linear equations class 10 by aryan kathuria
TRANSCRIPT
Linear Equations
Definition of a Linear Equation
A linear equation in two variable x is an equation that can be written in the form ax + by + c = 0, where a ,b and c are real numbers and a and b is not equal to 0.
An example of a linear equation in x is .
Let ax + by +c = O , where a ,b , c are real numbers such that a and b ≠ O. Then, any pair of values of x and y which satisfies the equation ax + by +c = O, is called a solution of it.
A Pair of Linear Equation In Two VariablesCan Be Solved –i. By Graphical Methodii. By Algebraic Method
SOLUTIONS OF A LINEAR EQUATION
GRAPHICAL SOLUTIONS OF A LINEAR EQUATION
Let us consider the following system of linear equations in two variable
I. 2x-y=-1II. 3x+2y=9
Now solution of each equation are to be taken by-i. Firstly express one variable in terms of other.ii. Now assign any value to one variable and then
determine the value of other variable.Then plot the equations and determine the solutionsOf system of linear equation in two variables
For Eq. 1 2x - y= -1 y = 2x + 1Put x = 0y = 2(0) + 1 y = 1Put x = 2y = 2(2) + 1 y = 5
For Eq. 23x + 2y = 9y = 9 - 3x/2
Put x = 3y = 9 - 3(3) /2 y = 0
Put x = -1y = 9 - 3(-1) /2 y = 6
X 0 2
y 1 5
X 3 -1
y 0 6
1 2 3 4 -4 -3 -2 -1
54321
-1-2-3-4-5-6
( 2,5 )( -1,6 )
( 3,0 )( 0,1 )
X = 1Y = 3
Is the solution for system of equations
TYPES OF SOLUTIONS of system of equations
UNIQUE SOLUTION
consistent a1/a2 ≠ b1/b2 Lines intersecting at
a point
INFINITE SOLUTIONSconsistent
a1/a2 = b1/b2 = c1/c2 Coincident Lines
NO SOLUTION
Non consistent
a1/a2 = b1/b2 ≠ c1/c2 Parallel Lines
TYPES OF METHOD:- TO SOLVE A PAIR OF LINEAR EQUATION IN TWO VARIABLE
BY ALGEABRIC METHODS
I. Elimination MethodII. Substitution Method
III.Cross-Multiplication Method
• Make one variable equal in both the equations and then either add or subtract the equations to eliminate the variable.
• The resulting equation in one variable is solved and then by substituting this value of the variable in either of the given equation, the value of the other variable is also obtained.
ELIMINATION METHOD
Q. Solve using the method of Elimination
I. 2x + y = 8II. 2(x + 6y = 15)For making the coefficients of x in eq.(1) and eq.(2) equal, we multiply eq.(2) by 2 and get 2x + y = 8 -(3) 2x + 12y = 30 -(4)Subtracting Eq.(3) from eq.(4), we have 11y = 22 y = 2Put this value of y in eq.(1) 2x + 2 = 8 2x = 6 x = 3
• Find the value of one variable in the terms of other variable. Substitute it in other equation and we will get value of one of the variable.
Then put the value of variable in any one of the equation and the value of other variable is also obtained.
SUBSTITUTION METHOD
Q. Solve using the method of Substitution
I. x + 2y = -1II. 2x - 3y = 12
From Eq.(1), we have x = -1 – 2y - (3) Substituting this value of x in eq.(2), we have 2(-1 – 2y) – 3y =12 - 2 - 4y – 3y = 12 - 7y = 14 y = -2Put this value of y in Eq.(3), we have x = -1 -2(-2) x = -1 + 4 x = 3
Q. Solve using the method of cross-Multiplication
5x + 3y = 35 5x + 3y - 35 = 02x + 4y = 28 2x + 4y - 28 = 0 a1 = 5 b1 = 3 c1= -35a2 = 2 b2 = 4 c2= -28
___x___ = ___y___ = ___1___ (3)(-28)-(4)(-35) (-35)(2)-(-28)(5) (5)(4)-(2)(3) -84 + 140 -70 + 140 20 – 6 56/14 70/14 14 x = 4 and y = 5