Question 121939
#1



Start with the given system

{{{y=2x+6}}}
{{{y=-x-3}}}




{{{(-x-3)=2x+6}}}  Plug in {{{y=-x-3}}} into the first equation. In other words, replace each {{{y}}} with {{{-x-3}}}. Notice we've eliminated the {{{y}}} variables. So we now have a simple equation with one unknown.



{{{(-x-3)=2x+6}}} Distribute



{{{-x-3=2x+6}}} Combine like terms on the left side



{{{-x=2x+6+3}}}Add 3 to both sides



{{{-x-2x=6+3}}} Subtract 2x from both sides



{{{-3x=6+3}}} Combine like terms on the left side



{{{-3x=9}}} Combine like terms on the right side



{{{x=(9)/(-3)}}} Divide both sides by -3 to isolate x




{{{x=-3}}} Divide





Now that we know that {{{x=-3}}}, we can plug this into {{{y=-x-3}}} to find {{{y}}}




{{{y=-(-3)-3}}} Substitute {{{-3}}} for each {{{x}}}



{{{y=0}}} Simplify



So our answer is {{{x=-3}}} and {{{y=0}}} which also looks like *[Tex \LARGE \left(-3,0\right)]




Notice if we graph the two equations, we can see that their intersection is at *[Tex \LARGE \left(-3,0\right)]. So this verifies our answer.



{{{ graph( 500, 500, -5, 5, -5, 5, 2x+6, -x-3) }}} Graph of {{{y=2x+6}}} (red) and {{{y=-x-3}}} (green)



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#2

*[invoke solving_linear_system_by_substitution -3,1,9,1,2,4] 




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#3


{{{y + 4 = 2x}}} Start with the first equation



{{{y= 2x-4}}} Solve for y by subtracting 4 from both sides





Start with the given system

{{{6x-3y=12}}}
{{{y=2x-4}}}




{{{6x-3(2x-4)=12}}}  Plug in {{{y=2x-4}}} into the first equation. In other words, replace each {{{y}}} with {{{2x-4}}}. Notice we've eliminated the {{{y}}} variables. So we now have a simple equation with one unknown.



{{{6x-6x+12=12}}} Distribute



{{{12=12}}} Combine like terms on the left side



{{{0=12-12}}}Subtract 12 from both sides



{{{0=0}}} Combine like terms on the right side



{{{0=0}}} Simplify


Since this equation is always true for any x value, this means x can equal any number. So there are an infinite number of solutions.