Solving fractional equations

The merger of fractional expressions in an algebraic equation by the Parāvartya Sutra and Sūnyam Samuccaye. (For special cases when the denominators are first-degree binomials in one variable.)

Test one: If the sum of the numerators on the left equals the single numerator on the right then the Parāvartya Sutra applies. That means if N₁+N₂ = N_r then the Parāvartya Sutra (the merging formula) applies.

Test two (when the x-term in the denominators has a coefficient larger than one): Divide each numerator by the x-coefficient in the denominator . Then add those quotients on the LHS and compare to the RHS. If they are equal, then the sutra applies.

Proceed by merging the fraction on the RHS into the LHS, so that only two rational expressions remain. To merge the RHS into the LHS, we subtract the constant term of the RHS denominator from each constant term in the binomials (in the denominator) on the LHS and multiply that difference by the numerator of that term (rational expression) on the left. Set the LHS equal to zero. Thus, the process is complete.

Example one

$$\frac{3}{(x+1)}$$ + $\frac{4}{(x+2)}$ = $\frac{7}{(x+3)}$

Test one, the sum of the numerators: Here, 3+4=7. Yes. Step one: We set down the two denominators to be retained, x+1 and x+2. Step two: As the 3 from the RHS is to be merged, we subtract that 3 from the 1 in the first denominator. 1-3=-2. Negative two is the remainder. Multiply the (-2) by the first numerator, 3, to obtain -6 as the product. Put this down as the new numerator of the new first fractional expression, -6/(x+1). Step three: Do the same thing for the second term. We take 2-3=-1. (-1)(4)=-4. Set -4 as the new numerator of the second fractional expression on the LHS. -4/(x+2). Step four: Set the RHS equal to zero to see the derived equation.

$$\frac{-6}{(x+1)}$$ - $\frac{4}{(x+2)}$ = 0 Step five: Solve for the value of x. Transpose a fraction. Cross multiply to remove fractions. Then transpose terms and coefficients to solve for x.

$$\frac{-6}{(x+1)}$$ = $\frac{4}{(x+2)}$

Therefore, 4x+4 = -6x-12 
And         10x = -16 
And so,       x = -16/10 
              x = -8/5. 

==General Algebraic Proof merging fractions when N₁+N₂ = N_r== A Generalized (Special Case) Fractional Equation:

$$\frac{p}{(x+a)}$$ + $\frac{q}{(x+b)}$ = $\frac{p+q}{(x+c)}$

Split the RHS into separate fractions.

$$\frac{p}{(x+a)}$$ + $\frac{q}{(x+b)}$ = $\frac{p}{(x+c)}$ + $\frac{q}{(x+c)}$

Transpose fractions to group fractions of same N.

$$\frac{p}{(x+a)}$$ - $\frac{p}{(x+c)}$ = $\frac{q}{(x+c)}$ - $\frac{q}{(x+b)}$

Therefore, we convert each side to its LCD.

$$\frac{p(x+c-x-a)}{(x+a)(x+c)}$$ = $\frac{q(x+b-x-c)}{(x+c)(x+b)}$

Hence, by simplifying the numerators, $\frac{p(c-a)}{(x+a)}$ = $\frac{q(b-c)}{(x+b)}$

Or as the derived equation: $\frac{q(b-c)}{(x+b)}$ + $\frac{p(a-c)}{(x+a)}$ = 0

Set the cross-products equal. p(c-a)(x+b) = q(b-c)(x+a)

Hence, when we gather both x-terms on LHS. x[p(c-a)+q(c-b)] = bp(a-c) + aq(b-c)

Hence, when we solve for x. x = $\frac{bp(a-c) + aq(b-c)}{(p(c-a) + q(c-b))}$

Example two: In disguise,

$$\frac{5}{(x-2)}$$ + $\frac{2}{(3-x)}$ = $\frac{3}{(x-4)}$

By transposition, we have the form that clearly satisfies the sutra. Numerators test, 2+3=5, yes, the sutra applies.

$$\frac{2}{(x-3)}$$ + $\frac{3}{(x-4)}$ = $\frac{5}{(x-2)}$

Therefore,

$$\frac{-2}{(x-3)}$$ + $\frac{-6}{(x-4)}$ = 0

    -2x+8 = 6x-18 
      -8x = -26 
And so, x = (-26)/(-8) 
        x = 13/4.  

