NCTB Class 6 Math solution – Linear Equation

The article is about the NCTB class 6 math solution chapter 9. Chapter 9 is describe about the Linear Equation. Moreover, This is a mathematical statement and equality. So, Mathematical statement with the sign of equality is called an Equation. Here we say, the unknown quantity x is a variable. Usually, the small letters in the English alphabets like as( a,b,d,x,y,z,p,q) are used for the variable or the unknown quantity and simple equation with unknown quantity or variable is known as a Linear Equation.

NCTB Class 6 Math solution – Linear Equation

In addition to, The process of finding the unknown quantity or the variable in an equation is called the solution of the equation. So, The value of the solution is the root of the equation. If the value of the root is substituted on both sides of the equation, then the left hand side and right hand side are equal.

Things to know for finding solutions of equations

  1. If a quantity is added to each equal quantity of an equation, then the sums
    will be equal to each other.
  2. If a quantity is subtracted from each equal quantity, then subtracted results
    will be equal to each other
  3. Though, each equal quantity of an equation is multiplied by a quantity, then the
    products will be equal.
  4. If each equal quantity is divided by a non-zero quantity, then the quotients
    will be equal.
    When you are going to solve the problem of Linear Equation then, you have to know about the above rules. Class six students want to solve about the chapter 9 of Linear Equation. I hope, you get here all solution of class 6 math book 2023. Moreover, you see the new curriculum in 2023. So, You have to know about the new curriculum math solution 2023. We are trying to class 6 math solve by easiest method.
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NCTB Class 6 Math solution – Linear Equation

CHAPTER 1 : The Story of Numbers
CHAPTER 2 : The Story of Two Dimensional Objects
CHAPTER 3 : Information Investigation and Analysis
CHAPTER 4 : Trees of Prime Factors
CHAPTER 5 : Measurement of Length
CHAPTER 6 : The World of Integers
CHAPTER 7 : The Game of Fractions part 1
The Game of Fractions : part 2
CHAPTER 8: World of Unknown expressions
CHAPTER 9: Linear Equation
CHAPTER 10 : Story of Three Dimensional Objects
CHAPTER 11 : Unitary Method, Percentages and Ratio
Chapter 12 : Set the Formula, get the Formula

Below, you will see chapter 9 solution.

1 number math solution class 6 chapter 9

NCTB class 6 math solution chapter 9
NCTB class 6 math solution chapter 9

NCTB Class 6 Math solution – Linear Equationv

Collect your number 1 math from the image.

3 number math solution


i) 2x+5= 15
Or, 2x = 15-5
Or, 2x = 10
Or, x= 10/2
Or, x = 5
So, answer is 5

So, x= 10.
Then, lefthand side= 2.10+5=20 + 5 =25,
Whether, is not equal to the right side.
Again, x= -5,
Then, Lefthand side = 2.(-5)+5=-10+5=-5
So, Which is not equal to the right-hand side.
That’s why, 10 and -5 are not base value of 2x+5=15
ii) 5 – y=7
Or, -y=7-5
Or, -y=2
Or, y=2
So main base value = -2

NCTB Class 6 Math solution - Linear Equation
NCTB Class 6 Math solution – Linear Equation

Now,y=12 Then, Lefthand side 5-12 = -7
Which is not equal to right-hand side.
Again, y=2 then,5-2=3
Whether, lefthand side and right-hand side are not equal.
That’s why, 12 and 2 are not the base value of 5-y=7
iii) 5x-2= 3x+8
Or, 5x,-3x=8+2
So,2x= 10


Or, x= 10/5
So, x= 2
So, the main base value =2

NCTB Class 6 Math solution – Linear Equation

Now, x= 1
Then, Lefthand side
=5.1-2= 3.1 + 8
0r,= 3=11, Which is not equal to the right-hand side.
X= -5 then, Lefthand side = 5, (-5) -2= -25-2= -27
Right-hand side =3.-5) + 8 = -7
Whether, two side are not equal
That’s why 1 and -5 are base value of 5x-2= 3x+8
iv) 2y+2= 16
Or, 2y= 16-2
Or, 2y=14
Or, y= 14/2
So,y = 7
So, main base value = 7
Now, y= 18 Then,
Lefthand side, 2.18 + 2 =36+2=38
So, which is not equal to the right-hand side.
Again, x= 9 Then, left hand side= 2. 9 + 2 =18 + 2 =20
So, which is not equal to the right side.
That’s why, 18 and 9 are not base value of 2y+2= 16

v) 4z – 5 = 2z + 19
Or, 4z – 2z = 19+5
Or, 2z = 24
Or, z= 24/2
Or, z = 12
So, right base value= 12
Now, z=7 Then, Lefthand side = 4.7 -5=28-5=23
Right-hand side = 2.7 + 19 =14+19=33
So, two part are not equal
Again, z=4
Then, lefthand side = 4.4-5= 16-5=11
Right-hand side = 2.4+19=8+19=27
So, two sides are not equal.
That’s why, 7 and 4 are not base value of 4z-5=2z+19

