Solve by the elimination method

3r-7s=-28
7r+3s=70

What is the solution of the system
r=7 and s=7

I didn't solve the equations but I substituted your answers into both equations and each value satisfied each equation.

To solve the system of equations using the elimination method, you need to eliminate one variable by adding or subtracting the two equations. Let's start with the given equations:

Equation 1: 3r - 7s = -28
Equation 2: 7r + 3s = 70

To eliminate one variable, we can multiply Equation 1 by 3 and Equation 2 by 7 to make the coefficients of 's' equal:

3 * Equation 1: 9r - 21s = -84
7 * Equation 2: 49r + 21s = 490

Now, we can add the two equations together:

(9r - 21s) + (49r + 21s) = -84 + 490
9r + 49r - 21s + 21s = 406
58r = 406

Next, divide both sides of the equation by 58 to solve for 'r':

58r / 58 = 406 / 58
r = 7

Now that we have the value of 'r', we can substitute it back into either Equation 1 or Equation 2. Let's use Equation 1:

3(7) - 7s = -28
21 - 7s = -28
-7s = -28 - 21
-7s = -49

Lastly, divide both sides of the equation by -7 to solve for 's':

-7s / -7 = -49 / -7
s = 7

Therefore, the solution to the system of equations is r = 7 and s = 7, which matches the values you provided.