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,

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).

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

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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