Linear equations in two variables

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x y LINEAR EQUATION IN TWO VARIABLES 1 3 Z

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DON BOSCO GIRLS COLLEGE SUBMITTED TO : - MS .MANIKA RASTOGI SUBMITTED BY :- GROUP PYTHAGORUS

ACKNOWLEDGEMENT:

ACKNOWLEDGEMENT We owe our heartily thanks to Ms. Manika who helped us in this project . We are also thankful to our principal Ms. Aruna Khanna who gave us the permission to do this project . thanks a lot…….

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CONTENT INTRODUCTION TO LINEAR EQUATION SOLUTIONS FOR LINEAR EQUATIONS GRAPHICAL METHOD OF SOLVING LINEAR EQUATION ALGEBRAIC METHOD OF SOLVING LINEAR EQUATION SUBSTITUTION METHOD ELEMINATION METHOD CROSS MULTIPLICATION METHOD

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Linear equations Linear equations are functions which have two variables . They have an independent and dependent variable. INTRODUCTION

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A pair of linear equations in two variables can be represented and solved by the 1) Graphical method 2) Algebraic method

Graphical method:

Graphical method The graph of the pair Of linear equations in two variables is represented by two lines on the graph ….

Parallel lines :

If the lines on graph are parallel , then they have no solution . These kind of equations are inconsistent . Parallel lines

Intersecting lines :

Intersecting lines If the lines on graph intersect , then they have a unique solution. These kind of equations are consistent .

Coincident lines :

Coincident lines If the lines on graph are coincident , there are infinite solutions . These kind of equations are consistent .

Algebraic method :

Algebraic method There are several algebraic methods to solve linear equations . Some of them are :- Substitution method Elimination method Cross multiplication method

Substitution method:

Substitution method This one of the algebraic method to solve simultaneous equation in this method the equation the solved by taking the value of each variables separately. Lets see the steps to solve the substitution method.

Steps :

Steps These are the steps to solve the linear equation (simultaneous equation) by substitution method . STEP 1:- Find the value of one variable say y in term of other , i.e. x from either equation which is convenient . STEP 2 :- Substitute the value of y in the other equation and reduce it to an equation with one variable. STEP 3 :- Substitute the value of x or y obtain in step two used in step 1 to obtain the value of other variable .

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7x -15y = 2 –(i) X +2y =3 –(ii) Solution : X + 2y = 3 x = 3 - 2y Now substitute the value of x in eq. (i) 7(3 – 2y )– 15y = 2 21 – 14y – 15y = 2 -29y = -19 , y = 19/29 X = 3 – 2 ( 19/29) = 49/29 Therefore the solution is x = 49/29 and y = 19/29 ……… Example

Elimination method :

Elimination method Elimination method is the second method for solving linear equations by algebraic method . This method is sometimes more convenient than the substitution method .

STEPS :

STEPS STEP 1:- first multiply both the equations by some suitable non-zero constants to make the coefficients of one variable either x or y . STEP 2:- then add or subtract both equation so that one variable get eliminated if you get one variable go to step 3 ……..

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STEP 3:- Solve the equation in one variable x or y so obtained to get its value . STEP 4:- substitute the value of x in the either of original equations to get the value of the other variable .

Example:

Example Question:- The ratio of incomes of two persons is 9:7 and ratio of their expenditure is 4:3.If each of them manages to save Rs.2000per month , find the monthly income .

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Let us denote the income of the persons Rs. 9x and 7x and their expenditure by 4y and 3y respectively .Then the equations formed in the given situations are 9x-4y = 2000 – (i) 7x-3y = 2000 - (ii) Now multiply eq. (i) by 3 , and eq. (ii) by 4 , to make the coefficients of y equal….

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27x – 12y = 6000 -(i) 28x – 12y = 8000 - (ii) subtract eq. (i) & (ii) , (28x – 27x) – (12y – 12y) = 8000-6000 x = 2000 Substituting value of x in eq. (i) we get 9 (2000) – 4y = 2000 y = 4000 so the solution of equation is x = 2000 , y = 4000 so their monthly incomes are Rs. 18,000 and Rs. 14,000 respectively .

CROSS MULTIPLICATION METHOD :

CROSS MULTIPLICATION METHOD In cross multiplication method two fractions or rational expressions, can be cross-multiply  to simplify the equation or determine the value of a variable.

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Lets consider the general form of a pair of linear equations. a 1 x + b 1 y + c 1 = 0 and a 2 x + b 2 y + c 2 = 0  Recall that when a1 divided by a2 is not equal to b1 divided by b2 , the pair of linear equations will have a unique solution. To solve this pair of equations for x  and y  using cross-multiplication, we’ll arrange the variables x  and y  and their coefficients a1 , a2 ,  b1 and b2 , and the constants c1   and c2  as shown in the next slide ….

Example:

Example Ques. From a bus stand in Bangalore . If we buy 2 tickets to Malleswaram and 3 tickets to Yeshwanthpur , the cost is Rs. 46 ; but if we buy 3 tickets to Malleswaram and 5 tickets to Yeshwanthpur the total cost is Rs. 74 . Find the fares from the bus stand to Malleswaram , and to Yeshwanthpur . SOL :- Let Rs. x be the fare from the bus stand in Bangalore to Malleshwaram and Rs. Y to Yeshwanthpur from the given information we have 2x + 3y = 46 i.e. 2x + 3y – 46 = 0 3x + 5y = 74 i.e. , 3x + 5y – 74 = 0 to solve this eq. by cross multiplication method , we draw the diagram as given in the next slide

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x y 1 -46 2 3 5 -74 3 5 Then, x y 1 = = (3)(-74) –(5)(-46) (-46)(3)- (-74)(2) (2)(5)- (3)(3) i.e., x y 1 = = -222+230 -138+148 10-9

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x y 1 = = 8 10 1 x 1 y 1 = and = 8 1 10 1 x=8 and y=10

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Made by : Aditi Chaudhary , Kajal Gupta , Shefali Solomon & Ashmita Singh

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