using the elimination method ...does this problem have an answer?...if so how do you get it...
5r - 4s = -21
4r + 5s = 57
I think I got it...can someone please tell me if I am right?...
is it (r,s)(3,9)?
yes you are correct.
Yes, the system of equations given can be solved using the elimination method. Here's how to get the answer:
1. Multiply the first equation by 4 and the second equation by 5 to make the coefficients of "r" in both equations the same.
4(5r - 4s) = 4(-21) => 20r - 16s = -84
5(4r + 5s) = 5(57) => 20r + 25s = 285
2. Now, we can eliminate the variable "r" by subtracting the first equation from the second equation. Subtract the two equations:
(20r + 25s) - (20r - 16s) = 285 - (-84)
20r + 25s - 20r + 16s = 285 + 84
41s = 369
3. Simplify and solve for "s":
41s = 369
s = 369 / 41
s ≈ 9
4. Substitute the value of "s" back into either of the original equations to solve for "r". Let's use the first equation:
5r - 4(9) = -21
5r - 36 = -21
5r = -21 + 36
5r = 15
r = 15 / 5
r = 3
Therefore, the solution to the system of equations is r = 3 and s ≈ 9.