Tiêu đề 03-Linear equation in two variables (Part-1)(1).pdf ✅

3 CHAPTER LINEAR EQUATION IN TWO VARIABLES CONTENTS • Linear equation in two variables • General form of pair of linear equation • Graph of linear equation in two variables • Graphical representation of pair of linear equation • Types of solutions • Methods of Solving system of linear equation • Word problems ➢ LINEAR EQUATIONS IN TWO VARIABLES ➢ A statement of equality of two algebraic expressions, which involve one or more unknown quantities is known as an equation. If there are two unknown quantities then equation is called linear equation in two variables. A linear equation is an equation which involves linear polynomials. A value of the variable which makes the two sides of the equation equal is called the solution of the equation. Same quantity can be added/subtracted to/from both the sides of an equation without changing the equality. Both the sides of an equation can be multiplied/divided by the same non-zero number without changing the equality. Note :- To find value of variables in any equation we required number of equation equal to number of variables in equation. ➢ GENERAL FORM OF PAIR OF LINEAR EQUATION    + + = + + = a x b y c 0 a x b y c 0 2 2 2 1 1 1 where a1, b1, c1 & a2, b2, c2 are constants. GRAPH OF LINEAR EQUATION ax + by + c = 0 IN TWO VARIABLES, WHERE a  0, b  0 ➢ (i) Step I : Obtain the linear equation, let the equation be ax + by + c = 0. (ii) Step II : Express y in terms of x to obtain y = –       + b ax c (iii) Step III : Give any two values to x and calculate the corresponding values of y from the expression in step II to obtain two solutions, say (1 , 1 ) and (2 , 2 ). If possible take values of x as integers in such a manner that the corresponding values of y are also integers. (iv) Step IV : Plot points (1 , 1 ) and (2 , 2 ) on a graph paper. (v) Step V : Join the points marked in step IV to obtain a line. The line obtained is the graph of the equation ax + by + c = 0. ❖ EXAMPLES ❖ Ex.1 Draw the graph of the equation y – x = 2. Sol. We have, y – x = 2  y = x + 2 When x = 1, we have : y = 1 + 2 = 3 When x = 3, we have : y = 3 + 2 = 5 Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation. x 1 3 y 3 5
Plotting the points (1, 3) and (3, 5) on the graph paper and drawing a line joining them, we obtain the graph of the line represented by the given equation as shown in Fig. (1, 3) (3,5) 9 8 7 6 5 4 3 2 1 –1 –2 –3 –4 –5 –6 –7 –8 –9 –7 –6 –5 –4 –3 –2 –1 1 2 3 4 5 6 7 X  X Y Y y – x = 2 Ex.2 Draw a graph of the line x – 2y = 3. From the graph, find the coordinates of the point when (i) x = – 5 (ii) y = 0. Sol. We have x – 2y = 3  y = 2 x − 3 When x = 1, we have : y = 2 1− 3 = –1 When x = –1, we have : y = 2 −1− 3 = –2 Thus, we have the following table : x 1 –1 y –1 –2 Plotting points (1, –1) & (–1, –2) on graph paper & joining them, we get straight line as shown in fig. This line is required graph of equation x – 2y = 3. 9 8 7 6 5 4 3 2 1 –1 –2 –3 –4 –5 –6 –7 –8 –9 –7 –6 –5 –4 –3 –2 –1 1 2 3 4 5 6 7 X Y Y x – 2y = 3 (3, 0) (1, –1) X  (–1, –2) To find the coordinates of the point when x = –5, we draw a line parallel to y-axis and passing through (–5, 0). This line meets the graph of x – 2y = 3 at a point from which we draw a line parallel to x-axis which crosses y- axis at y = –4. So, the coordinates of the required point are (–5, –4). Since y = 0 on x-axis. So, the required point is the point where the line meets x-axis. From the graph the coordinates of such point are (3, 0). Hence, required points are (–5, –4) and (3, 0). GRAPHICAL REPRESENTATION OF PAIR OF LINEAR EQUATIONS ➢ ➢ Let the system of pair of linear equations be a1 x + b1 y = c1 ....(1) a2 x + b2 y = c2 ....(2) We know that given two lines in a plane, only one of the following three possibilities can happen - (i) The two lines will intersect at one point. (ii) The two lines will not intersect, however far they are extended, i.e., they are parallel. (iii) The two lines are coincident lines. Y X 1 2 P() 0 Y X 1 and 2 0 Y X 1 2 X' 0 Y' ❖ EXAMPLES ❖ Ex.3 The path of highway number 1 is given by the equation x + y = 7 and the highway number 2 is given by the equation 5x + 2y = 20. Represent these equations geometrically. Sol. We have, x + y = 7  y = 7 – x ....(1) In tabular form
