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Linear Equation in two
Variables
Definition:
A pair of linear equations in two variables is
said to form a system of simultaneous linear
equations.
Eg: a1x + b1y + c1=0
a2x + b2y + c2=0
where a1, a2, b1, b2, c1, c2 are real
numbers.
Consistent system:
A system of simultaneous linear equations is
said to be consistent, if it has at least one
solution.
In-consistent system:
A system of simultaneous linear equations is
said to be in-consistent, if it has no solution.
Solutions
To solve a pair of linear equations in two
variables can be solved by the :
1) Graphical method
2) Algebraic method
Graphical method
Unique Solution: If a1/a2 = b1/b2
y
x’ x
a1x+b1y+c1=0 a2x+b2y+c2=0
Intersecting solution y’
No solution: a1/a2 = b1/b2 = c1/c2
x
a1x+b1y+c1=0
y’ y
a2x+b2y+c2=0
Parallel lines x’
Infinitely many solution: if
a1/a2=b1/b2=c1/c2
y
a1x+b1y+c1=0
x’ x
a2x+b2y+c2=0
Coincident lines
y’
Algebraic method
1) Substitution method.
2) Elimination method.
3) Cross-multiplication method.
Substitution Method
In this method, we express one of the
variables in terms of the other
variables from either of the two
equations and then this expression is
put in the other equation to obtain an
equation in one variable as explained
in the following algorithm.
Algorithm
Step I obtain the two equations. Let the equations be
a1x+b1y+c1=0
and a2x+b2y+c2=0
Step II Choose either of the two equations, say (i), and find the
value of one variable, say y, in terms of the other,
i.e. x.
Step III Substitute the value of y, obtained in step II, in the other
equation i.e. (ii) to get an equation in x.
Step IV Solve the equation obtained in step III to get the value of
x.
Step V Substitute the value of x obtained in step IV in the
expression for y in terms of x obtained in step II to get
the value of y.
Step VI The values of x and y obtained in steps IV and V
respectively constitute the solution of the given system
of two linear equation.
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 one of the
given equations, the value of the other
variable can be obtained. Following
algorithm explains the procedure.
Algorithm
Step I Obtain the two equations.
a1x+b1y+c1=0
a2x+b2y+c2=0
Step II Multiply the equation so as to make the coefficients of
the variable to be eliminated equal.
Step III Add or subtract the equations obtained in step II
according as the terms having the same coefficients are
opposite or of the same sign.
Step IV Solve equation in one variable obtained in step III.
Step V substitute the value found in step IV in any one of the
equations and find the value of the other variable.
The values of the variables in steps IV and V constitute the
solution of the given system of equations.
Cross-multiplication Method
Theorem
Let a1x+b1y+c1=0
a2x+b2y+c2=0
be a system of linear equation in two variables x and y
such that a1/a2 = b1/b2 i.e. a1b2 – a2b1 = 0. Then the
system has a unique solution given by:
(b1c2 – b2c1) (c1a2 – c2a1)
x = and y =
(a1b2 – a2b1) (a1b2 - a2b1)

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Maths project

  • 1. Linear Equation in two Variables
  • 2. Definition: A pair of linear equations in two variables is said to form a system of simultaneous linear equations. Eg: a1x + b1y + c1=0 a2x + b2y + c2=0 where a1, a2, b1, b2, c1, c2 are real numbers.
  • 3. Consistent system: A system of simultaneous linear equations is said to be consistent, if it has at least one solution. In-consistent system: A system of simultaneous linear equations is said to be in-consistent, if it has no solution.
  • 4. Solutions To solve a pair of linear equations in two variables can be solved by the : 1) Graphical method 2) Algebraic method
  • 6. Unique Solution: If a1/a2 = b1/b2 y x’ x a1x+b1y+c1=0 a2x+b2y+c2=0 Intersecting solution y’
  • 7. No solution: a1/a2 = b1/b2 = c1/c2 x a1x+b1y+c1=0 y’ y a2x+b2y+c2=0 Parallel lines x’
  • 8. Infinitely many solution: if a1/a2=b1/b2=c1/c2 y a1x+b1y+c1=0 x’ x a2x+b2y+c2=0 Coincident lines y’
  • 9. Algebraic method 1) Substitution method. 2) Elimination method. 3) Cross-multiplication method.
  • 10. Substitution Method In this method, we express one of the variables in terms of the other variables from either of the two equations and then this expression is put in the other equation to obtain an equation in one variable as explained in the following algorithm.
  • 11. Algorithm Step I obtain the two equations. Let the equations be a1x+b1y+c1=0 and a2x+b2y+c2=0 Step II Choose either of the two equations, say (i), and find the value of one variable, say y, in terms of the other, i.e. x. Step III Substitute the value of y, obtained in step II, in the other equation i.e. (ii) to get an equation in x. Step IV Solve the equation obtained in step III to get the value of x. Step V Substitute the value of x obtained in step IV in the expression for y in terms of x obtained in step II to get the value of y. Step VI The values of x and y obtained in steps IV and V respectively constitute the solution of the given system of two linear equation.
  • 12. 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 one of the given equations, the value of the other variable can be obtained. Following algorithm explains the procedure.
  • 13. Algorithm Step I Obtain the two equations. a1x+b1y+c1=0 a2x+b2y+c2=0 Step II Multiply the equation so as to make the coefficients of the variable to be eliminated equal. Step III Add or subtract the equations obtained in step II according as the terms having the same coefficients are opposite or of the same sign. Step IV Solve equation in one variable obtained in step III. Step V substitute the value found in step IV in any one of the equations and find the value of the other variable. The values of the variables in steps IV and V constitute the solution of the given system of equations.
  • 14. Cross-multiplication Method Theorem Let a1x+b1y+c1=0 a2x+b2y+c2=0 be a system of linear equation in two variables x and y such that a1/a2 = b1/b2 i.e. a1b2 – a2b1 = 0. Then the system has a unique solution given by: (b1c2 – b2c1) (c1a2 – c2a1) x = and y = (a1b2 – a2b1) (a1b2 - a2b1)