subtract the second equation from the first...
I am not certain on your second equation, an equals is missing.
I tried and verified the equation with the following assumption to make it work.
3r - 5s = -14
3r +3s = 56
subtract the second equation from first equation:
You will get -8s = -70
s = 70/8
substitute s in any of the original problem equations to get the value of r.
3r = -14 + 5s
3r = -14 + 5 (70/8)
= (-112 + 350)/8
3r = 119/4
r = 119/12
To verify the values - apply the values to one of the equations to hold true.
3r -5s should be equal to -14
= 3 (119/12) - 5 (70/8)
= 119/4 - 350/8
= (238 - 350) / 8
So we are good!
Additional note regarding posting of your question:
So, your problem statement should be
3r + 3s = 56
Hope this clarifies!
algebra - how do you solve this problem by elimination method 5r-3s=14 3r+5s=56
Algebra - Solve by elimination method.. 3r - 5s = 14 5r + 3s = 46.
algebra - elimination method 3r-5s=-14 5r+35-56
Algebra - add. (5r^4-3r^3+3r^2+14r-8)+(r^5+8r^3+7r^2-3r+5)+(-8r^4+r^2-8r-9)the ...
algebra - if (6r - x) is on factors of 18r^2 + 12ry - 3xr - 2xy, what is the ...
algerbra - 3-3r=4 Do you wish to solve for r? Put r on one side and everything ...
algebra - 3r-5s = -14
alberga - 3r-5s=-4 5r+3s=16 elimination method
algebra - Process of eliminiation method ? Is there a solution, infinity or no ...
Math116 - Solve by elimination 5r-3s=13 3r+5s=69