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Linear equations Class 10 by aryan kathuria

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Linear equations Class 10 by aryan kathuria

  1. 1. Linear Equations Definition of a Linear Equation A linear equation in two variable x is an equation that can be written in the form ax + by + c = 0, where a ,b and c are real numbers and a and b is not equal to 0. An example of a linear equation in x is .
  2. 2. Let ax + by +c = O , where a ,b , c are real numbers such that a and b ≠ O. Then, any pair of values of x and y which satisfies the equation ax + by +c = O, is called a solution of it. A Pair of Linear Equation In Two Variables Can Be Solved – i. By Graphical Method ii. By Algebraic Method SOLUTIONS OF A LINEAR EQUATION
  3. 3. GRAPHICAL SOLUTIONS OF A LINEAR EQUATION Let us consider the following system of linear equations in two variable I. 2x-y=-1 II. 3x+2y=9 Now solution of each equation are to be taken by- i. Firstly express one variable in terms of other. ii. Now assign any value to one variable and then determine the value of other variable. Then plot the equations and determine the solutions Of system of linear equation in two variables
  4. 4. For Eq. 1 2x - y= -1 y = 2x + 1 Put x = 0 y = 2(0) + 1 y = 1 Put x = 2 y = 2(2) + 1 y = 5 For Eq. 2 3x + 2y = 9 y = 9 - 3x/2 Put x = 3 y = 9 - 3(3) /2 y = 0 Put x = -1 y = 9 - 3(-1) /2 y = 6 X 0 2 y 1 5 X 3 -1 y 0 6
  5. 5. 1 2 3 4-4 -3 -2 -1 5 4 3 2 1 -1 -2 -3 -4 -5 -6 ( 2,5 ) ( -1,6 ) ( 3,0 ) ( 0,1 ) X = 1 Y = 3 Is the solution for system of equations
  6. 6. TYPES OF SOLUTIONS of system of equations UNIQUE SOLUTION consistent a1/a2 ≠ b1/b2 Lines intersecting at a point INFINITE SOLUTIONS consistent a1/a2 = b1/b2 = c1/c2 Coincident Lines NO SOLUTION Non consistent a1/a2 = b1/b2 ≠ c1/c2 Parallel Lines
  7. 7. TYPES OF METHOD:- TO SOLVE A PAIR OF LINEAR EQUATION IN TWO VARIABLE BY ALGEABRIC METHODS I. Elimination Method II.Substitution Method III.Cross-Multiplication Method
  8. 8. • Make one variable equal in both the equations and then either add or subtract the equations to eliminate the variable. • The resulting equation in one variable is solved and then by substituting this value of the variable in either of the given equation, the value of the other variable is also obtained. ELIMINATION METHOD
  9. 9. Q. Solve using the method of Elimination I. 2x + y = 8 II. 2(x + 6y = 15) For making the coefficients of x in eq.(1) and eq.(2) equal, we multiply eq.(2) by 2 and get 2x + y = 8 -(3) 2x + 12y = 30 -(4) Subtracting Eq.(3) from eq.(4), we have 11y = 22 y = 2 Put this value of y in eq.(1) 2x + 2 = 8 2x = 6 x = 3
  10. 10. • Find the value of one variable in the terms of other variable. Substitute it in other equation and we will get value of one of the variable. Then put the value of variable in any one of the equation and the value of other variable is also obtained. SUBSTITUTION METHOD
  11. 11. Q. Solve using the method of Substitution I. x + 2y = -1 II. 2x - 3y = 12 From Eq.(1), we have x = -1 – 2y - (3) Substituting this value of x in eq.(2), we have 2(-1 – 2y) – 3y =12 - 2 - 4y – 3y = 12 - 7y = 14 y = -2 Put this value of y in Eq.(3), we have x = -1 -2(-2) x = -1 + 4 x = 3
  12. 12. CROSS- MULTIPLICATION METHOD
  13. 13. Q. Solve using the method of cross-Multiplication 5x + 3y = 35 5x + 3y - 35 = 0 2x + 4y = 28 2x + 4y - 28 = 0 a1 = 5 b1 = 3 c1= -35 a2 = 2 b2 = 4 c2= -28 ___x___ = ___y___ = ___1___ (3)(-28)-(4)(-35) (-35)(2)-(-28)(5) (5)(4)-(2)(3) -84 + 140 -70 + 140 20 – 6 56/14 70/14 14 x = 4 and y = 5

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