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LINEAR EQUATIONS IN TWO VARIABLES

LINEAR EQUATIONS IN TWO VARIABLES. Introduction :

margaret-donovan
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LINEAR EQUATIONS IN TWO VARIABLES

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  1. LINEAR EQUATIONS IN TWO VARIABLES

  2. Introduction : • A simple linear equation is an equality between two algebraic expressions involving an unknown value called the variable. In a linear equation the exponent of the variable is always equal to 1. The two sides of an equation are called Right Hand Side (RHS) and Left-Hand Side (LHS). They are written on either side of equal sign. LHS= RHS EX 4x+3 = 15 2x+5y = 0 -2x+3y= 6

  3. System of equations or simultaneous equations – A pair of linear equations in two variables is said to form a system of simultaneous linear equations. For Example, 2x – 3y + 4 = 0 x + 7y – 1 = 0 Form a system of two linear equations in variables x and y.

  4. GENERAL FORM The general form of a linear equation in two variables xand y is ax + by + c = 0 , a =/= 0, b=/=0, where a, b and c being real numbers. A solution of such an equation is a pair of values, one for x and the other for y, which makes two sides of the equation equal. Every linear equation in two variables has infinitely many solutions which can be represented on a certain line.

  5. TO CHECK IF THE PAIR OF EQUATIONS HAVE SOLUTION • Firstly we have to know that whether the equations can be solved or not. For this we have the rules shown below :- 1. a1 =/= b1 (UNIQUE SOLUTION) a2 b2 2. a1 = b1 = c1 (INFINITE SOLUTIONS) a2 b2 c2 3. a1 = b1 =/= c1 (NO SOLUTION) a2 b2 c2

  6. ALGEBRAIC METHODS OF SOLVING SIMULTANEOUS LINEAR EQUATIONS The most commonly used algebraic methods of solving simultaneous linear equations in two variables are *Method of substitution *Method of elimination *Method of Cross- multiplication

  7. SUBSTITUTION METHOD STEPS Obtain the two equations. Let the equations be a1x + b1y + c1 = 0 ----------- (i) a2x + b2y + c2 = 0 ----------- (ii) Choose either of the two equations, say (i) and find the value of one variable , say ‘y’ in terms of x Substitute the value of y, obtained in the previous step in equation (ii) to get an equation in x

  8. SUBSTITUTION METHOD Solve the equation obtained in the previous step to get the value of x. Substitute the value of x and get the value of y. Let us take an example x + 2y = -1 ------------------ (i) 2x – 3y = 12 -----------------(ii)

  9. SUBSTITUTION METHOD x + 2y = -1 x = -2y -1 ------- (iii) Substituting the value of x in equation (ii), we get 2x – 3y = 12 2 ( -2y – 1) – 3y = 12 - 4y – 2 – 3y = 12 - 7y = 14 , y = -2 ,

  10. SUBSTITUTION METHOD Putting the value of y in eq (iii), we get x = - 2y -1 x = - 2 x (-2) – 1 = 4 – 1 = 3 Hence the solution of the equation is ( 3, - 2 )

  11. ELIMINATION METHOD • In this method, we eliminate one of the two variables to obtain an equation in one variable which can easily be solved. Putting the value of this variable in any of the given equations, the value of the other variable can be obtained. • We eliminate one variable first , to get a linear equation in one variable.

  12. Step 1. first multiply both the equation by some suitable non-zero constants to make the coefficients of one variable numerically equal. Step 2. then add or subtract one equation from the other so that one variable gets eliminated. If you get an equation in one variable, go to step Step 3. solve equation in one variable so obtained to get its value. For example: we want to solve, 3x + 2y = 11 2x + 3y = 4 ELIMINATION METHOD

  13. Let 3x + 2y = 11 --------- (i) 2x + 3y = 4 ---------(ii) Multiply 3 in equation (i) and 2 in equation (ii) and subtracting eqn. iv from iii, we get 9x + 6y = 33 ------ (iii) 4x + 6y = 8 ------- (iv) 5x = 25 => x = 5

  14. putting the value of y in equation (ii) we get, 2x + 3y = 4 2 x 5 + 3y = 4 10 + 3y = 4 3y = 4 – 10 3y = - 6 y = - 2 Hence, x = 5 and y = -2

  15. CROSS MULTIPLICATION METHOD • This is a method very useful for solving the linear equation in two variables • Let us consider two equations:-a1x + b1y + c1 = 0 a2x + b2y + c2 = 0 x y 1 b1 c1 a1 b1 b2 c2 a2 b2

  16. CROSS MULTIPLICATION METHOD • x = y = 1 b1c2 – b2c1 c1a2 – c2a1 a1b2 – a2b1 By this way the equations are solved and the values are obtained. We have to put the values of the known and get the values of the unknown. We can write this as given below also

  17. CROSS MULTIPLICATION METHOD x = b1c2 - b2c1 a1b2 - a2b1 y = c1a2 - c2a1 a1b2 - a2b1

  18. REDUCIBLE EQUATIONS The equations which cannot be solved simply are converted to the reduced forms and then solved such as :- • 2/x + 3/y = 13  2(1/x) + 3(1/y) = 13 • 5/x – 4/y = -2  5(1/x) – 4(1/y) = -2 Let (1/x) = p & (1/y) = q, then the equations :- 2p + 3q = 13 & 5p – 4q = -2 can be solved by any of the three methods mentioned above.

  19. FOMATIVE ASSESSMENT(MCQ) • 1. Which of the following is the solution of the pair of linear equations 3x – 2y = 0, 5y – x = 0 (a) (5, 1) (b) (2, 3) (c) (1, 5) (d) (0, 0) • 2. One of the common solution of ax + by = c and y-axis is _____ (a) (0, c/b) (b) (0,b/c ) (c) , 0 , (c/ b ) (d) (0, c/ b) • 3. If the value of x in the equation 2x – 8y = 12 is 2 then the corresponding value of y will be (a) –1 (b) +1 (c) 0 (d) 2

  20. ACTIVITY • To verify graphically • That the pair of linear equations x+y-5=0 ,2x+2y-6=0 in which a1/a2=b1/b2 =/= c1/c2 gives a pair of parallel straight lines. x + y = 5 …..(1) 2x +2y = 6…(2) • SOLUTION FOR X+Y=5 X 0 5 Y 5 0 • SOLUTION FOR 2X+2Y=6 X 0 3 Y 3 0 • After plotting the points on graph we get

  21. Which represent the pair of parallel lines

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