Linear Equation in One Variable – Class 7 Math Solution – Chapter 12 ( 234 and 236 page individual task)

Class 7 math solution from NCTB book. Here, you will be get the solution about Linear Equation in One Variable of chapter 12 math solution. Today, We are discuss about the chapter 12. Here, you get Linear Equation in One Variable Class 7 Math Solution Chapter 12 ( 234 and 236 page task) . Though, it is your individual task but you need for help. So, we are providing the solutions of Linear Equation in One Variable Class 7 Math Solution Chapter 12 ( 234 and 236 page individual work). Moreover, you can read the whole post of Linear Equation in One Variable Class 7 Math Solution Chapter 12 ( 234 and 236 page individual work).

Read more : chapter 12: Linear equation in one variable

231 page individual task chapter 12

Linear Equation in One Variable Class 7 Math Solution Chapter 12 ( 234 and 236 page individual task)

1 Number question solution
Suppose, the number is = x
So, 5 added with 2 times x than the equation is = 2x + 5
According to question, Balance weight equilibrium is, One side 2x + 5 and another side 25. Now, from this method we can calculate the x,

Linear Equation in One Variable - Class 7 Math Solution -
Linear Equation in One Variable – Class 7 Math Solution –

So, Number is = 10
Now, the mathematical solve from the equation
2x + 5 = 25
Or, 2x + 5 – 5 = 25 – 5
Or, 2x = 20
So, x ‌= 20/2
X = 10
So, the number is = 10

2 number solution


Suppose, biggest number is = x
And, smaller number is = ( 55 – x)
According to question, 5x = 6 ( 55 – x)

So, the equilibrium weight in the balance is. One side 5x and another side is 6 ( 55 – x).

Linear Equation in One Variable - Class 7 Math Solution -
Linear Equation in One Variable – Class 7 Math Solution –

So, biggest number is = 30
And, smaller number is = 55 – 30= 25
So, the numerical solution is
5x = 330 – 6x
Or, 5x = 330 – 6x
Or, 5x + 6x = 330
Now, 11 x = 330
Or, x = 330/11
Or, x = 30
So, biggest number is = 30
And, smaller number is = 55 – 30= 25

3 number math solution

Suppose, Rita has x taka
So, Gita has x – 6 tk and mita has= ( x + 12) taka.
According to question,
X + (x-6) + ( x +12) = 180

3 number math solution
3 number math solution

Now, one side is X + (x-6) + ( x +12) and another side is 180 for equilibrium balance weight.
Now, Rita has 58 taka.
And, Gita has ( 58 + 12) = 70 taka
So, the numerical solution is
X + (x-6) + ( x +12) = 180
Or, 3x + 6 = 180
Or, 3x = 180 – 6
Now, 3x = 174
Or, x = 174/3
Or, x = 58
Now, Rita has 58 taka.
And, Gita has ( 58 + 12) = 70 taka

Individual task page 236

i) 3x – 2×2 = 7
Or, 3x – 2×2- 7 = 0
Or, – 2×2 +3x -7 = 0
So, 2×2 -3x + 7 = 0
So, the ideal equation is = 2×2 -3x + 7
And a, b, c = 2, -3, 7
ii) ( x – 7) (x +7) =3x
Or, x2 -7x +7x – 49 = 3x
Or, x2 – 49 = 3x
So, x2 – 3x – 49 = 0
And, a, b, c = 1, -3, -49

iii) 5 + 2z2= 6z
Or, 5 + 2z2 -6z = 0p
Or, 2z2 – 6z + 5 = 0
And, a, b, c, = 2, -6, 5

iv) 2x ( x – 3) = 15
Or, 2×2 – 6x = 15
Or, 2×2 – 6x – 15 = 0
And, a, b, c, = 2, -6, -15,
v) 5w ( 7w – 2) = 10w +1
Or, 35w2 – 10w = 10w + 1
Or, 35w2 – 10w – 10w -1 = 0
So, 35w2 – 20w -1 = 0
And, a, b, c,= 35, _20, – 1

vi) 4y – 3y (y) =9
Or, 4y – 3y2 =9
Or, 4y – 3y2 – 9 = 0
So, – 3y2 + 4y – 9 =0
Or, 3y2 – 4y + 9 = 0
And, a, b, c, = 3, -4, 9

vii) a + 2a2 – 19 = 5a2
Or, a + 2a2 – 19 – 5a2= 0
Or, a – 3a2 -19= 0
So, -3a2 + a – 19 = 0
Or, 3a2 – a + 19 = 0
And a, b, c = 3, -1, 19

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