Question 1155480
{{{x + 2y + 3z = 10}}}.......eq.1
{{{2x - y + z = 9}}}.......eq.2
{{{3x - z = 3}}}.......eq.3
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{{{3x - z = 3}}}.......eq.3...solve for {{{z}}}

{{{z=3x - 3}}}.........eq.3a


{{{2x - y + z = 9}}}.......eq.2...solve for {{{z}}}

{{{ z =-2x+y+ 9}}}.......eq.2a


from eq.3a and eq.2a we have


{{{3x - 3=-2x+y+ 9}}}......solve for {{{y}}}

{{{y=3x - 3+2x-9}}}

{{{y=5x - 12}}}........eq.b


go to


{{{x + 2y + 3z = 10}}}.......eq.1, substitute {{{y}}} and {{{z}}}

{{{x + 2(5x - 12) + 3(3x - 3) = 10}}}......solve for {{{x}}}

{{{x + 10x - 24 + 9x - 9 = 10}}}

{{{20x - 33 = 10}}}

{{{20x = 10+33}}}

{{{20x = 43}}}

{{{highlight(x = 43/20)}}}


go to

{{{y=5x - 12}}}........eq.b, substitute {{{x}}}

{{{y=5(43/20) - 12}}}

{{{y=43/4 - 12}}}

{{{y=43/4 - 48/4}}}

{{{highlight(y= - 5/4)}}}


go to


{{{z=3x - 3}}}.........eq.3a,  substitute {{{x}}}

{{{z=3(43/20) - 3}}}

{{{z=129/20 - 60/20}}}

{{{highlight(z=69/20 )}}}