠A+∠C=180∘
and ∠B+∠D=180∘
⇒ (4y + 20) – 4x = 180°
and (3y – 5) + (–7x + 5) = 180°
⇒ – 4x + 4y – 160 = 0
and – 7x + 3y – 180 = 0
⇒ – x + y – 40 = 0
and, – 7x + 3y – 180 = 0
Multiplying (1) by 3, we get
– 3x + 3y – 120 = 0
Subtracting (3) from (2), we get
– 4x – 60 = 0
⇒ x = – 15
Putting x = – 15 in (1), we get
15 + y – 40 = 0
⇒ y – 25 = 0
⇒ y = 25
Hence, ∠A = 4y + 20 = 4 × 25 + 20
= 100 + 20 = 120°
∠B=3y−5=3×25−5
=75−5=70∘
∠C=−4x=(−4)×(−15)=60∘
and, ∠D=−7x+5=−7×(−15)+5
=105+5=110∘
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