Question 169624



Start with the given system of equations:


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




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


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



{{{-y=4-2x}}}  Subtract {{{2x}}} from both sides



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



{{{y=(-2x+4)/(-1)}}} Divide both sides by {{{-1}}}



{{{y=((-2)/(-1))x+(4)/(-1)}}} Break up the fraction



{{{y=2x-4}}} Reduce




---------------------


Since {{{y=2x-4}}}, we can now replace each {{{y}}} in the second equation with {{{2x-4}}} to solve for {{{x}}}




{{{2x-highlight((2x-4))=3}}} Plug in {{{y=2x-4}}} into the second 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.




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



{{{4=3}}} Combine like terms on the left side



{{{0=3-4}}}Subtract 4 from both sides



{{{0=-1}}} 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 the system is inconsistent