elimination method really confuses me.

solve by elimination method
5r -3s = 17
3r + 5s =17

A. solution is ______ (ordered pair, integer or fraction)
B. Infinitely many
C. No Solution

try to get either the x or the y coefficients to be alike

in this case I will get the y's to have a 15 in front

so multiply the first by 5 and the second by 3 to get

25r - 15s = 85
9r + 15s = 51

add them
34r = 136
r = 136/34 = 4

back in second, (or first)
3(4) + 5s = 17

12 + 5s = 17
5s = 5
s = 1

thank you so much, i was not even close

4x-3y=-4. 3x=2y-4

2r+3s=29,3r+2s=16

To solve a system of equations using the elimination method, we want to eliminate one of the variables by adding or subtracting the equations so that we end up with a new equation containing only one variable. Here's how you can solve the given system of equations:

1. Start by multiplying the first equation by 3 and the second equation by -5 to make the coefficients of either 'r' or 's' in both equations equal:

Multiply the first equation by 3:
3(5r - 3s) = 3(17)
15r - 9s = 51 (equation 1)

Multiply the second equation by -5:
-5(3r + 5s) = -5(17)
-15r - 25s = -85 (equation 2)

2. Now, add equation 1 and equation 2 to eliminate the 'r' variable:

(15r - 9s) + (-15r - 25s) = 51 + (-85)
-34s = -34
Divide both sides of the equation by -34 to solve for 's':
s = 1

3. With the value of 's', substitute it back into either of the original equations to solve for 'r'. Let's use the first equation:

5r - 3(1) = 17
5r - 3 = 17
Add 3 to both sides of the equation:
5r = 20
Divide both sides of the equation by 5 to solve for 'r':
r = 4

4. Now we have the values of 'r' and 's', which are 'r = 4' and 's = 1'. So the solution to the system of equations is the ordered pair (4, 1).

Since we have a specific solution, the correct answer is A. solution is (4, 1) (ordered pair).

It's important to note that if the variables were eliminated but didn't result in a consistent solution, we would have either infinitely many solutions (answer B) or no solution (answer C). But in this case, we have a single ordered pair solution.