Question 1124158


let’s  three numbers be {{{x}}}, {{{y}}}, and {{{z}}}


if the sum of three numbers is {{{10}}}, we have

{{{x+y+z=10}}}........eq.1



if the sum of {{{twice}}} the first​ number, {{{4}}} times the second​ number, and {{{5}}} times the third number is {{{33}}}, we have

{{{2x+4y+5z=33}}}........eq.2



if  the {{{difference}}} between {{{6}}} times the first number and the second number is {{{19}}}, we have

{{{6x-y=19}}}..............eq.3 ....solve for {{{y}}}

{{{y=6x-19}}}..........eq.a 



go to {{{2x+4y+5z=33}}}........eq.2, substitute {{{y}}}

{{{2x+4(6x-19)+5z=33}}}
{{{2x+24x-76+5z=33}}}
{{{26x+5z=33+76}}}
{{{26x+5z=109}}}....solve for {{{z}}}
{{{5z=109-26x}}}
{{{z=(109-26x)/5}}}.........eq.b



go to {{{x+y+z=10}}}........eq.1  substitute {{{y}}} and   {{{z}}}


{{{x+6x-19+(109-26x)/5=10}}}....solve for {{{x}}}


{{{5x+30x-19*5+(109-26x)=10*5}}}

{{{5x+30x-95+109-26x=50}}}

{{{35x-26x=50+95-109}}}

{{{9x=145-109}}}

{{{x=36/9}}}

{{{highlight(x=4)}}}



go to {{{y=6x-19}}}..........eq.a , plug in {{{4}}} for {{{x}}}

{{{y=6*4-19}}}
{{{y=24-19}}}
{{{highlight(y=5)}}}



go to {{{z=(109-26x)/5}}}.........eq.b , plug in {{{4}}} for {{{x}}}



{{{z=(109-26*4)/5}}}

{{{z=(109-104)/5}}}

{{{z=5/5}}}

{{{highlight(z=1)}}}