Question 125998
Let's solve this system by using substitution

Start with the given system of equations:


{{{system(-6x+2y=4,3x-y=-4)}}}




Now in order to solve this system by using substitution, we need to solve (or isolate) one variable. I'm going to solve for y.





So let's isolate y in the first equation


{{{-6x+2y=4}}} Start with the first equation



{{{2y=4+6x}}} Add {{{6x}}} to both sides



{{{2y=+6x+4}}} Rearrange the equation



{{{y=(+6x+4)/(2)}}} Divide both sides by {{{2}}}



{{{y=((+6)/(2))x+(4)/(2)}}} Break up the fraction



{{{y=3x+2}}} Reduce




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Since {{{y=3x+2}}}, we can now replace each {{{y}}} in the second equation with {{{3x+2}}} to solve for {{{x}}}




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




{{{3x-3x-2=-4}}} Distribute the negative



{{{-2=-4}}} Combine like terms on the left side



{{{0=-4+2}}}Add 2 to both sides



{{{0=-2}}} Combine like terms on the right side




Since this equation is <font size=4><b>never</b></font> true for any x value, this means there are no solutions. So this also means that the system is inconsistent.