Note, that the LHS without the independent terms: x2 + 2x x(x + 2) ------- = -------- x2 + 3x x(x + 3) And so, is in the same ratio as the entire RHS. By the Upasūtra, x + 2 7 ----- = - x + 3 5 or, 5x + 10 = 7x + 21 or, -2x = 11 or, x = -11/2 Thus, the solution is: x = -11/2 |
Note, that the LHS without the independent terms: 2x2 + 3x x(2x + 3) -------- = --------- 3x2 + 4x x(3x + 4) And so, is in the same ratio as the entire RHS. By the Upasūtra, 2x + 3 10 ------ = -- 3x + 4 14 or, 28x + 42 = 30x + 40 or, -2x = -2 or, x = 1 Thus, the solution is: x = 1 |
Before re-arranging, let us find the sum of binomials to check if the condition of this Upasūtra is met: For LHS, (x+1)+(x+2)+(x+9) = 3x + 12 For RHS, (x+3)+(x+4)+(x+5) = 3x + 12 Hence, this Upasūtra is applicable. Re-arranging, we get: (x+1)(x+2) (x+3) ---------- = ----- (x+4)(x+5) (x+9) or, x2 + 3x + 2 x + 3 ------------ = ----- x2 + 9x + 20 x + 9 Note, that the LHS without the independent terms: x2 + 3x x(x + 3) ------- = -------- x2 + 9x x(x + 9) And so, is in the same ratio as the entire RHS. By the Upasūtra, x + 3 2 ----- = -- x + 9 20 or, 20x + 60 = 2x + 18 or, 18x = -42 or, x = -7/3 Thus, the solution is: x = -7/3 |
Before re-arranging, let us find the sum of binomials to check if the condition of this Upasūtra is met: For LHS, (x+2)+(x+3)+(x+11) = 3x + 16 For RHS, (x+4)+(x+5)+(x+7) = 3x + 16 Hence, this Upasūtra is applicable. Re-arranging, we get: (x+2)(x+3) (x+5) ---------- = ------ (x+4)(x+7) (x+11) or, x2 + 5x + 6 x + 5 ------------- = ------ x2 + 11x + 28 x + 11 Note, that the LHS without the independent terms: x2 + 5x x(x + 5) -------- = --------- x2 + 11x x(x + 11) And so, is in the same ratio as the entire RHS. By the Upasūtra, x + 5 6 ------ = -- x + 11 28 or, 28x + 140 = 6x + 66 or, 22x = -74 or, x = -37/11 Thus, the solution is: x = -37/11 |
Consider an equation of the form, Ac + d A ------ = - Bc + e B Then, B(Ac + d) = A(Bc + e) or, ABc + Bd = ABc + Ae or, Bd = Ae or, A/B = d/e Hence, solving this equation would provide the solution for the original equation. |
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