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Real Numbers
Pair of Linear Equations in Two Variables
- Linear Equation in Two Variables
- Graphical Method of Solution of a Pair of Linear Equations
- Substitution Method
- Elimination Method
- Cross - Multiplication Method
- Equations Reducible to a Pair of Linear Equations in Two Variables
- Consistency of Pair of Linear Equations
- Inconsistency of Pair of Linear Equations
- Algebraic Conditions for Number of Solutions
- Simple Situational Problems
- Pair of Linear Equations in Two Variables
- Relation Between Co-efficient
Arithmetic Progressions
Quadratic Equations
- Quadratic Equations
- Solutions of Quadratic Equations by Factorization
- Solutions of Quadratic Equations by Completing the Square
- Nature of Roots of a Quadratic Equation
- Relationship Between Discriminant and Nature of Roots
- Situational Problems Based on Quadratic Equations Related to Day to Day Activities to Be Incorporated
- Application of Quadratic Equation
Polynomials
Circles
- Concept of Circle - Centre, Radius, Diameter, Arc, Sector, Chord, Segment, Semicircle, Circumference, Interior and Exterior, Concentric Circles
- Tangent to a Circle
- Number of Tangents from a Point on a Circle
- Concept of Circle - Centre, Radius, Diameter, Arc, Sector, Chord, Segment, Semicircle, Circumference, Interior and Exterior, Concentric Circles
Triangles
- Similar Figures
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- Basic Proportionality Theorem (Thales Theorem)
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Introduction to Trigonometry
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- Surface Area and Volume of Three Dimensional Figures
notes
3) Cross Multiplication Method- To understand this method properly we shall take help of a example
`4x+2y-7=0` ......eq1
`5x+3y-8=0` ......eq2
While practicing this method if we want to find value of a particular variable we will take that variable as a numerator and the denominator will be the substraction between the cross multiplication of coefficients of the remaining two variables. Like in this case it will be like
`x/"(2×-8)-(-7×3)"`
`y/"(4×-8)-(-7×5)"`
`1/"(4×3)-(2×5)"`
All of them are equal to each other, so if you to find value of x then take,
`x/"(2×-8)-(-7×3)" = 1/"(4×3)-(2×5)"`
`x/"-16+21"=1/"12-10"`
`x/5= 1/2`
`x=5/2`
To find the value of y,
`y/"(4×-8)-(-7×5)"= 1/"(4×3)-(2×5)"`
`y/"-32+35" = 1/"12-10"`
`y/3 = 1/2`
`y= 3/2`
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