SOLUTION: Simplify:
(y-z)/yz - (z-x)/zx - (x-y)/xy
This posted answer is 2(y-z)/yz
But I don't know how this was arrived at.
I keep getting 0/yzx
Appreciate any help.
Algebra.Com
Question 234462: Simplify:
(y-z)/yz - (z-x)/zx - (x-y)/xy
This posted answer is 2(y-z)/yz
But I don't know how this was arrived at.
I keep getting 0/yzx
Appreciate any help.
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
Original equation is:
(y-z)/yz - (z-x)/zx - (x-y)/xy
Multiply numerator and denominator by (yz*zx*xy) to get:
(yz*zx*xy)*(y-z)/yz) - (yz*zx*xy)*(z-x)/zx) - (yz*zx*xy)*x-y)/xy)
This becomes:
( (zx*xy)*(y-z) - (yz*xy)*(z-x) - (yz*zx)*x-y) ) / (yz*zx*xy)
This is equivalent to:
( x^2yz * (y-z) - xy^2z * (z-x) - xyz^2 * (x-y) ) / (x^2y^2z^2)
Simplify by multiplying out all the factors to get:
( x^2y^2z - x^2yz^2 - xy^2z^2 + x^2y^2z - x^2yz^2 + xy^2z^2 ) / (x^2y^2z^2) *****
Combine like terms to get:
(2x^2y^2z - 2x^2yz^2) / (x^2y^2z^2)
Factor the numerator to get:
2x^2yz * (y - z) / (x^2y^2z^2)
x^2 in numerator and denominator cancel out.
y in numerator and y^2 in denominator become y in denominator
z in numerator and z^2 in denominator become z in denominator.
You are left with:
2(y - z) / yz
It's a real eyesore.
Putting the x and y and z in order helps to see it as I did above.
Also the signs might very easily have thrown you off.
The numerator above before combining like terms was:
( x^2y^2z - x^2yz^2 - xy^2z^2 + x^2y^2z - x^2yz^2 + xy^2z^2 ) *****
These terms added together:
( x^2y^2z - x^2yz^2 - xy^2z^2 + x^2y^2z - x^2yz^2 + xy^2z^2 ) *****
and these terms added together:
( x^2y^2z - x^2yz^2 - xy^2z^2 + x^2y^2z - x^2yz^2 + xy^2z^2 ) *****
and these terms canceled out:
( x^2y^2z - x^2yz^2 - xy^2z^2 + x^2y^2z - x^2yz^2 + xy^2z^2 ) *****
RELATED QUESTIONS
prove that for x,y,z>=0,... (answered by robertb)
x^2+xy+yz+zx=30
y^2+xy+yz+zx=15
z^2+xy+yz+zx=18
find... (answered by richwmiller,ikleyn)
x^2+y^2+z^2- xy -yz-zx-yz=0 then prove that x=y=z
(answered by Edwin McCravy)
Factoring(xy+yz+zx)^2+(x+y+z)^2(x^2+y^2+z^2) (answered by solver91311)
if x²+y²+z²-64/xy-yz-zx=-2 and x+y=3z,than value of z... (answered by Edwin McCravy)
Let x = 2012(a - b), y = 2012(b - c), and z = 2012(c - a), where a,b,c are real numbers,... (answered by Edwin McCravy,ikleyn)
Solve for x, y,z
x +yz = 13
y + zx = 17
z + xy =... (answered by Alan3354)
the value of expression x^2 + y^2 + z^2 -xy -yz- zx ,if x= 97 ,y =98 ,... (answered by MathLover1,lokesh mishra)
If x = 2015, y = 2014 and z = 2013, then the value of
x² + y² +z² – xy – yz – zx... (answered by ikleyn)