SOLUTION: How would I use the elimination method to for this equation? 5r+2t=5 3r-7t=3

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Question 938455: How would I use the elimination method to for this equation?
5r+2t=5
3r-7t=3
Found 3 solutions by ewatrrr, MathLover1, Edwin McCravy:
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
7(5r+2t = 5)
2(3r-7t) = 3) Re TY, it is 35r + 2(3r) = 41r
...
35r + 14t = 35
6r - 14t = 6
41r = 41
41r = 41
r = 1 and t = 0 5%2B2t+=+5

Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

5r%2B2t=5
3r-7t=3
I am going to use this solver which uses variables x and y, and your variables are r and t; so, you just take x=r and y=t
Solved by pluggable solver: Solving a System of Linear Equations by Elimination/Addition


Lets start with the given system of linear equations

5%2Ax%2B2%2Ay=5
3%2Ax-7%2Ay=3

In order to solve for one variable, we must eliminate the other variable. So if we wanted to solve for y, we would have to eliminate x (or vice versa).

So lets eliminate x. In order to do that, we need to have both x coefficients that are equal but have opposite signs (for instance 2 and -2 are equal but have opposite signs). This way they will add to zero.

So to make the x coefficients equal but opposite, we need to multiply both x coefficients by some number to get them to an equal number. So if we wanted to get 5 and 3 to some equal number, we could try to get them to the LCM.

Since the LCM of 5 and 3 is 15, we need to multiply both sides of the top equation by 3 and multiply both sides of the bottom equation by -5 like this:

3%2A%285%2Ax%2B2%2Ay%29=%285%29%2A3 Multiply the top equation (both sides) by 3
-5%2A%283%2Ax-7%2Ay%29=%283%29%2A-5 Multiply the bottom equation (both sides) by -5


So after multiplying we get this:
15%2Ax%2B6%2Ay=15
-15%2Ax%2B35%2Ay=-15

Notice how 15 and -15 add to zero (ie 15%2B-15=0)


Now add the equations together. In order to add 2 equations, group like terms and combine them
%2815%2Ax-15%2Ax%29%2B%286%2Ay%2B35%2Ay%29=15-15

%2815-15%29%2Ax%2B%286%2B35%29y=15-15

cross%2815%2B-15%29%2Ax%2B%286%2B35%29%2Ay=15-15 Notice the x coefficients add to zero and cancel out. This means we've eliminated x altogether.



So after adding and canceling out the x terms we're left with:

41%2Ay=0

y=0%2F41 Divide both sides by 41 to solve for y



y=0 Reduce


Now plug this answer into the top equation 5%2Ax%2B2%2Ay=5 to solve for x

5%2Ax%2B2%280%29=5 Plug in y=0


5%2Ax%2B0=5 Multiply



5%2Ax=5-0 Subtract 0 from both sides

5%2Ax=5 Combine the terms on the right side

cross%28%281%2F5%29%285%29%29%2Ax=%285%29%281%2F5%29 Multiply both sides by 1%2F5. This will cancel out 5 on the left side.


x=1 Multiply the terms on the right side


So our answer is

x=1, y=0

which also looks like

(1, 0)

Notice if we graph the equations (if you need help with graphing, check out this solver)

5%2Ax%2B2%2Ay=5
3%2Ax-7%2Ay=3

we get



graph of 5%2Ax%2B2%2Ay=5 (red) 3%2Ax-7%2Ay=3 (green) (hint: you may have to solve for y to graph these) and the intersection of the lines (blue circle).


and we can see that the two equations intersect at (1,0). This verifies our answer.

Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
We want to eliminate a letter.

5r+2t = 5
3r-7t = 3

Let's choose r to eliminate.

The coefficient of r in the first equation is 5.
The coefficient of r in the second equation is 3.
The least common multiple of 5 and 3 is 15.  
We want the terms in r to be equal in magnitude
but opposite in sign.  To do that, we multiply both
sides of the first equiation by -3 and both sides
of the second equation by 5.  We get

-3(5r+2t) = -3(5)
 5(3r-7t) =  5(3)

 -15r- 6t = -15
  15r-35t =  15

Next we add them vertically, term by term:

 -15r- 6t = -15
  15r-35t =  15
--------------
   0r-41t =   0
     -41t =   0
    
Divide both sides by -41

 %28-41t%29%2F%28-41%29 = 0%2F%28-41%29

        t = 0

Substitute 0 for t in either one of the 

    5r+2t = 5  
  5r+2(0) = 5
     5r+0 = 5
       5r = 5

Divide both sides by 5

    5r%2F5 = 5%2F5
        
        r = 1

Solution (r,t) = (1,0)

Edwin