Solving an Algebraic EquationDate: 12/21/95 at 6:57:2 From: Anonymous Subject: 7th grade math If x = 2y = 3z =1, then what is x(y-z)squared + y(z-x)squared + z(x-y)squared? The answer I ended up was 7/18, But my teacher said it was 1/3. I don't get how it is done. Date: 12/23/95 at 18:16:38 From: Doctor Elise Subject: Re: 7th grade math Hi! I'm going to write "^2" instead of "squared" to save space. Here we go: >If x=2y=3z=1, then what is x(y-z)square + y(z-x)square + z(x-y)square?? x = 1 2y = 1, so y = 1/2 3z = 1, so z = 1/3 So, x(y-z)^2 = 1(1/2 -1/3)^2 = (1/6)^2 = 1/36, and y(z-x)^2 = 1/2(1/3 - 1)^2 = 1/2(-2/3)^2 = 1/2(4/9) = 4/18 = 2/9 and z(x-y)^2 = 1/3(1-1/2)^2 = 1/3(1/2)^2 = 1/3(1/4) = 1/12 Finally, 1/36 + 2/9 + 1/12 = 1/36 + 8/36 + 3/36 = 12/36 = 1/3 Ta da! -Doctor Elise, The Geometry Forum |
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