Question 126655



Start with the given system of equations:


{{{system(x+5y=10,-2x-10y=-20)}}}




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+5y=10}}} Start with the first equation



{{{5y=10-x}}}  Subtract {{{x}}} from both sides



{{{5y=-x+10}}} Rearrange the equation



{{{y=(-x+10)/(5)}}} Divide both sides by {{{5}}}



{{{y=((-1)/(5))x+(10)/(5)}}} Break up the fraction



{{{y=(-1/5)x+2}}} Reduce




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Since {{{y=(-1/5)x+2}}}, we can now replace each {{{y}}} in the second equation with {{{(-1/5)x+2}}} to solve for {{{x}}}




{{{-2x-10highlight(((-1/5)x+2))=-20}}} Plug in {{{y=(-1/5)x+2}}} into the first equation. In other words, replace each {{{y}}} with {{{(-1/5)x+2}}}. Notice we've eliminated the {{{y}}} variables. So we now have a simple equation with one unknown.




{{{-2x+(-10)(-1/5)x+(-10)(2)=-20}}} Distribute {{{-10}}} to {{{(-1/5)x+2}}}



{{{-2x+(10/5)x-20=-20}}} Multiply



{{{(5)(-2x+(10/5)x-20)=(5)(-20)}}} Multiply both sides by the LCM of 5. 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>)




{{{-10x+10x-100=-100}}} Distribute and multiply the LCM to each side




{{{-100=-100}}} Combine like terms on the left side



{{{0=-100+100}}}Add 100 to both sides



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





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