SOLUTION: how do I solve the system of equations 3r-4s+0 and 2r+5s=23?

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Question 175701: how do I solve the system of equations 3r-4s+0 and 2r+5s=23?
Found 2 solutions by jim_thompson5910, Mathtut:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!

Start with the given system of equations:
system%283r-4s=0%2C2r%2B5s=23%29


5%283r-4s%29=5%280%29 Multiply the both sides of the first equation by 5.


15r-20s=0 Distribute and multiply.


4%282r%2B5s%29=4%2823%29 Multiply the both sides of the second equation by 4.


8r%2B20s=92 Distribute and multiply.


So we have the new system of equations:
system%2815r-20s=0%2C8r%2B20s=92%29


Now add the equations together. You can do this by simply adding the two left sides and the two right sides separately like this:


%2815r-20s%29%2B%288r%2B20s%29=%280%29%2B%2892%29


%2815r%2B8r%29%2B%28-20s%2B20s%29=0%2B92 Group like terms.


23r%2B0s=92 Combine like terms.


23r=92 Simplify.


r=%2892%29%2F%2823%29 Divide both sides by 23 to isolate r.


r=4 Reduce.


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


15r-20s=0 Now go back to the first equation.


15%284%29-20s=0 Plug in r=4.


60-20s=0 Multiply.


-20s=0-60 Subtract 60 from both sides.


-20s=-60 Combine like terms on the right side.


s=%28-60%29%2F%28-20%29 Divide both sides by -20 to isolate s.


s=3 Reduce.


So our answer is r=4 and s=3.


This means that the system is consistent and independent.

Answer by Mathtut(3670) About Me  (Show Source):
You can put this solution on YOUR website!
there are a few ways but lets use elimination method which means you want to manipulate the equation so the point where one set of terms are eliminated
:
I assumed that the +0 in eq 1 was really =0
:
3r-4s=0......eq 1
2r+5s=23.....eq 2
:
multiply all terms in eq 1 by 5 and all the terms in eq 2 by 4 and then add the 2 equations together. I will do the math then rewrite the equations next to one another so you can see what is taking place.
:
5(3r-4s=0)---->15r-20s=0 (revised eq 1)
4(2r+5s=23)---->8r+20s=92 (revised eq 2)
:
can you see what happens to the s terms when we add the two equations together. They are eliminated because -20s+20s=0. We are left with 15r+8r=92+0
:
23r=92
:
highlight%28r=4%29
:
now take r's found value and plug it back into any equation. I am going to use eq 1
:
3(4)-4s=0--->12-4s=0--->4s=12
:
highlight%28s=3%29