
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

Chapter Notes – Linear Equation
(1) An equation in the form ax + by + c = 0, where a, b and c are real numbers, and a≠ 0 and b ≠ 0, is called a linear equation in two variables x and y.
For Example: 2x + 3y + 7 = 0, where a = 2, b = 3, c =5 are real numbers. So, given equation is a linear equation in two variables.
(2) Each solution (x, y) of a linear equation in two variables, ax + by + c = 0, corresponds to a point on the line representing the equation, and vice versa.
For Example: 2x + 3y = 5 has (1, 1) as its solution. So,(1, 1) will lie on the line 2x + 3y = 5.
(3) The general form for a pair of linear equations in two variables x and y is a_{1}x + b_{1} y + c_{1 }= 0 and a_{2}x + b_{2} y + c_{2} = 0, where a_{1} , b_{1} , c_{1} , a_{2} , b_{2} , c_{2} are all real numbers and a_{1}^{2} + b_{1}^{2} ≠ 0, a_{2}^{2} + b_{2}^{2} ≠ 0.
For Example: 2x + 3y – 7 = 0 and 9x – 2y + 8 = 0 forms a pair of linear equations.
(4) A pair of linear equations which has no solution is called an inconsistent pair of linear equations. In this case, the lines may be parallel a_{1}/a_{2} = b_{1}/b_{2} ≠ c_{1}/c_{2}.
For Example: x + 2y – 4 = 0 and 2x + 4y – 12 = 0 are parallel lines.(5) A pair of linear equations in two variables, which has a solution, is called a consistent pair of linear equations. In this case, the lines may intersect in a single point and a_{1}/a_{2} ≠ b_{1}/b_{2}.
For Example: x – 2y = 0 and 3x + 4y – 20 intersects each other at unique point (4, 2).(6) A pair of linear equations which are equivalent has infinitely many distinct common solutions. Such a pair is called a dependent pair of linear equations in two variables. In this case, the lines may be coincident and a_{1}/a_{2} = b_{1}/b_{2} = c_{1}/c_{2}.
For Example: 2x + 3y – 9 = 0 and 4x + 6y – 18 = 0 are coincident lines.
(7) Algebraic Methods of Solving a Pair of Linear Equations:
(i) Substitution Method
Follow steps given below to understand Substitution Method:
Step 1: Find the value of one variable, say y in terms of the other variable, i.e., x from either equation, whichever is convenient.
Step 2: Substitute this value of y in the other equation, and reduce it to an equation in one variable, i.e., in terms of x, which can be solved. Sometime, one can get statements with no variable. If this statement is true, you can conclude that the pair of linear equations has infinitely many solutions. If the statement is false, then the pair of linear equations is inconsistent.
Step 3: Substitute the value of x (or y) obtained in Step 2 in the equation used in Step 1 to obtain the value of the other variable.
For Example: Solve the following pair of equations by substitution method: 7x – 15y = 2 and x + 2y = 3
We can rewrite x + 2y = 3 as x = 3 – 2y – (1)
Substituting value of x in 7x – 15y = 2, we get,
7(3 – 2y) – 15y = 2
21 – 14y – 15y = 2
29y = 19
Thus, y = 19/29.
Now, substituting value of y in (1), we get,
x = 3 – 2(19/29) = 49/29.
(ii) Elimination Method:
Follow steps given below to understand Elimination Method:
Step 1: First multiply both the equations by some suitable nonzero constants to make the coefficients of one variable (either x or y) numerically equal.
Step 2: Then add or subtract one equation from the other so that one variable gets eliminated. If you get an equation in one variable, go to Step 3. If in Step 2, we obtain a true statement involving no variable, then the original pair of equations has infinitely many solutions. If in Step 2, we obtain a false statement involving no variable, then the original pair of equations has no solution, i.e., it is inconsistent.
Step 3: Solve the equation in one variable (x or y) so obtained to get its value.
Step 4: Substitute this value of x (or y) in either of the original equations to get the value of the other variable.
For Example: Solve the following pair of equations by elimination method: 9x – 4y = 2000 and 7x – 3y = 2000.
Multiplying 9x – 4y = 2000 by 3 and 7x – 3y = 2000 by 4, we get,
27x – 12y = 6000 and 28x – 12y = 8000
Subtracting both these equations, we get,
(28x – 27x) – (12y – 12y) = 8000 – 6000
x = 2000
Substituting value of x in 9x – 4y = 2000, we get,
9(2000) – 4y = 2000
y = 4000.
(iii) Cross Multiplication Method:
Follow steps given below to understand Cross Multiplication Method:
Step 1: Write the given equations in the form a_{1}x + b_{1} y + c_{1 }= 0 and a_{2}x + b_{2}y + c_{2} = 0.
Step 2: Take the help of the diagram belowAnd write the equations as shown below
Step 3: Find x and y, provided a_{1}b_{1} – a_{2}b_{1} ≠ 0.
For Example: Solve the following pair of equations by cross multiplication method: 2x + 3y – 46 = 0 and 3x + 5y – 74 = 0.
Given equations are in form a_{1}x + b_{1}y + c_{1} = 0 and a_{2}x + b_{2}y + c_{2} = 0.Now, with the help of diagram we can rewrite equations as
x/((3)(74) – (5)(46) = y/((46)(3) – (74)(2) = 1/((2)(5) – (3)(3)
x/(222 + 230) = y/(138 + 148) = 1/(10 – 9)
x/8 = y/10 = 1/1
x/8 = 1/1 and y/10 = 1/1
Thus, x = 8 and y = 10.
(8) Equations Reducible to a Pair of Linear Equations in Two Variables:
Let us understand it by an example:
For Example: Solve 2/x + 3/y = 13 and 5/x – 4/y = 2.
We can rewrite the given equations as,
2(1/x) + 3(1/y) = 13 and 5(1/x) – 4(1/y) = 2
Let us substitute 1/x a = p and 1/y = q, so we get,
2p + 3q = 13 and 5p – 4q = 2
On solving these equations, we get,
p = 2 and q = 3.
We know, p = 1/x = 2 and q = 1/y = 3.
Thus, x = ½ and y = 1/3.
5 Comments
How we access live tutorial classes please tell me
For any information regarding live class please call us at 8287971571
I am not able to access any test of coordinate geometry, can you pls resolve the issue?
We have fixed the issue. Please feel free to call us at 8287971571 if you face such type of issues.
Hi!
I am a student in class 10th and ive noticed that you dont give full courses on your youtube channel. I mean, its understandable. But we request you to at least put full course of only one chapter on youtube so that we can refer to that. Only one from the book Because your videos are very good and they enrich our learning.
Please give this feedback a chance and consider our request.
Dronstudy lover,
Ananyaa