ax+by=c and a2x+b2y=c2 by elimination method
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General form of a linear equation in two unknown quantities: ax + by + c = 0, (a, b ≠ 0) Two such equations can be written as: a1x + b1y + c1 = 0 ----------- (i) a2x + b2y + c2 = 0 ----------- (ii) Let us solve the two equations by the method of elimination, multiplying both sides of equation (i) by a2 and both sides of equation (ii) by a1, we get: a1a2x + b1a2y + c1a2 = 0 a1 a2x + a1b2y + a1c2 = 0 Subtracting, b1a2y - a1b2y + c1a2 - c2a1 = 0 or, y(b1 a2 - b2a1) = c2a1 - c1a2 Therefore, y = (c2a1 - c1a2)/(b1a2 - b2a1) = (c1a2 - c2a1)/(a1b2 - a2b1) where (a1b2 - a2b1) ≠ 0 Therefore, y/(c1a2 - c2a1) = 1/(a1b2 - a2b1), ------------- (iii) Again, multiplying both sides of (i) and (ii) by b2 and b1respectively, we get; a1b2x + b1b2y + b2c1 = 0 a2b1x + b1b2y + b1c2 = 0 Subtracting, a1b2x - a2b1x + b2c1 - b1c2 = 0 or, x(a1b2 - a2b1) = (b1c2 - b2c1) or, x = (b1c2 - b2c1)/(a1b2 - a2b1) Therefore, x/(b1c2 - b2c1) = 1/(a1b2 - a2b1) where (a1b2 - a2b1) ≠ 0 -------------- (iv) From equations (iii) and (iv), we get: x/(b1c2 - b2c1) = y/(c1a2) - c2a1 = 1/(a1b2 - a2b1) where (a1b2 - a2b1) ≠ 0 So, the formula for cross-multiplication and its use in solving two simultaneous equations can be presented as: If (a1b2 - a2b1) ≠ 0 from the two simultaneous linear equations a1x + b1y + c1 = 0 ----------- (i) a2x + b2y + c2 = 0 ----------- (ii) we get, by the cross-multiplication method: x/(b1c2 - b2c1) = y/(c1a2 - c2a1) = 1/(a1b2 - a2b1) ---------- (A) That means, x = (b1c2 - b2c1)/(a1b2 - a2b1) y = (c1a2 - c2a1)/(a1b2 - a2b1)