LINEAR EQUATION IN TWO VARIABLES

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LINEAR EQUATION IN TWO VARIABLES

PREPARED BY BHAVYA KUMAR SHARMA TGT MATHS KV
PAURI
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• 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 4x3 15 2x5y 0 -2x3y 6

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• 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.

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• 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.

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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

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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

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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

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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)

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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 ,

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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 )

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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.

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• 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
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• Let 3x 2y 11 --------- (i)
• 2x 3y 4 ---------(ii)
• Multiply 3 in equation (i) and 2 in equation (ii)
  and subtracting eq iv from iii, we get
• 9x 6y 33 ------ (iii)
• 4x 6y 8 ------- (iv)
• 5x 25
• gt x 5

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• 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

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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

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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

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CROSS MULTIPLICATION METHOD

• x b1c2 - b2c1
• a1b2 - a2b1
• y c1a2 - c2a1
• a1b2 - a2b1

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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.

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• 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

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• ACTIVITY
• To verify graphically
• That the pair of linear equations xy-50
  ,2x2y-60 in which a1/a2b1/b2 / c1/c2 gives a
  pair of parallel straight lines.
• x y 5 ..(1)
• 2x 2y 6(2)
• SOLUTION FOR XY5
• X 0 5
• Y 5 0
• SOLUTION FOR 2X2Y6
• X 0 3
• Y 3 0
• After plotting the points on graph we get

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Which represent the pair of parallel lines
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THANKS

• PREPARED BY
• BHAVYA KUMAR SHARMA
• TGT (MATHS)
• K V PAURI