Example three

$$\frac{4}{(2x+1)}$$ + $\frac{9}{(3x+2)}$ = $\frac{15}{(3x+1)}$

Test one: 4+9 is not 15. But as the coefficients of x are different in the three denominators, we can try the LCM (of the x=terms) conversion and test the numerators again.

$$\frac{12}{(6x+3)}$$ + $\frac{18}{(6x+4)}$ = $\frac{30}{(6x+2)}$

Since N₁+N₂=N_r or (12)+(18)=(30), the sutra applies.

Applying the sutra, we have the derived equation.

$$\frac{12}{(6x+3)}$$ + $\frac{36}{(6x+4)}$ = 0

And: 72x+48 = -216x-108 
       288x = -156 
          x = -156/288 
          x = -13/24 

Summarizing the conversion and test of applicability of the sutra, we have test two: divide each numerator by the x-coefficient in the denominator then add those quotients on the LHS and compare their sum to the RHS quotient. If they are equal, then the sutra applies. Thus, on the LHS, 4/2 + 9/3 = 2+3 = 5; and on the RHS, 15/3 = 5. Yes, the Parāvartya Sutra applies. Merge the fractions. Solve for x.

An extension method for multiple mergers when (on the LHS) N₁+N₂+N₃ = N_r (on the RHS).

Example four:

$$\frac{1}{(x+2)}$$ + $\frac{3}{(x+3)}$ + $\frac{5}{(x+5)}$ = $\frac{9}{(x+4)}$ Test numerators: 1+3+5=9, yes. Merge RHS.

$$\frac{-2}{(x+2)}$$ + $\frac{-3}{(x+3)}$ + $\frac{5}{(x+5)}$ = 0 Test numerators: -2-3+5=0, yes. Transpose fractions.

$$\frac{2}{(x+2)}$$ + $\frac{3}{(x+3)}$ = $\frac{5}{(x+5)}$ Test again, 2+3=5, yes. Merge the RHS fractional expression.

$$\frac{-6}{(x+2)}$$ + $\frac{-6}{(x+3)}$ = 0 By the Sūnyam Samuccaye remove numerators. By the basic formula from a derived equation:

m/(x+a) + n/(x+b) = 0 
          m/(x+a) = -n/(x+b)
          -n(x+a) = m(x+b) 
           -nx-na = mx+mb
           -mx-nx = mb+na 
          -x(m-n) = mb+na         (m minus n) ??
                x = (-mb-na)/(m+n)
                x = (18+12)/(-6-6) 
                x = 30/-12 
                x = -2.5 

These two successive steps of merging can be combined into one step by multiplying N1 first by (2-4) and then by (2-5) or by 6. Similarly multiplying N₂ first by (3-4) and then by (3-5) or by 2. Proceed as before.

Algebraic Proof of the Parāvartya (multiple merger) method.

$$\frac{m}{(x+a)}$$ + $\frac{n}{(x+b)}$ + $\frac{p}{(x+c)}$ = $\frac{m+n+p}{(x+d)}$

Merge the RHS fraction. Therefore,

$$\frac{m(a-d)}{(x+a)}$$ + $\frac{n(b-d)}{(x+b)}$ + $\frac{p(c-d)}{(x+c)}$ = 0

Transpose the third fraction to the RHS.

$$\frac{m(a-d)}{(x+a)}$$ + $\frac{n(b-d)}{(x+b)}$ = $\frac{p(d-c)}{(x+c)}$

Merge the RHS fraction. Therefore, the derived equation is:

$$\frac{m(a-d)(a-c)}{(x+a)}$$ + $\frac{n(b-d)(b-c)}{(x+b)}$ = 0

Similarly, the merger formula can be extended to any number of terms.

