Vedic Math-1 Dubai UAE Vedic Math-1 Algebra Equations Basic Equations: Type 1: Example: 5x+2 = 3x + 6 => x = (6-2)/(5-3) = 4/2 = 2 Type 2: Example: (x+1)(x+2)= (x+3)(x+4) => x= (12-2)/(1+2-3-4) = 10/(-4) = -5/2 Type 3: Example: (3x+2)/(2x+1) = 4/3 x = (4-6)/(9-8) = -2/1 = -2 Type 4: Example: 1/(x+2) + 2/(x+1) = 0 => x = (-1-4)/3 = -5/3 First Method: A term which occurs as a common factor in all the terms is equated to 0. Examples: 12x+3x = 4x + 3x … As x is common factor both sides.So, x = 0. 9(x+3) = 4(x+3) … As (x+3) is common term both sides. So, x+3 = 0. x = -3. Second Method: The product of the independent terms is same both sides then equated to 0. Examples: (x+5)(x+4) = (x+2)(x+10) As product of independent terms (non-x terms): 5 x 4 = 2 x 10 , is same on both sides. Therefore, x=0. Third Method: The sum of the Denominators of two fractions having the same numerical numerator is equated to 0. Examples: 1/(2x-1) + 1/(4x-1) = 0 Therefore, sum of denominators: 2x-1 + 4x-1 = 0 On solving x = 1/3 Fourth Method: The sum of the Numerators and the sum of the Denominators is the same, then that sum equated to 0. Examples: (2x +9)/ (2x +7) = (2x +7)/ (2x +9) Here, Addition of both numerators = Addition of both Denominators. Thus, 2x + 9 + 2x + 7 = 0 Hence 4x + 16 = 0 hence x = -4 If there is a numerical factor in the algebraic sum, then we remove that factor (3x +4)/ (6x +7) = (x +1)/ (2x +3) Here, Addition of both numerators = 4x +5 Addition of both Denominators = 8x + 10 =2(4x +5) where, 2 is the numerical factor. So remove it 4x +5 =0. Hence x= -5/4 NOTE: In above both examples, when we do cross multiplication the x2 term is getting cancelled. So instead of being quadratic eq and having 2 roots, it becomes simple linear equation and THUS has single root. But if x2 term is not getting cancelled then it becomes quadratic equation and thus will have 2 roots/values. See next Meaning. Fifth Method: Same Meaning as that of Type 4 Examples: (3x +4)/ (6x +7) = (5x +6)/ (2x +3) Here as well like previous meaning, Addition of both numerators = Addition of both Denominators. So x = -5/4 BUT on cross multiplication x2 term is not getting cancelled so it is quadratic equation and thus will have to find the 2nd root as well, Calculated as |D1 – D2| = |N1 – N2| = 2x + 2 = 0(On removal of numerical factor) So x = -1. So Values of x are -5/4 and -1. Simultaneous equations-2 ax + by = p cx + dy = q Solving, x = (bq – pd) / (bc – ad) y = (cp – aq) / (bc – ad) Notice that for calculation of numerators (x any y) cyclic method is used and Denominators remains same for both x and y. Examples: 2x + 3y =6 3x + 4y = 3 Applying above formula: x = (9 – 24)/ (9 – 8) = -15 y = (18 – 6) (9 – 8) = 12 -3x + 5y = 2 4x + 3y = -5 Applying above formula: x = (-25 -6) / (20+9) = -31/29 y = (8-15) / (20+9) = -7/29