This document provides an introduction to linear equations for 6th grade mathematics. It discusses key concepts like equality, equations, linear equations involving one variable, and properties of equations. Examples of forming equations from statements and vice versa are provided. Methods for solving different types of linear equations are described, including transposing terms to one side and using properties of equations. Learning objectives focus on understanding concepts like equality and equations, solving one-variable linear equations, and representing word problems with equations. Learning outcomes emphasize solving systems of linear equations and translating between real-world situations and algebraic representations.

1. SUB: MATHEMATICS
STD - VI
TOPIC: LINEAR EQUATIONS

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of chapter linear equations

3. LEARNING OBJECTIVES
Students will be able to
i. understand the concept of equality
ii. investigate the meaning of an equation
iii. solve first degree equations in one variable with
coefficients
iv. investigate what equation can represent a
particular problem.

4. INTRODUCTION

5. Imagine that you are taking a taxi while on vacation.
You know that the taxi service charges ₹90 to pick
your family up from your hotel and another ₹15 per
km for the trip. After the completion of your trip
the taxi driver gave you a bill of ₹1440. If we take
“x" to represent the distance in km travelled, the
linear equation would be:
15x + 90 = 1440

6. A statement which states that two algebraic
expressions are equal is called an equation.
3x  2y  8
6  x  x  9
3X2
 5  X 5
Some more examples:

7. The equation involving only one variable in first order is
called a linear equation in one variable.
3x 5 0
8y  2
7a153a

8. A LINER EQUATION IS ALSO CALLED A FIRST DEGREE
EQUATION AS THE
HIGHEST POWER OF VARIABLE IS 1.
EXAMPLE OF LINER EQUATIONS :
x + 4 = - 2
2x + 5 = 10
5 – 3x = 8

9. Formation of Equation From
Statement
A number increased by 8 is equal to 15 .Find the number.
Solution: let be a number ‘x’
Given ,the number increased by 8 equal 15.
=>x+8 = 15
=>x= 15 -8
=> x=7

10. Formation of statement from equation
2x+5 = 10
Two times a number increased by 5 is equals to 10
x-5 = 12
Five subtracted from a number is equals to 12

11. ACTIVITY (Art Integration) Click to view

12. PROPERTIES OF AN EQUATION
•If same quantity is added to both sides of the equation, the sums are equal.
Thus: x=7 => x+a=7+a
•If same quantity is subtracted from both sides of an equation, the
differences are equal
Thus: x=7 => x-a=7-a
•If both the sides of an equation are multiplied by the same quantity, the
products are equal.
Thus: x=7 => ax=7a
•If both the sides of an equation are divided by the same quantity, the quotients
are equal.
Thus: x=7 => x÷a=7÷a

13. 3.To solve an equation of the form ax=b
Q. Solve 3x=9
Solution: 3x=9
=> 3x/3= 9/3
=> x = 3
4. To solve an equation of the form x/a=b
Q. Solve x/2 = 6
Solution: x/2 = 6
=> x = 6x2
=> x = 12

14. SHORT- CUT METHOD (SOLVING AN EQUATION BY TRANSPOSING TERMS)
1.In an equation, an added term is transposed (taken) from one side to the other, it is
subtracted.
i.e., x+4=10
=> x=10-4=6 (4 is transposed)
2. In an equation, a subtracted term is transposed to the other side, it is added.
i.e., y-6=5
=>y=5+6=11
3. In an equation, a term in multiplication is transposed to the other side, it is divided.
3x=12
=>x=12/3=4
4. In an equation a term in division is taken to the other side it is multiplied. i.e
=> y=6×4=24 (4 is transposed)
(6 is transposed)
(3 is transposed)

15. Solve: 3x+8=14
Solution: 3x=14-8 (transposing 8)
=> 3x=6(transposing 3)
=> x=6/3
=>x=2
To Solve a Linear Equation by using more than one
property

16. (2) 2a-3=5
Solution: 2a-3=5
=> 2a=5+3 (transposing 3)
=> 2a=8
=> a= 8/2 (transposing2)
=> a=4

17. Transpose the terms containing the variable, to one side and the
constants to the other side.
E.g.:(1) Solve 10y-3=7y+9
Solution: 10y-7y = 9+3 (transposing 7y to the left & 3 to the right)
=> 3y = 12
=> y = 12/3
=> y = 4
Solving Equation with variables in both sides

18. (2) Solve: 2(x-5) + 3(x-2) = 8+7(x-4)
Solution:
2x-10+3x-6 = 8+7x-28(removing the brackets)
=> 5x-16 = 7x-20
=> 5x-7x = -20+16
=>
=>
=>
-2x = -4
x = -4/-2
x = 2

19. EXAMPLES
Q1. Solve x+4=10 (2 Marks)
Solution: x+4=10
x+4-4=10-4 (subtracting 4 from both the sides) [1]
x=6 [1]
Q2. Solve y-6=5 (2 Marks)
Solution: y-6=5
y-6+6=5+6 (Adding 6 in both sides) [1]
=> y=11 [1]

20. Q3. Solve 3(x+5)=18 (4 Marks)
Solution: 3(x+5)=18
=> 3x+15= 18 (Transposing 15 ) [1]
=> 3x = 18 -15 [1]
=> 3x = 3 [1]
=> x= 3/3 (Transposing 3) [0.5]
=> x = 1 [0.5]

21. LEARNING OUTCOMES
As a result of studying this topic, students will be
able to:
1.solve system of linear equation by substitution/ elimination
method.
2.Translate word problems into algebraic expression and equation.
3.Expand the given expression using distributive property. 4.Find out
linear equation in one variable has one solution
5.Create equations based on real life situation and solve a system of
linear equation
6.corelate real world situations and solve algebraic problems using
their knowledge on linear equation.

22. CONCEPT MAP

23. THANK YOU