Question 123356



Start with the given system of equations:


{{{system(4x-12y=5,-x+3y=-1)}}}




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


{{{4x-12y=5}}} Start with the first equation



{{{-12y=5-4x}}}  Subtract {{{4x}}} from both sides



{{{-12y=-4x+5}}} Rearrange the equation



{{{y=(-4x+5)/(-12)}}} Divide both sides by {{{-12}}}



{{{y=((-4)/(-12))x+(5)/(-12)}}} Break up the fraction



{{{y=(1/3)x-5/12}}} Reduce




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Since {{{y=(1/3)x-5/12}}}, we can now replace each {{{y}}} in the second equation with {{{(1/3)x-5/12}}} to solve for {{{x}}}




{{{-x+3highlight(((1/3)x-5/12))=-1}}} Plug in {{{y=(1/3)x-5/12}}} into the first equation. In other words, replace each {{{y}}} with {{{(1/3)x-5/12}}}. Notice we've eliminated the {{{y}}} variables. So we now have a simple equation with one unknown.




{{{-x+(3)(1/3)x+(3)(-5/12)=-1}}} Distribute {{{3}}} to {{{(1/3)x-5/12}}}



{{{-x+(3/3)x-15/12=-1}}} Multiply



{{{(12)(-1x+(3/3)x-15/12)=(12)(-1)}}} Multiply both sides by the LCM of 12. This will eliminate the fractions  (note: if you need help with finding the LCM, check out this <a href=http://www.algebra.com/algebra/homework/divisibility/least-common-multiple.solver>solver</a>)




{{{-12x+12x-15=-12}}} Distribute and multiply the LCM to each side




{{{-15=-12}}} Combine like terms on the left side



{{{0=-12+15}}}Add 15 to both sides



{{{0=3}}} Combine like terms on the right side



{{{0=3}}} Simplify


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