CLASS 7MATHEMATICS : CLASS 7MATHEMATICS SIMPLE EQUATIONS
An Equation : An Equation What is an equation?
An equation is a condition on a variable.
A variable is something that can vary , i.e; change. A variable takes on different numerical values ; its value is not fixed.
Variables are usually denoted by letters of the alphabet , such as x,y,z,l,m,n,etc;
From variables , we form expressions.
The expressions are formed by performing operations like addition,subtraction,multiplication and division on the variables.
In an equation there is always an equality sign.
in short : : in short : An equation is a condition on a variable. The condition is that two expressions should have equal value. One of the two expressions must contain the variable.
E.g. 4x + 5 = 6x - 25
Write the following statements in the form of equations : Write the following statements in the form of equations The sum of three times x and eleven is 32.
Five times a number p is 20.
Add 7 to three times n to get 1.
Taking away 5 from x gives 9. 3x + 11 = 32
5p = 20
3n + 7 = 1
x – 5 = 9
Setting up an equation for word problems : Setting up an equation for word problems E.g. Raju’s father’s age is 5 years more than3 times Raju’s age. Raju’s father is 44 years old. Set up an equation to find Raju’s age.
We don’t know Raju’s age. Let us take it to be y years.
Three times Raju’s age is 3y years.
Raju’s father’s age is 5 years more than 3y.
Raju’s father’s age would be (3y+5) years old.
It is given that Raju’s father’s age is 44 yrs.
Therefore, 3y + 5 = 44
This is an equation in y.
It will give Raju’s age when solved.
Solving an equation : Solving an equation Adding or subtracting on both sides
(i) 3p -10 = 5
Add 10 to both the sides
3p -10+10 = 5+10
3p =15
(ii) 5x +12 = 27
Subtract 12 from both the sides
5x+12-12 = 27-12
5x = 15 Transposing
(i) 3p–10 = 5
Transpose (-10) from L.H.S to R.H.S
(On transposing -10 becomes +10)
3p = 5+10 or 3p = 15
(ii) 5x + 12 =27
Transposing +12
(on transposing +12 becomes -12)
5x = 27-12 or 5x = 15
Slide 7 : Time for some examples to solve
5a + 28 = 10 : 5a + 28 = 10 To solve this equation
Let us subtract 28 from both the sides of the equation
5a + 28 – 28 = 10 – 28
5a = - 18
a = -18/ 5
4p + 10 = -2 : 4p + 10 = -2 To solve this equation
Let us subtract 10 from both the sides of the equation
4p + 10 – 10 = -2 + 10
4p = 8
p = 8/4
p = 2
2 ( x + 4 ) = 12 : 2 ( x + 4 ) = 12 To solve this equation
Let us divide by 2 on both the sides of the equation.
2(x+4) / 2 = 12/2
x + 4 = 6
Subtract 4 from both the sides
x+4-4 = 6-4
x = 2
Solve the following word problem: : Solve the following word problem: The teacher tells the class that the highest marks obtained by a student in her class is twice the lowest marks plus 7. The highest score is 87.What is the lowest score?
Slide 12 : Let the lowest marks be m.
Twice the lowest marks = 2m
Highest marks obtained = 87
From the given condition:
2m + 7 = 87
To solve the equation , let us subtract 7 from both the sides
2m + 7 – 7 = 87 -7
2m = 80
m = 80/2
m = 40
The lowest score in the class is 40.
Let us take one more example : Let us take one more example Irfan says that he has 7 marbles more than 5 times the marbles Parmit has. Irfan has 37 marbles. How many marbles does Parmit have?
Slide 14 : Let the number of marbles Parmit has be y.
5 times the marbles Parmit has = 5y
Number of marbles Irfan has = 37
From the given condition:
5y + 7 = 37
Subtract 7 from both the sides of the equation
5y + 7 – 7 = 37 -7
5y = 30
Dividing by 5 on both the sides
y = 6
Parmit has 6 marbles with him.
Slide 15 : ANY DOUBTS ???
Slide 16 : THANK YOU STUDENTS !!