Re: FindInstance question? Any ideas to solve this equality.

*To*: mathgroup at smc.vnet.net*Subject*: [mg128563] Re: FindInstance question? Any ideas to solve this equality.*From*: Bill Rowe <readnews at sbcglobal.net>*Date*: Sun, 4 Nov 2012 00:44:22 -0400 (EDT)*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com*Delivered-to*: l-mathgroup@wolfram.com*Delivered-to*: mathgroup-newout@smc.vnet.net*Delivered-to*: mathgroup-newsend@smc.vnet.net

On 11/2/12 at 11:51 PM, lrebanks at gmail.com (Lea Rebanks) wrote: >FindInstance question: >Hi All, >I have evaluated the following equality with FindInstance function >and there appears to be no results. >Please could someone review the following equality to see if my >result is TRUE or Is it possible to find one or more Integer results >to this equality. >radiusV1 = 145.98630072715727; >radiusV2 = 95.62128455180272; >slopeV1Deg = 9.462322208025617; >slopeV2Deg = 18.43494882292201; >FindInstance[(360*radiusV1)/(slopeV1Deg + 360*timeV1x) == >(360*radiusV2)/(slopeV2Deg + 360*timeV2x) && >(360*(radiusV1*2))/(slopeV1Deg*2 + 360*(timeV1x*2)) == >(360*(radiusV2*2))/(slopeV2Deg*2 + 360*(timeV2x*2)) && >timeV1x > 0 && timeV2x > 0, {timeV1x, timeV2x}, >Integers, 2] >{} It appears there are no integral solutions. That is I get the same result as you even after replacing all machine precision values with exact values and using other techniques. Note, you do not accomplish anything by multiplying top/bottom of one equation by a constant and adding this as a "new" equation. Doing so, provides no additional information about your variables. All that this can accomplish is make Mathematica do more work. Specifically, In[1]:= eqns = List @@ ((360*radiusV1)/(slopeV1Deg + 360*timeV1x) == (360* radiusV2)/(slopeV2Deg + 360*timeV2x) && (360*(radiusV1*2))/(slopeV1Deg*2 + 360*(timeV1x*2)) == (360*(radiusV2*2))/(slopeV2Deg*2 + 360*(timeV2x*2)) && timeV1x > 0 && timeV2x > 0); In[2]:= SameQ @@ Simplify[eqns[[;; 2]]] Out[2]= True So, your first two conditions are not linearly independent and are really the same condition