4 number math solution chapter 9


a)
1 dozen = 12
1 pen price = x tk
12 pen price =12x tk
So, 100-12x=40
Or, -12x= 40-100
Or,-12x= -60
Or, 12 x= 60
Or, x= 60/12
Or, x= 5
So, Every pen price =5 tk
b) 1 exercise book price= 12 tk
Y exercise book price =12y tk
So,
40-12y=4
Or, -12y= 4-40
Or, -12y = -36
Or, 12y =36
Or, y= 36/12
Or,y=3
So, mina purchase exercise book=3

NCTB Class 6 Math solution - Linear Equation
NCTB Class 6 Math solution – Linear Equation

5 number NCTB math solution class 6 chapter 9

Suppose, Mr Karim invest x tk by 10% profit
So, Mr Karim invest tk (56000 – x) by 12% profit.
Given that,
(56000-x) ×12%+x×10%=6400
Or, (56000-x) ×12/100+x×10/100 =6400
Or, (56000-x) ×12+x×10= 6400 ×100 ( multiplying by 100 both side)
Or, 56000 × 12 -12x +10x= 640000
Or, -2x = 640000 – 672000
Or, -2x = -32000
Or, 2x= 32000
Or, x= 32000/2
So, x= 16000
So, Mr Karim invest 16000. Tk by 10% profit

6 number math solution class 6 chapter 9

We know that, A Cricket match century means 100 runs and double century means 200 runs.
So, The total runs of Sakib and Mushfiqur = (200-2)= 198 runs
Suppose, Runs of Mushfiqur = x
So, run of sakib = 2x
According to question,
X+2x= 198
Or, 3x = 198
Or, x= 198/3
Or, x= 66
So, Run of Musfiqur = 66 and Run of Shakib run= 66×2 = 132 run

8 number math solution class6 chapter 9

8) a.

So, The number of water bottle = x
1 water bottle weight = 150 gm
Weight of bag = 50 gm
Weight of water bottle + weight of bag= y
So,
1 water bottle weight is = 150gm
So, x water bottle weight = 150x gm
We know that,
Bottle weight + Bag weight = y
So, 150x + 50= y
So, the equation is = 150x + 50= y

B) From’ a ‘ we get,
150x + 50 = y
Or, y= 150x + 50
Or, y= 150×15 + 50
So, y= 2300
C) we get from ‘a’,
150x + 50 = y
Or, 150x + 50 = 1100
Or, 150x = 1100 – 50
Or, 150x = 1050
Or, x= 1050/150
So, x= 7


9 number math solution class 6 chapter 9


9) a.
1 packet biscuit price = 20 tk
So, X packet biscuit price = 20x tk
Now, x packet biscuit price + 1 bottle drink price = y tk
Or, 20x + 15 = y
So, The relations of x and y= 20x + 15 = y

B,
Or, 20‌×25 +15= y
Or, 500 + 15 = y
So, y, = 515

C.
So, From ‘a’ we get,
20x + 15 = y
Or, 20x + 15 = 255
Or, 20x = 255 – 15
Or, 20x = 240
Or, x= 240/20
So, x= 12

10 number math solution class 6 chapter 9

  1. A) Given that,
    Width of field = x merer
    Length of field = x + 16 meter
    So, Perimeter of the field
    = 2×( length + width)
    = 2 × ( x + 16 + x) meter
    = 2 × ( 2x + 16) meter
    = 4x + 32 meter
    So, The perimeter is = 4x + 32 meter
  2. So, Below the number B

B)
Given that, perimeter of field = 120 meter
Now, from a we get, The perimeter is = 4x + 32
So, 4x + 32 = 120
Or, 4x = 120 – 32
Or, 4x = 88
Or, x = 88/4
So, x = 22


Width of field = 22 meter
So, length of field = 22 + 16 meter.
So, length = 38 meter

So, Area of field
= (length × width) Square
= ( 38 × 22) Square Meter
So, answer is = 836 Square meter

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