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 can help you with both the elimination and substitution methods to solve these systems of equations.
1. Solve by elimination method:
To solve the system of equations:
2r - 5s = -33
5r + 2s = 48
First, we need to eliminate one variable, either r or s, by multiplying one or both of the equations by appropriate constants so that the coefficients in front of either r or s become equal in magnitude but opposite in sign.
Let's eliminate the variable "r" by multiplying the first equation by 5 and the second equation by 2:
(5)(2r - 5s) = (5)(-33) --> 10r - 25s = -165
(2)(5r + 2s) = (2)(48) --> 10r + 4s = 96
Next, add the two equations together to eliminate the "r" variable:
(10r - 25s) + (10r + 4s) = -165 + 96
10r + 10r - 25s + 4s = -69
20r - 21s = -69
Now, we have a new equation with just one variable, "s". Let's solve for "s" in this new equation:
20r - 21s = -69 --> 20r = 21s - 69 --> r = (21s - 69) / 20
We can now substitute this expression for "r" back into one of the original equations. Let's substitute into the first equation:
2r - 5s = -33
2((21s - 69) / 20) - 5s = -33
(42s - 138) / 20 - 5s = -33
42s - 138 - 100s = -660 (multiplying through by 20 to eliminate the fractions)
-58s - 138 = -660
-58s = -522
s = (-522) / (-58) --> s = 9
Now that we have found the value for "s", we can substitute it back into one of the original equations to solve for "r". Let's substitute into the first equation:
2r - 5s = -33
2r - 5(9) = -33
2r - 45 = -33
2r = -33 + 45
2r = 12
r = 12 / 2
r = 6
Therefore, the solution to the system of equations is r = 6 and s = 9.
2. Solve by substitution method:
To solve the system of equations:
5x + 6y = 5
x - 9y = 40
We will solve one of the equations for one variable and then substitute that expression into the other equation.
Let's solve the second equation for "x":
x - 9y = 40
x = 9y + 40
Now, substitute this expression for "x" into the first equation:
5x + 6y = 5
5(9y + 40) + 6y = 5
45y + 200 + 6y = 5
51y + 200 = 5
51y = 5 - 200
51y = -195
y = -195 / 51 --> y = -3.82 (approx)
Now that we have found the value for "y", substitute it back into one of the original equations to solve for "x". Let's substitute into the second equation:
x - 9y = 40
x - 9(-3.82) = 40
x + 34.38 = 40
x = 40 - 34.38
x = 5.62 (approx)
Therefore, the solution to the system of equations is x = 5.62 (approx) and y = -3.82 (approx).
I hope this helps you understand how to solve systems of equations using both the elimination and substitution methods!