Disguised Problems (I)
We come across many problems which seem to bear no relation to the above sutras. But by simple manipulation, we can arrive at a form which is already well known to us.
Consider the following examples :
Example 1 :
1 - 1 = 1 - 1
x - 5 x - 3 x + 9 x + 11
Transposing the equations, we get
1 + 1 = 1 + 1
x - 5 x + 11 x + 9 x - 3
This is of the form discussed in sixth meaning.
Hence x = -3
Example 2 :
1 - 1 = 1 + 1
x + a x + a + b x + c - b x + c
Interchanging the terms,
1 + 1 = 1 + 1
x + a x + c x + c - b x + a + b
Hence, 2x + a + c = 2x + a + c
==> x = -(a + c)/2
Disguised Problems (II)
In this type we divide the numerator by the denominator and after canceling the common terms the problem is converted to any one of the forms discussed earlier.
Example 1 : Consider
x - 1 + x - 4 = x - 3 - x - 2
x - 2 x - 5 x - 4 x - 3
This can be simplified to obtain
x - 2 + 1 + x - 5 + 1 = x - 4 + 1 + x - 3 + 1
x - 2 x - 5 x - 4 x - 3
==> 1 + 1 + 1 + 1 = 1 + 1 + 1 + 1
x - 2 x - 5 x - 4 x - 3
==> 1 + 1 = 1 + 1
x - 2 x - 5 x - 4 x - 3
Hence according to the sutra,
2x - 7 = 0 ==> x = 7/2
Example 2 :
3x - 5 + 2x - 13 = x - 3 - 4x - 19
x - 2 x - 7 x - 4 x - 5
This can be simplified to obtain
3x - 6 + 1 + 2x - 14 + 1 = x - 4 + 1 + 4x - 20 + 1
x - 2 x - 7 x - 4 x - 5
==> 3 + 1 + 2 + 1 = 1 + 1 + 4 + 1
x - 2 x - 5 x - 4 x - 3
==> 1 + 1 = 1 + 1
x - 2 x - 5 x - 4 x - 3
Hence according to the sutra,
2x - 9 = 0 ==> x = 9/2
Disguised Problems (III)
In this case the LCM of the numerators is obtained and multiplied and divided with each term. On observation we find that this is one of the forms we have already come across.
Example 1 : Consider
3 + 4 = 2 + 12
3x + 4 4x + 7 2x + 6 12x + 1
The LCM of the terms in the numerators is 12. Multiplying and dividing each term by 12 we get,
12 + 12 = 12 + 12
12x + 16 12x + 21 12x + 36 12x + 1
==> 12x + 16 + 12x + 21 = 12x + 36 + 12x + 1
==> 24x + 37 = 0
According to the sutra x = -37/24
NOTE : We cannot always be sure of the possibility of the sum of denominators on each side being equal.
Therefore to be sure we multiply N1 by D2 and N2 by D1 on one hand and sum them up and multiply N3 by D4 and N4 by D3 and after summing, check for equality. If equal, we continue with the above method. Also the sum of ratio of numerator to coefficient of x in each term on both sides should be equal.
Example 2 :
3x - 5 + 2x - 13 = x - 3 - 4x - 19
x - 2 x - 7 x - 4 x - 5
-2 + 3 + 2 + 1 = - 4 + 3 + 4 + 1
3x - 1 x + 3 3x - 7 x + 5
Taking the LCM of the numerator terms after canceling out the common terms on both the sides, we get
3 + 3 = 3 + 3
3x + 9 3x - 1 3x + 15 3x - 7
According to the sutra, 6x + 8 = 0 ==> x = -4/3