$$\frac{m}{(x+a)}$$ + $\frac{n}{(x+b)}$ + $\frac{p}{(x+c)}$ + $\frac{q}{(x+d)}$ + $\frac{r}{(x+e)}$ + ...$\frac{w}{(x+j)}$ = $\frac{m+n+p+q+r+...+w}{(x+k)}$

Therefore, the general formula is:

$$\frac{m(a-w)(...)(a-e)(a-d)(a-c)}{(x+a)}$$ + $\frac{n(b-w)(...)(b-e)(b-d)(b-c)}{(x+b)}$ = 0

Complex mergers in special fractional equations

This special, harder, complex fractional equation can be tackled by the Parāvartya Sutra and the Sūnyam Sutra,

Example five:

$$\frac{10}{(2x+1)}$$ + $\frac{3}{(3x-2)}$ = $\frac{2}{(2x-3)}$ + $\frac{15}{(3x+2)}$

Test (1) ratios of coefficients: 10/2 + 3/3 = 2/2 + 15/3, yes. Test (2) The numerators of the LCM form: (10)(3)=(2)(15) or as proportions: 10:15::2:3 or 10:2::15:3, yes.

Transposing fractions:

$$\frac{10}{(2x+1)}$$ - $\frac{15}{(3x+2)}$ = $\frac{2}{(2x-3)}$ - $\frac{3}{(3x-2)}$

Converting to the LCM x-term coefficient:

$$\frac{30}{(6x+3)}$$ - $\frac{30}{(6x+4)}$ = $\frac{6}{(6x-9)}$ - $\frac{6}{(6x-4)}$

Cross multiply and simplify:

$$\frac{30}{(6x+3)(6x+4)}$$ = $\frac{30}{(6x-9)(6x-4)}$

Test (3) N₁ = N₂ in the derived equation. That is, the numerator of the final derived equation must be the same on both sides. If yes, then by the Sūnyam Samuccaye remove the numerators and set the denominators equal to each other.

Hence, (6x+3)(6x+4) = (6x-9)(6x-4)
      Therefore, 6x = (36-12)/(3+4+9+4) = 24/20 = 6/5. 
      And         x = 1.2 

"This gives us the necessary clue, namely, that, after putting up the L.C.M. coefficient for x in all the denominators, (D₁)(D₂) = (D₃)(D₄). As the transposition, the L.C.M. etc., can be done mentally, this clue amounts to a solution of the equation at sight.

General Formula

When the four fractional expressions pass these tests: N₁=N₂=D₄-D₃ and N₃=N₄=D₂-D₁ then the general formula is applicable.

$$\frac{(m-n)}{(x+p)}$$ + $\frac{(p-q)}{(x+n)}$ = $\frac{(m-n)}{(x+q)}$ + $\frac{(p-q)}{(x+m)}$

By transposing terms we see that the tests are satisfied,

(m-n)=(m-n)=(x+m)-(x-n) and (p-q)=(p-q)=(x+p)-(x+p):

$$\frac{(m-n)}{(x+p)}$$ - $\frac{(m-n)}{(x+q)}$ = $\frac{(p-q)}{(x+m)}$ - $\frac{(p-q)}{(x+n)}$

Therefore, (m-n)(1/(x+p) - 1/(x+p)) = (p-q)(1/(x+m) - 1/(x+n))

Therefore,

$$\frac{(m-n)(q-p)}{(x+p)(x+q)}$$ = $\frac{(p-q)(n-m)}{(x+m)(x+n)}$

As the numerators are equivalent, the sutra applies. Therefore, the denominators are set equal. (x+p)(x+q) = (x+m)(x+n) and so, x = (mn-pq)/(p+q-m-n)

References

Vedic Mathematics: Sixteen Simple Mathematical Formulae from the Vedas, by Swami Sankaracarya (1884-1960), Motilal Banarsidass Indological Publishers and Booksellers, Varnasi, India, 1965; reprinted in Delhi, India, 1975, 1978. 367 pages.

Internal Link

Author, Venkatraman Shastri (Jagadguru Swami Sri Bhārati Kŗşņa Tīrthaji Mahāraja Sankaracarya) (1884-1960) Jagadguru Swami Sri Bharati Krishna Tirthaji Maharaja