Question 934937

{{{2x+y+z=14}}}.....eq.1
{{{-x-3y+2z=-2}}}.....eq.2
{{{4x-6y+3z=-5}}}.....eq.3
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start with
{{{2x+y+z=14}}}.....eq.1
{{{-x-3y+2z=-2}}}.....eq.2...both sides multiply by {{{2}}}
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{{{2x+y+z=14}}}.....eq.1
{{{-2x-6y+4z=-4}}}.....eq.2
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{{{cross(2x)+y+z-cross(2x)-6y+4z=14-4}}}

{{{5z-5y=10}}}

{{{5(z-y)=10}}}

{{{z-y=10/5}}}

{{{z-y=2}}}....solve for {{{z}}}

{{{z=y+2}}}...............1a


{{{-x-3y+2z=-2}}}.....eq.2...both sides multiply by {{{4}}}
{{{4x-6y+3z=-5}}}.....eq.3
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{{{-4x-12y+8z=-8}}}.....eq.2
{{{4x-6y+3z=-5}}}.....eq.3
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{{{-cross(4x)-12y+8z+cross(4x)-6y+3z=-8-5}}}

{{{-18y+11z=-13}}}...solve for {{{z}}}

{{{11z=18y-13}}}

{{{z=18y/11-13/11}}}.............1b

1a and 1b have equal left sides, so right sides must be equal too

{{{y+2=18y/11-13/11}}}.....solve for {{{y}}}

{{{11y+22=18y-13}}}

{{{13+22=18y-11y}}}

{{{35=7y}}}

{{{35/7=y}}}

{{{highlight(y=5)}}}

now find {{{z}}}

{{{z=y+2}}}...............1a

{{{z=5+2}}}

{{{highlight(z=7)}}}

go to one of the given equations, plug in values for {{{y}}} and {{{z}}} and find {{{x}}}

{{{2x+y+z=14}}}.....eq.1

{{{2x+5+7=14}}}

{{{2x+12=14}}}

{{{2x=14-12}}}

{{{2x=2}}}

{{{highlight(x=1)}}}

so, your solutions are: {{{highlight(x=1)}}},{{{highlight(y=5)}}}, and 
{{{highlight(z=7)}}}


this is another way to solve this system:


*[invoke cramers_rule_3x3 2, 1, 1, 14, -1, -3, 2, -2, 4, -6, 3, -5]