how do you solve this problem by elimination method 5r-3s=14 3r+5s=56
5r-3s=14 times 3
3r+5s=56 times 5
15 r - 9 s = 42
15 r + 25s = 280
-----------------
0 r - 34 s = - 238
so
s = 238/34 which is 7
go back and get r
To solve the given system of equations by the elimination method, follow these steps:
Step 1: Multiply the first equation by 3 and the second equation by 5. This will give us equations with the same coefficient for either 'r' or 's', which will allow us to eliminate one variable.
(3 * 5r - 3 * 3s = 3 * 14) -> (15r - 9s = 42)
(5 * 3r + 5 * 5s = 5 * 56) -> (15r + 25s = 280)
Step 2: Now, subtract the first equation from the second equation. This will eliminate the 'r' variable.
(15r + 25s) - (15r - 9s) = 280 - 42
15r - 15r + 25s + 9s = 238
34s = 238
Step 3: Solve for 's' by dividing both sides of the equation by 34.
34s / 34 = 238 / 34
s = 7
Step 4: Substitute the value of 's' (which we just found) into any of the original equations. Let's substitute it into the first equation:
5r - 3(7) = 14
5r - 21 = 14
Step 5: Solve for 'r' by isolating the variable 'r':
5r = 14 + 21
5r = 35
Step 6: Divide both sides of the equation by 5 to solve for 'r':
5r / 5 = 35 / 5
r = 7
Step 7: Check the solution by substituting the values of 'r' and 's' into the second equation:
3r + 5s = 56
3(7) + 5(7) = 56
21 + 35 = 56
56 = 56
Since the values of 'r' = 7 and 's' = 7 satisfy both equations, the solution to the system of equations is r = 7 and s = 7.