solve the elimination method 7r-3s=0

3r+7s=58

what is the solution of the system

From the first equation, we know that r and s are in the ratio 3:7.

Trial substitution of 3 and 7 into the second equation gives 58, which means that 3 and 7 are the solutions for r and s respectively.

By elimination:
7r-3s=0...(1)
3r+7s=58...(2)

Use linear combination to eliminate r: 7(2)-3(1)
21r+49s-(21r-9s)=58*7
58s=58*7
s=7
Substitute s=3 in (1)
7r-3(7)=0
r=3(7)/7=3
this gives
r=3, s=7.

-rs^3-r^3s=

To solve the system of equations using the elimination method, we aim to eliminate one variable by adding or subtracting the equations. Let's go through the steps:

1. Multiply the first equation by 3 and the second equation by 7 to make the coefficient of 'r' in both equations equal.

Equation 1: 3 * (7r - 3s) = 0 --> 21r - 9s = 0
Equation 2: 7 * (3r + 7s) = 58 --> 21r + 49s = 406

2. Now, we can subtract the modified Equation 1 from Equation 2 to get rid of the 'r' terms.

(21r + 49s) - (21r - 9s) = 406 - 0
21r - 21r + 49s + 9s = 406
58s = 406

3. Solve for 's' by dividing both sides of the equation by 58.

s = 406 / 58
s = 7

4. Substitute the value of 's' back into either original equation. Let's use Equation 1.

7r - 3(7) = 0
7r - 21 = 0
7r = 21
r = 21 / 7
r = 3

Therefore, the solution to the system of equations is r = 3 and s = 7.