3r-5s=-4 5r+3s=16 elimination method
3r-5s= -4
5r+3s= 16
Multiply the 1st equation by 5 and the second by 3...
15r-25s= -20
15r+9s=48
Subtract both
-34s= -68
divide by -34
s=2
substitute
3r-10= -4
3r=6
r=2
sixty is 5/6 of what number
60*6
DUH!
To solve the system of equations using the elimination method, we need to eliminate one of the variables by adding or subtracting the equations. Let's go step-by-step to solve the given system:
1. Multiply the first equation by 5 and the second equation by 3 to make the coefficients of 'r' the same in both equations:
Equation 1: 5(3r - 5s) = 5(-4) --> 15r - 25s = -20
Equation 2: 3(5r + 3s) = 3(16) --> 15r + 9s = 48
2. Now, we will subtract Equation 1 from Equation 2 to eliminate 'r':
(15r + 9s) - (15r - 25s) = 48 - (-20)
15r + 9s - 15r + 25s = 48 + 20
34s = 68
Divide both sides by 34: s = 2
3. Substitute the value of 's' into any of the original equations. Let's use the first equation:
3r - 5(2) = -4
3r - 10 = -4
Add 10 to both sides: 3r = 6
Divide both sides by 3: r = 2
4. Therefore, the solution to the system of equations is r = 2 and s = 2.
By following these steps, we were able to solve the system of equations using the elimination method.