Question 1021035

{{{x + y + z = 110}}}..........(1)
{{{x = 4y}}}............(2)
{{{z = x + 2}}}...............(3)

{{{4y + y + z = 110}}}......substitute {{{x}}} from (2)

{{{5y + z = 110}}}

{{{z = 110-5y}}}........since {{{x = 4y}}}->{{{y=x/4}}}


{{{z = 110-5(x/4)}}}.........(a)

from (3) and (a) we have

{{{x + 2=110-5(x/4)}}} .......solve for {{{x}}}

{{{x + 5(x/4)=110-2}}}

{{{4x/4 + 5x/4=108}}}

{{{9x/4=108}}}

{{{9x=108*4}}}

{{{x=432/9}}}

{{{highlight(x=48)}}}

go to {{{z = x + 2}}}...............(3) substitute {{{48}}} for {{{x}}}

{{{z = 48 + 2}}}

{{{highlight(z = 50)}}}

go to {{{x = 4y}}} substitute {{{48}}} for {{{x}}}
->{{{y=48/4}}}

{{{highlight(y=12)}}}