3r-5s=-14 3r+3s-56
subtract the second equation from the first...
3r-3r-5s-3s=-14+?56
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
= 238/8
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
= -112/8
= -14
So we are good!
Additional note regarding posting of your question:
So, your problem statement should be
3r + 3s = 56
Hope this clarifies!
To solve the system of equations:
1. First, rewrite the equations in the standard form, which is "Ax + By = C".
Equation 1: 3r - 5s = -14
Equation 2: 3r + 3s = 56
2. Next, we can use the method of elimination to solve the system of equations. We want to eliminate one of the variables, so that we can solve for the remaining variable.
To eliminate the "r" variable, we need to multiply Equation 1 by 3 and Equation 2 by -3, such that the coefficients of "r" in both equations are equal.
Multiply Equation 1 by 3: 3 * (3r - 5s) = 3 * (-14)
9r - 15s = -42
Multiply Equation 2 by -3: -3 * (3r + 3s) = -3 * 56
-9r - 9s = -168
3. Now, add the two new equations together to eliminate the "r" variable:
(9r - 15s) + (-9r - 9s) = -42 + (-168)
-24s = -210
4. Divide both sides of the equation by -24 to solve for "s":
-24s / -24 = -210 / -24
s = 8.75
5. Substitute the value of "s" back into either of the original equations to solve for "r". Let's use Equation 1:
3r - 5(8.75) = -14
3r - 43.75 = -14
3r = -14 + 43.75
3r = 29.75
r = 9.92
6. Therefore, the solution to the system of equations is r ≈ 9.92 and s ≈ 8.75.