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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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โˆ™ 2009-05-16 15:47:55
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โˆ™ 2020-04-27 16:25:36

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Q: 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?
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