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

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) (Show Source): You can put this solution on YOUR website!

Start with the given system of equations:

Multiply the both sides of the first equation by 5.

Distribute and multiply.

Multiply the both sides of the second equation by 4.

Distribute and multiply.

So we have the new system of equations:

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

Group like terms.

Combine like terms.

Simplify.

Divide both sides by to isolate .

Reduce.

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

Now go back to the first equation.

Plug in .

Multiply.

Subtract from both sides.

Combine like terms on the right side.

Divide both sides by to isolate .

Reduce.

So our answer is and .

This means that the system is consistent and independent.

Answer by Mathtut(3670) (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
:

:
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
:

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