अध्याय 4: सरल समीकरण Class 7 Math
Simple Equations Class 7 Notes: Chapter 4
Introduction to Simple Equations
Variables and Expressions
Variable is a quantity that can take any value, its value is not fixed. It is a symbol for a number whose value is unknown yet.
Expressions are formed by performing operations like addition, subtraction, multiplication and division on the variables.
Example: 6x – 3 is an expression in variable x.
Algebraic Equation
An equation is a condition on a variable such that two expressions in the variable should have equal value.
Example: 8x−8=16 is an equation.
The value of the variable in an equation for which the equation is satisfied is called the solution of the equation.
Example: The solution for the equation 2x−3=5 is x=4.
Mathematical Operations on Expressions
- Addition of variables: (3x+4z)+(5y+6)
- Subtraction of variables: (4x−7y)−(6y+5)
- Multiplication of variables: (5xy+6)×7x
- Division of variables: (8xz+5z)/(5x-6y)
Solving an Equation
Solving an equation involves performing the same operations on the expressions on either side of the “=” sign so that the value of the variable is found without disturbing the balance.
Example : Solve 2x+4=10
Consider 2x+4=10
⇒2x+4−4=10−4 [Subtracting 4 from both LHS and RHS] ⇒2x=6
⇒2x/2=6/2 [Dividing both LHS and RHS by 2] ⇒x=3
Methods of Solving an Equation
Method 1: performing the same operations on the expressions on either side of the “=” sign so that the value of the variable is found without disturbing the balance.
Opertions involve Adding, subtracting, multipling or dividing on both sides.
Example: x+2=6
Subtract 2 from LHS and RHS
⇒ LHS: x+2−2=x
⇒ RHS: 6−2=4
But LHS = RHS
⇒ x = 4
Method 2: Transposing
It involves moving the terms to one side of the equation to find out the value of the variable.
When terms move from one side to another they change their sign.
Example: x+2=6
Transpose (+2) from LHS to RHS
⇒x=6−2
⇒x=4
Applying Equations
Forming Equation from Solution
Given a solution, many equations can be constructed.
Example: Given solution: x = 3
Multiply both sides by 4,
⇒ 4x=4×3
Add -5 to both sides,
⇒ 4x−5=12−5
⇒ 4x−5=7
Similarly, more equations can be constructed.
Applications (Word problem)
Example: Ram’s father is 3 times as old as his son Ram. After 15 years, he will be twice the age of his son. Form an equation and solve it.
Solution: Let Ram’s age be x.
⇒ His father’s age is 3x.
After 15 years:
3x+15=2(x+15)
On solving,
3x+15=2x+30
3x−2x=30−15
x=15
∴ Ram’s age is 15 and his dad’s age is 45.