Question 134454

Start with the given system of equations:


{{{system(x-3y=5,-2x+6y=-10)}}}




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


{{{x-3y=5}}} Start with the first equation



{{{-3y=5-x}}}  Subtract {{{x}}} from both sides



{{{-3y=-x+5}}} Rearrange the equation



{{{y=(-x+5)/(-3)}}} Divide both sides by {{{-3}}}



{{{y=((-1)/(-3))x+(5)/(-3)}}} Break up the fraction



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




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




{{{-2x+6highlight(((1/3)x-5/3))=-10}}} Plug in {{{y=(1/3)x-5/3}}} into the first equation. In other words, replace each {{{y}}} with {{{(1/3)x-5/3}}}. Notice we've eliminated the {{{y}}} variables. So we now have a simple equation with one unknown.




{{{-2x+(6)(1/3)x+(6)(-5/3)=-10}}} Distribute {{{6}}} to {{{(1/3)x-5/3}}}



{{{-2x+(6/3)x-30/3=-10}}} Multiply



{{{-2x+2x-10=-10}}} Reduce



{{{-30=-30}}} Combine like terms on the left side



{{{0=-30+30}}}Add 30 to both sides



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




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