Maths ppt For Linear Equation In One Variable Class 10th!!
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Transcript of Maths ppt For Linear Equation In One Variable Class 10th!!
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Created By AbhishekClass-10
Roll No-03
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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 2x – 3y + 4 = 0.
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A pair of linear equations in two variables can be
solved by the:
(i) Graphically method
(ii) Algebraic method
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GRAPHICAL SOLUTIONS OF A LINEAR EQUATION
Let us consider the following system of two simultaneous linear equations in
two variable.
2x – y = -1
3x + 2y = 9
Here we assign any value to one of the two variables and then determine the value of the other variable from the
given equation.
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xxx 00
xY
xY
0 2
1 5
3 -10 6
For the equation
2x –y = -1 ---(1) 2x +1 = y Y = 2x + 1
3x + 2y = 9 --- (2)2y = 9 – 3x 2Y = 9- 3x
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Types of Solutions of Systems of Equations
• One solution – the lines cross at one point
• No solution – the lines do not cross
• Infinitely many solutions – the lines coincide
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Algebraic method TO SOLVE A PAIR
OF LINEAR EQUATION IN TWO
VARIABLE
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To solve a pair of linear equations in two variables algebraically, we have
following methods:-
(i) Substitution method
(ii) Elimination method
(iii) Cross-multiplication method
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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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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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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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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.
• For example: we want to solve,
3x + 2y = 11
2x + 3y = 4
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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
x = 5
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• putting the value of X 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 METHODLet’s consider the general form of a pair of linear equations.
To solve this pair of equations for and using cross-multiplication, we’ll arrange the variables and their
coefficients
, and , and the constants and
We can convert non linear equations in to linear equation by a suitable substitution
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CROSS MULTIPLICATION METHOD
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