Question 408161
The sum of three integers is no more than {{{110}}}. 


let integers be: first- {{{x}}}, second-{{{y}}} and third-{{{z}}}


{{{x+y+z<100}}}


If one integer is {{{9}}} less than twice the {{{first}}} 


{{{y+9<2x}}}


and the third is {{{one}}}{{{ fourth}}} the {{{first}}} 


{{{z=(1/4)x}}}


what are the integers?



{{{x+y+z<100}}}....replace {{{y}}} with {{{y+9<2x}}}.or {{{y<2x-9}}}


{{{x+(2x-9)+z<100}}}...replace {{{z}}} with {{{z=(1/4)x}}}


{{{x+(2x-9)+(1/4)x<100}}}.......solve for {{{x}}}


{{{x+2x-9+(1/4)x<100}}}


{{{3x+(1/4)x<100+9}}}


{{{3x+(1/4)x<109}}}.........multiply all by {{{4}}}


{{{4*3x+4(1/4)x<4*109}}}


{{{12x+x<436}}}


{{{13x<436}}}


{{{x<436/13}}}


{{{x<33.5}}}........it could be equal to 33.4,so

{{{x=33.4}}}



{{{y<2x-9}}


{{{y<2(33.5)-9}}


{{{y<58.08}}....it could be equal to 58.07...so,

{{{y=58.07}}



{{{z=(1/4)33.4}}}

{{{z=8.35}}}



check:


{{{x+y+z<100}}}

{{{33.4+58.07+8.35<100}}}

{{{99.82<100}}}