1) The document discusses equations, including defining what an equation is, the difference between constants and variables, how to form equations from statements and vice versa, and how to solve simple equations by adding, subtracting, multiplying, or dividing both sides of the equation by the same quantity. 2) It provides examples of forming equations from word problems and statements, and solving equations through arithmetic operations like subtraction, division, and transposition (moving a term to the other side of the equation). 3) Solving equations involves adding or subtracting the same quantity to both sides of the equation or multiplying/dividing both sides by the same non-zero quantity while maintaining the equality. This allows for isolating the variable

1. Class-7
Mathematics
SimpleEquation
By: FaiyazulAhmad

2. Contents t o be covered
2
• Equation
• Constants & Variables
• Forming an equation
• Solving an equation
• Understanding transpose

3. Introduction

4. What is an equation
An equation is formed when two
expressions are equal to each other .
An equation helps us in finding the
unknown quantity in the expression.
Value of expression on LHS is always
equal to value of expression RHS. It
doesn’t change if the expression are
interchanged.
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5. Constants Va r ia b les
• Constants: A constant term is a term in an
algebraic expression that has a value that is
constant or cannot change, because it does not
contain any modifiable variables. For example, in
the quadraticpolynomial.the 3 isa constantterm
• Variables : A variable is a quantity that may
change within the context of a mathematical
problem or experiment. Typically, we use a single
letter to represent a variable. The letters x, y, and z
arecommongenericsymbolsusedforvariables.
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6. Setting up an Equation
• Think of any number.
• Thenhisfriend askedhim to multiply the number he thought
with4.
• Later on he was asked to add five to it(the number he got on
multiplying with 4).
• Thenhisfriend askedhim what answerhe got, the friend replied
65.
• Sothe equation with this descriptionformed is4x +5=65
• Letsput the different value of xie
x=1=>4*1 +5 = 9
x=5 =>4*5 +5 =25
x=15 =>4*15 +5 =65
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7. Forming equation from statement
• Equationscanalso be formed from the
statements.Example,
• Thesum of3 times xand 11 is 32.
• Sowe can write it as:
3 times x=3x
Sumof 3 times xand 11 =3x+11
Sumof 3 times xand 11 is 32=
3x+11 =32
• Sothe equation formedis3x+11 =32
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8. Forming Statement from an Equation
Statement canalso be formedfrom
the given equation.Example,
• x- 5 =9
Sowe canwrite the statementas:
Takingaway five fromxgives9
• 5p =20
Sowe canwrite the statementas:
5 timesof a numberpisequalto20
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9. Solving an equation
• A balanced equation is like a weighing
machine with equal weights on its both
the pan .That is if we add same weight on
both the pan the arms remains horizontal.
Similarly if we remove if we remove same
weight from both the pan the arms
remains horizontal.
• Similarly in an equation if we add or
subtract the same number from both the
sides, the equation remains unchanged.
Case is same in multiplication and division
too.
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10. Example o f solving o f equation
Letstake an example:
x+3 = 8
Now we shall subtract 3 from both the sides
(on subtracting 3 the equation on LHS is
reduced to x),ie
x+3 – 3 =8 – 3
=x= 5
We can check the answer by putting the value
of xas5, sowe get 5 +3 =8 =>8 =8
=>LHS=RHS.
Hence the value of x is correct.
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11. Example o f solving o f equation
Letstake an example:
5y =35
Now we shall divide 5 from both the sides
(on dividing5 the equation isreduced to y), ie
5y / 5 =35 / 5
=y = 7
Wecan check the answer by putting thevalue of
y as 7, so we get 5 x 7 =35 => 35 = 35
=>LHS=RHS.
Henceour valueof y is correct.
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12. Example o f solving o f equation
Letstake anexample:
12p– 5 = 25
s
ides
Now we s
hall add 5 on both the
(on adding5 the equationisreducedto 12p),ie
12p- 5 +5 =25 +5 =12p= 30
Now we divide 12 from both the sides (on dividing 12 the
equationisreducedto p),ie
=12p= 30
=12/12 p=30 /12 =>p=5 / 2
We can check the answer by putting the value of p as 5 / 2,
sowe get 12 x5 /2 -5 = 25 => 60 / 2 -5 = 25 => 50 /2 = 25 =>
25 =25=>LHS=RHS.
Henceour value of piscorrect.
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13. Transpose
Transposemeanstransferring/ changingthe
sideof the numberin theequation.
We haveanequation12p – 5 = 25
Ontransposing5 ie changingthe sidefrom LHSto
RHSwe get, 12p =25 +5
 12p =30
 p= 30/12
 p= 5/2
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14. Lets brush up with an example
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15. Closure
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