Using Trial SolutionsDate: 07/28/99 at 15:12:15 From: Edward Subject: Matrix and Multiple Unknowns How do you do this? x - sqrt(yz) = 42 y - sqrt(xz) = 6 z - sqrt(xy) = -30 It seems easy but it is really HARD! I think that it's kind of like this: x - sqrt(yz) + y - sqrt(zx) = 48 (add eq'n. 1 and 2) x + y - sqrt(z)[sqrt(y) + sqrt(z)] = 48 (factor sqrt(z)) sqrt(z) = (48-x-y)/-[sqrt(y)+sqrt(z)] Now, should I substitute my new equation into equation 1 or 2, OR should I square the equation and then substitute into equation 3? Date: 07/28/99 at 18:13:50 From: Doctor Anthony Subject: Re: Matrix and Multiple Unknowns Write the equations in the form x-42 = sqrt(yz) -> (x-42)^2 = yz y-6 = sqrt(xz) -> (y-6)^2 = xz z+30 = sqrt(xy) -> (z+30)^2 = xy Trial solutions are probably the quickest method for solving these equations. Looking at the equations it is probable that all three of x, y, z are multiples of 6. From third equation one at least of x or y will be greater than 30 and z is probably small. From the first equation x is greater than 42. If we try z = 6 we obtain (x-42)^2 = 6y (y-6)^2 = 6x 36^2 = xy and multiplying the first two of these equations (x-42)^2 (y-6)^2 = 36 xy taking square roots of both sides (x-42)(y-6) = 6 (36) = 216 xy - 42y - 6x + 252 = 216 1296 - 42y - 6x + 36 = 0 6x + 42y = 1332 x + 7y = 222 and xy = 1296 so x = 1296/y 1296/y + 7y = 222 1296 + 7y^2 = 222y 7y^2 - 222y + 1296 = 0 222 +- sqrt(222^2 - 36288) 222 +- 114 y = -------------------------- = ---------- 14 14 = 24 or 7.71428 So take y = 24 Then x = 1296/24 = 54 And so the three roots are x = 54, y = 24, z = 6 If you check these values in any of the original equations you will find that they satisfy those equations. - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/ |
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