SOLUTION: I have tried this but I am having trouble, can you please help. Use substitution to solve each system. If it does not have one solution then put no solution or indinitely many so

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Question 181876: I have tried this but I am having trouble, can you please help. Use substitution to solve each system. If it does not have one solution then put no solution or indinitely many solutions. X-5Y=36 2X+Y=-16
Answer by jim_thompson5910(35256)   (Show Source): You can put this solution on YOUR website!

Start with the given system of equations:





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 second equation

Start with the second equation


Subtract from both sides


Rearrange the equation


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

Since , we can now replace each in the second equation with to solve for



Plug in into the first equation. In other words, replace each with . Notice we've eliminated the variables. So we now have a simple equation with one unknown.



Distribute to


Multiply


Combine like terms on the left side


Subtract 80 from both sides


Combine like terms on the right side


Divide both sides by 11 to isolate x



Divide





-----------------First Answer------------------------------


So the first part of our answer is:









Since we know that we can plug it into the equation (remember we previously solved for in the first equation).



Start with the equation where was previously isolated.


Plug in


Multiply


Combine like terms



-----------------Second Answer------------------------------


So the second part of our answer is:









-----------------Summary------------------------------

So our answers are:

and

which form the point








Now let's graph the two equations (if you need help with graphing, check out this solver)


From the graph, we can see that the two equations intersect at . This visually verifies our answer.




graph of (red) and (green) and the intersection of the lines (blue circle).

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