Question 1134702

{{{2x+3y=11}}}.....eq.1
{{{4x+y=12}}}......eq.2
------------------------
start with

{{{4x+y=12}}}......eq.2...solve for {{{y}}}

{{{y=-4x+12}}}.......eq.2a...substitute in eq.1 to eliminate variable {{{y}}}



{{{2x+3(-4x+12)=11}}}.....eq.1

{{{2x-12x+36=11}}}...solve for {{{x}}}

{{{-10x+36=11}}}

{{{-11+36=10x}}}

{{{25=10x}}}

{{{25/10=x}}}

{{{x=5/2}}}

go back to eq.2a, substitute {{{5/2}}} for {{{x}}}

{{{y=-4*(5/2)+12}}}

{{{y=-10+12}}}

{{{y=2}}}


solution is intersection point ({{{x}}},{{{y}}})=({{{5/2}}},{{{2}}})



{{{3x+2y=24}}}.......eq.1
{{{2x-3y=42}}}.......eq.2
---------------------------

start with

{{{3x+2y=24}}}.......eq.1 ..solve for {{{y}}}

{{{2y=-3x+24}}}

{{{y=-3x/2+24/2}}}

{{{y=-3x/2+12}}}...........eq.1a

go to
{{{2x-3y=42}}}.......eq.2...substitute in {{{y}}} from eq.1a to eliminate variable {{{y}}}

{{{2x-3(-3x/2+12)=42}}}

{{{2x+9x/2-36=42}}}

{{{2x+9x/2=42+36}}}.........both sides multiply by {{{2}}}

{{{4x+9x=84+72}}}

{{{13x=156}}}

{{{x=156/13}}}

{{{x=12}}}

go back to eq.1a

{{{y=-3x/2+12}}}...........eq.1a substitute  {{{12}}} for {{{x}}} and solve for {{{y}}}

{{{y=-(3*12)/2+12}}}

{{{y=-(3*cross(12)6)/cross(2)+12}}}

{{{y=-3*6+12}}}

{{{y=-18+12}}}

{{{y=-6}}}


solution is intersection point ({{{x}}},{{{y}}})=({{{12}}},{{{-6}}})