SOLUTION: Solve the following systems of equations using the substitution method. What type of system is it? Name the solution if there is one. x-y=3 x+3y=6

Algebra ->  Systems-of-equations -> SOLUTION: Solve the following systems of equations using the substitution method. What type of system is it? Name the solution if there is one. x-y=3 x+3y=6      Log On


   

Question 905824: Solve the following systems of equations using the substitution method.
What type of system is it? Name the solution if there is one.
x-y=3
x+3y=6
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

x-y=3
x%2B3y=6
__________
Solved by pluggable solver: Solve the System of Equations by Graphing



Start with the given system of equations:


1x-y=3

1x%2B3y=6





In order to graph these equations, we need to solve for y for each equation.




So let's solve for y on the first equation


1x-y=3 Start with the given equation



-y=3-x Subtract +x from both sides



-y=-x%2B3 Rearrange the equation



y=%28-x%2B3%29%2F%28-1%29 Divide both sides by -1



y=%28-1%2F-1%29x%2B%283%29%2F%28-1%29 Break up the fraction



y=x-3 Reduce



Now lets graph y=x-3 (note: if you need help with graphing, check out this solver)



+graph%28+600%2C+600%2C+-10%2C+10%2C+-10%2C+10%2C+x-3%29+ Graph of y=x-3




So let's solve for y on the second equation


1x%2B3y=6 Start with the given equation



3y=6-x Subtract +x from both sides



3y=-x%2B6 Rearrange the equation



y=%28-x%2B6%29%2F%283%29 Divide both sides by 3



y=%28-1%2F3%29x%2B%286%29%2F%283%29 Break up the fraction



y=%28-1%2F3%29x%2B2 Reduce





Now lets add the graph of y=%28-1%2F3%29x%2B2 to our first plot to get:


+graph%28+600%2C+600%2C+-10%2C+10%2C+-10%2C+10%2C+x-3%2C%28-1%2F3%29x%2B2%29+ Graph of y=x-3(red) and y=%28-1%2F3%29x%2B2(green)


From the graph, we can see that the two lines intersect at the point (15%2F4,3%2F4) (note: you might have to adjust the window to see the intersection)


This tells us that the system of equations is consistent and independent.