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How to solve the system of equations 4x plus 3y plus 2z equals 34 and 2x plus 4y plus 3z equals 45 and 3x plus 2y plus 4z equals 47?
4x + 3y + 2z = 34; 2x + 4y + 3z = 45; 3x + 2y + 4z = 47 First, eliminate terms in z from 2 of the equations, by muliplying first equation by 2 and subtracting third equation from the answer: 8x + 6y + 4z = 68, subtract leaving 5x + 4y = 21 (equation 4) Similarly multiply the first equation by 3 and the second by 2 giving 12x + 9y + 6z = 102 and 4x + 8y + 6z = 90 Subtract again and we have 8x + y = 12 or y = 12 - 8x Substitute this in equation 4 gives 5x + 4(12 -8x) = 21 Simplify: 5x + 48 - 32x = 21 = -27x = -27 so x = 1 y = 12 - 8x so y = 4 and in one of the original equations 4 + 12 + 2z = 34, ie 2z = 34 -16 so z =9 Check: 2x + 4y + 3z = 2 +16 + 27 = 45 and 3x + 2y + 4z = 3 + 8 + 36 = 47 QED!
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