Solve by elimination method.

2r-5s=-33
5r+2s=48

Solve by substitution method.

5x+6y=5
x-9y=40

I always get these two confused... can someone please help me with these two problems?? I would really appreciate it..

Sure, I'd be happy to help you with both methods!

1. Solve by elimination method:

Step 1: Multiply each equation by coefficients such that the coefficients of one of the variables in the two equations are additive inverses (i.e., they add up to zero).

Multiply the first equation by 2 and the second equation by 5 to make the coefficients of 's' additive inverses:
Equation 1: 4r - 10s = -66
Equation 2: 25r + 10s = 240

Step 2: Add the modified equations to eliminate the 's' variable.
(4r - 10s) + (25r + 10s) = (-66) + 240
29r = 174
r = 6

Step 3: Substitute the value of 'r' back into either of the original equations to find the value of 's'. Let's use the first equation:
2r - 5s = -33
2(6) - 5s = -33
12 - 5s = -33
-5s = -45
s = 9

Therefore, the solution to the system of equations is r = 6 and s = 9.

2. Solve by substitution method:

Step 1: Solve one of the equations for one variable in terms of the other variable, then substitute it into the other equation.

Let's solve the second equation for x:
x - 9y = 40
x = 40 + 9y

Step 2: Substitute the expression for 'x' in terms of 'y' into the other equation.
5x + 6y = 5
5(40 + 9y) + 6y = 5
200 + 45y + 6y = 5
51y = -195
y = -3

Step 3: Substitute the found value of 'y' back into either of the original equations to find the value of 'x'. Let's use the second equation:
x - 9y = 40
x - 9(-3) = 40
x + 27 = 40
x = 13

So, the solution to the system of equations is x = 13 and y = -3.

I hope this helps! Let me know if you have any further questions.