Some more Examples :
Example # 3
Consider the following equation :
1
- 1 = 1
- 1
x + a + b
x + c
x + a
x - b + c
Here, we first need to put the equation in the standard form, so that the sutra can be applied.
Thus, re-arranging the terms, we get :
1 + 1
= 1
+ 1
x + a + b
x - b + c
x + a
x + c
Here, D1 + D2 = (x + a + b) + (x - b + c) = 2x + a + c
and D3 + D4 = (x + a) + (x + c) = 2x + a + c
Since, D1 + D2 = D3 + D4 = 2x + a + c
The solution is : 2x + a + c = 0 → 2x = -(a + c) → x = -(a + c)
÷ 2
Example # 4
Consider the following equation :
4x - 43 - 3x - 74
= 5x + 86
- 2x + 7
x - 11
x - 25
x + 17
x + 3
To convert the equation in the standard form, add and subtract the coefficients of x, of each numerator from each term. Thus, we get :
4x - 43 - 4
+ 4 - 3x - 74
- 3 +
3
= 5x + 86 -
5
+ 5 - 2x + 7
- 2 +
2
x - 11
x - 25
x + 17
x + 3
4x - 43 - 4x + 44 +
4
- 3x - 74 - 3x + 75
+ 3 = 5x + 86 - 5x - 85
+ 5 - 2x + 7 - 2x - 6
+ 2
x - 11
x - 25
x + 17
x + 3
→ 1
- 1
+ 7 = 1
- 1
+ 7
x - 11
x - 25 x + 17
x + 3
→ 1
- 1
= 1
- 1
x - 11
x - 25
x + 17
x + 3
Thus, rearranging the terms, we get :
→ 1
+ 1
= 1
+ 1
x - 11
x + 3
x + 17
x - 25
The equation is now in the standard form, so the sutra can be applied now :
Here D1 + D2 = (x - 11) + (x + 3) = 2x - 8
and D3 + D4 = (x + 17) + (x - 25) = 2x - 8
Since, D1 + D2 = D3 + D4 = 2x - 8
The solution is : 2x - 8 = 0 → 2x = 8
→ x = 8 ÷ 2 = 4
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