![the lines may intersect in a single point. In this case, the
pair of equations
has a unique solution (consistent pair of equations).
the lines may be parallel. In this case, the equations have
no solution
(inconsistent pair of equations).
the lines may be coincident. In this case, the equations have
infinitely many
solutions [dependent (consistent) pair of equations].](https://image.slidesharecdn.com/class10linearequ-241110082408-39b4cf14/85/Class-10-linear-equation-in-two-variables-5-320.jpg)
Class 10 linear equation in two variables
- 1. PAIR OF LINEAR EQUATIONS IN TWO VARIABLES
- 2. Equation which can be put in the form ax + by + c = 0, where a, b and c are real numbers, and a and b are not both zero, is called a linear equation in two variables x and y
- 3. The general form for a pair of linear equations in two variables x and y is a1x + b1y + c1 = 0 and a2 x + b2 y + c2 = 0, where a1 , b1 , c1 , a2 , b2 , c2 are all real numbers.
- 5. the lines may intersect in a single point. In this case, the pair of equations has a unique solution (consistent pair of equations). the lines may be parallel. In this case, the equations have no solution (inconsistent pair of equations). the lines may be coincident. In this case, the equations have infinitely many solutions [dependent (consistent) pair of equations].
- 6. A pair of linear equations which has no solution, is called an inconsistent pair of linear equations. A pair of linear equations in two variables, which has a solution, is called a consistent pair of linear equations. A pair of linear equations which are equivalent has infinitely many distinct common solutions. Such a pair is called a dependent pair of linear equations in two variables. Dependent pair of linear equations is always consistent.
- 7. ALGEBRAIC METHODS OF SOLVING A PAIR OF LINEAR EQUATIONS Substitution Method Elimination Method Cross - Multiplication Method
- 8. Solve the following pair of equations by substitution method: 7x – 15y = 2 (1) x + 2y = 3 (2) X+2y=3, x= 3-2y Sub x in equ 1 7(3-2y) – 15y = 2 21-14y-15y =2 21-29y=2 -29y= 2-21 -29y = -19 Y= 19/29 X= 3-2*19/29 = 49/29
- 9. Aftab 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. x-7=7(y-7) X-7=7y-49 X-7y=-49+7=-42 X-7y+42=0 (1) X+3= 3(y+3) X+3=3y+9 X-3y+3-9=0 X-3y-6=0 (2) X-3y-6=0 X=3y+6 x= 3*12+6= 42 3y+6-7y+42=0 -4y+48=0 -4y=-48 y= 12