Question 727145

{{{x=y-4}}}....eq.1
{{{3x+y=12}}}....eq.2
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to solve the linear system using substitution, substitute value of {{{x}}} from eq.1 in eq.2 

{{{3(y-4)+y=12}}}....eq.2..solve for {{{y}}}

{{{3y-12+y=12}}}

{{{3y+y=12+12}}}

{{{4y=24}}}

{{{y=24/4}}}

{{{highlight(y=6)}}}

go back to eq.1 and find {{{x}}}

{{{x=y-4}}}....eq.1...plug in {{{y=6}}}


{{{x=6-4}}}

{{{highlight(x=2)}}}