```Question 469016

{{{system(-3x+6y=21,2x-4y=14)}}}

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

{{{6y=21+3x}}} Add {{{3x}}} to both sides

{{{6y=+3x+21}}} Rearrange the equation

{{{y=(+3x+21)/(6)}}} Divide both sides by {{{6}}}

{{{y=((+3)/(6))x+(21)/(6)}}} Break up the fraction

{{{y=(1/2)x+7/2}}} Reduce

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Since {{{y=(1/2)x+7/2}}}, we can now replace each {{{y}}} in the second equation with {{{(1/2)x+7/2}}} to solve for {{{x}}}

{{{2x-4highlight(((1/2)x+7/2))=14}}} Plug in {{{y=(1/2)x+7/2}}} into the second equation. In other words, replace each {{{y}}} with {{{(1/2)x+7/2}}}. Notice we've eliminated the {{{y}}} variables. So we now have a simple equation with one unknown.

{{{2x+(-4)(1/2)x+(-4)(7/2)=14}}} Distribute {{{-4}}} to {{{(1/2)x+7/2}}}

{{{2x-(4/2)x-28/2=14}}} Multiply

{{{(2)(2x-(4/2)x-28/2)=(2)(14)}}} Multiply both sides by the LCM of 2. 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>)

{{{4x-4x-28=28}}} Distribute and multiply the LCM to each side

{{{-28=28}}} Combine like terms on the left side

{{{0=56}}} 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 consequently, there are no solutions for the system of equations.

So the system is inconsistent.
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