Soe using the eliminatin method.Show work.State if it has infinite solutions or no solutions...-15r+12s=24,30r-24s=48
multiply first equation by 2
-30r+24s=48
+30r-24s=48
hmm, if I add those, I get 0 = 96. a problem all right
try multiplying the second one by -1 then look
-30r+24s=48
-30r+24s=-48
that is two parallel lines with the same slope but they never cross (different intercepts). There are no solutions.
To solve the system of equations using the elimination method, we need to eliminate one of the variables by adding or subtracting the two equations.
Given the system of equations:
-15r + 12s = 24 ----(1)
30r - 24s = 48 ----(2)
To eliminate one variable, we can multiply one or both of the equations by any number(s) that will make the coefficients of either 'r' or 's' in both equations the same, but with opposite signs.
Let's multiply equation (1) by 2, and equation (2) by 1 to create oppositely signed coefficients for 'r':
-30r + 24s = 48 ----(3) (Multiply equation 1 by 2)
30r - 24s = 48 ----(2) (Multiplying equation 2 by 1)
Now, we can add equation (3) and equation (2) to eliminate the 'r' variable:
(-30r + 24s) + (30r - 24s) = 48 + 48
This simplifies to:
0s = 96
Since we are left with 0s = 96, we don't have any value for 's' that would satisfy this equation. Therefore, there are no solutions to this system of equations. Thus, the system has no solutions.