Po int s A B y 6 3 x 1 4 and 5x + 2y = 20  y = 2 20 − 5x ....(2) In tabular form Po int s C D y 5 0 x 2 4 7 6 5 4 3 2 1 1 2 3 4 5 6 7 0 X' Y' X Y A C B (4,0) D (4,3) (2,5) (1,6) Plot the points A (1, 6), B(4, 3) and join them to form a line AB. Similarly, plot the points C(2, 5). D (4, 0) and join them to get a line CD. Clearly, the two lines intersect at the point C. Now, every point on the line AB gives us a solution of equation (1). Every point on CD gives us a solution of equation (2). Ex.4 A father tells his daughter, “ Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.” Represent this situation algebraically and graphically. Sol. Let the present age of father be x-years and that of daughter = y years Seven years ago father’s age = (x – 7) years Seven years ago daughter’s age = (y – 7) years According to the problem (x – 7) = 7(y – 7) or x – 7y = – 42 ....(1) After 3 years father’s age = (x + 3) years After 3 years daughter’s age = (y + 3) years According to the condition given in the question x + 3 = 3(y + 3) or x – 3y = 6 ....(2) x – 7y = –42 Po int s A B C 6 7 8 7 x 42 y x 0 7 14 + = x – 3y = 6 Po int s D E F 0 2 4 3 x 6 y x 6 12 18 − = Plot the points A(0, 6), B(7, 7), C(14, 8) and join them to get a straight line ABC. Similarly plot the points D(6, 0), E(12, 2) and F(18, 4) and join them to get a straight line DEF. (0,6)(12,2)(18,4) (14,8) x–3y=6 5D10 15 20 25 30 35 40 45 15 10 A 5 E F B C (0,6) (7,7) (42, 12) 1 2 G 0 y x Ex.5 10 students of class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz. Sol. Let the number of boys be x and the number of girls be y. Then the equations formed are x + y = 10 ....(1) and y = x + 4 ....(2) Let us draw the graphs of equations (1) and (2) by finding two solutions for each of the equations. The solutions of the equations are given. Po int s A B y 10 x 10 2 x 0 8 x y 10 = − + = Po int s C D E y x 4 4 5 7 x 0 1 3 y x 4 = + = + Plotting these points we draw the lines AB and CE passing through them to represent the
equations. The two lines AB and Ce intersect at the point E (3, 7). So, x = 3 and y = 7 is the required solution of the pair of linear equations. y = x + 4 x + y = 10 x' x y (0,10) A E D C B (0, 4) (1, 5) (3, 7) (8, 2) i.e. Number of boys = 3 Number of girls = 7. Verification : Putting x = 3 and y = 7 in (1), we get L.H.S. = 3 + 7 = 10 = R.H.S., (1) is verified. Putting x = 3 and y = 7 in (2), we get 7 = 3 + 4 = 7, (2) is verified. Hence, both the equations are satisfied. Ex.6 Half the perimeter of a garden, whose length is 4 more than its width is 36m. Find the dimensions of the garden. Sol. Let the length of the garden be x and width of the garden be y. Then the equation formed are x = y + 4 ....(1) Half perimeter = 36 x + y = 36 ....(2) x y 4 Po int s A B y 4 0 x 0 4 = + − x y 36 Po int s C D y 36 x 26 16 x 10 20 + = = − Plotting these points we draw the lines AB and CD passing through them to represent the equations. x + y x = y = 36 + 4 x' x y C y' (10, 26) D(20, 16) A –4 0 4 8 12 16 20 24 28 24 20 16 12 8 4 B The two lines AB and CD intersect at the point (20, 16), So, x = 20 and y = 16 is the required solution of the pair of linear equations i.e. length of the garden is 20 m and width of the garden is 16 m. Verification : Putting x = 20 and y = 16 in (1). We get 20 = 16 + 4 = 20, (1) is verified. Putting x = 20 and y = 16 in (2). we get 20 + 16 = 36  36 = 36, (2) is verified. Hence, both the equations are satisfied. Ex.7 Draw the graphs of the equations x – y + 1 = 0 and 3x + 2y – 12 = 0. Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region. Sol. Pair of linear equations are : x – y + 1 = 0 ....(1) 3x + 2y – 12= 0 ....(2) In tabular form Po int s A B y x 1 1 5 x 0 4 = + In tabular form Po int s C D 6 3 2 12 3x y x 0 2 − = Plot the points A(0, 1), B(4, 5) and join them to get a line AB. Similarly, plot the points C(0, 6), D(2, 3) and join them to form a line CD.