Class: 10th Subject: Maths. Time: 35 Min Group No. : V

Topics Topic: Linear Equations in two Variables Sub Topic: Graphical solution of Linear Equations in two variables

General Objective To create interest in mathematics. To develop the abstract thinking and reasoning skill among the students To develop the habit of working in groups. To develop the quality of co-relating things with surroundings.

Specific objective How to solve Linear equations in two variables graphically.

Motivation and P. K. Testing 1. X+2=8 What could be the value of x?E.A A(2,3) What are the abscissa and ordinate in this point? E.A. 2&3 3.3x+2y=5 How many variables this equation have? E.A. Two 4. What could be the value of x in terms of y in x+y=3 E. A. x=3-y 5. How many solutions does a linear equation in two variables have E. A. Infinitely many 6. If the cost of three notebooks and two pens is Rs and that of one notebook and four pens is Rs , what will be the cost of one notebook and one pen each? E. A. will have nothing to answer.

Prompting Consider the cost of one notebook as x and that of one pen as y. The question gets transformed into algebraic equations as below: 3x+2y =160 x+4y =120 Now, how will you find out the values of x and y ?

Introduction In order to find a unique solution, we need two linear equations in two variables. A general linear equation system in two variables can be written as: a 1 x+b 1 y=c 1 a 2 x+b 2 y=c 2 where the coefficients of x & y are non zero real numbers Solution is unique only if a 1 /a 2 ≠ b 1 /b 2 There are two ways to solve such equations-GRAPHICAL and ALGEBRAIC

GRAPHICAL METHOD Step-1 First we write the equation system. Let us consider the equation system: x+3y=6 2x-3y=12

Step-2 Then we are to find at least two solutions of each equation and tabulate them. In the example at hand the tables can be: & x06 Y=(6-x)/320 x03 Y=(2x-12)/3-4-2

MATERIAL REQUIRED Graph paper Pencil Eraser & Sharpener scale

Step-3 Then we are to plot these points in Cartesian Co-ordinate system. We are to plot the points of first equation and join them. Similarly we are to plot the points of second equation and then join them. We will get two straight lines

x+ 3y = 6 2x – 3y = 12 Y Y' X' P (6,0)

Step-4 The co-ordinates of the point where the lines of the graph intersect each other is the solution of the equation system. In the example at hand, the point B(6,0) has x=6 and y=0 as co-ordinates. Therefore the solution of the equation system x+3y=6 2x-3y=12 Is x=6 y=0

Group assignments 1. x+y =5 & x-y =1EA x=3 & y=2 2. x+y =4 & x +2y =6EA x=2 & y=2 3. x+y =5 & 2x+y =9EA x=4 & y=1 4. x-y =2 & x+2y =5EA x=3 & y=1 5. 2x+3y=4 & x-5y=2EA x=2 & y=0

Home assignments 2x-3y=0 & -x+y=0EA: x=0 & y=0 3x-y=-2 & x+2y=-3EA: x=-1 & y=-1 4x-3y=-23 & -2x+y=11EA: x=-5 & y=1 X+3y=1 & 4x-7y=-15EA: x=x=-2 & y=1 3x+2y=-13 & x+4y=-11EA: x=-3 & y=-2

Recapitulation Write the equation system. Find at least two solutions of each equation and tabulate them. Plot graph relative to each equation. Read the point of intersection.

Prepared by 1. SURJEET DHIMAN TGT NM GSSS RAKKAR TEH RAKKAR 2. MS ANU TGT NM GHS GARLI 3. SH BHUVNESH DOGRA TGT NM GHS KALOHA TEH RAKKAR 4. SH RAJEEV SHARMA TGT NM GSSS PIR SALUHI TEH RAKKAR 5. SH DINESH KUMAR TGT NM GSSS SARAD DOGRI 6. SH SURESH KUMAR TGT NM GHS ALOH 7. SH CHANDER SHEKHAR TGT NM GSSS GARLI 8. SH DINESH KUMAR TGT NM GHS BHAROLI JADID 9. SH MANOHAR LAL TGT NM GHS ADHWANI 10. SH RIPU DAMAN GSSS TIHRI 11. SH NEERAJ SHARMA GHS SUDHANGAL 12. SH SANJEEV KAPOOR GSSS BHAROLI KOHALA 13. SMT VANDNA CHAUHAN TGTNM GHS KUHNA

GUIDED BY HONOURABLE R.P.’S  MR. ANIL SHARMA (Lect. Maths GSSS Dhaloon)  MR. VIKAS NAG (TGT N.M.GSSS Atyala Dai)

THANK YOU HAVE A NICE DAY