2r+3s=29
3r+2s=16
In elimination method
Arrrrg
Correction again
6r+9s=87...(*)
6r+4s=32...(**)
2r+3s=29...
3r+2s=16....
You have two choices eliminate r or s
I choose r
Multiply the first equation by 3
And the second by2
6r+9s=87.....(*)
9r+4s=32.....(**)
Subtract (**) from *
9s-4s=87-32
5s=55
s=11
To get r now from the very (*)
6r=87-9s
r=(87-9s)/6=?
Or
Assuming you want to do this all over again
This time eliminate s to get r what would you do??
To solve this system of equations using the elimination method, we want to eliminate one of the variables by multiplying one (or both) of the equations by a constant, so that when we add or subtract them, one of the variables will cancel out.
Let's start by multiplying the first equation by 3, and the second equation by 2. This will allow us to eliminate the "r" variable.
Multiplying the first equation by 3, we have:
3(2r + 3s) = 3(29)
Simplifying:
6r + 9s = 87
Multiplying the second equation by 2, we have:
2(3r + 2s) = 2(16)
Simplifying:
6r + 4s = 32
Now we have two new equations:
6r + 9s = 87
6r + 4s = 32
To eliminate the "r" variable, we can subtract the second equation from the first equation. This will give us a new equation with only the "s" variable:
(6r + 9s) - (6r + 4s) = 87 - 32
Simplifying:
6r - 6r + 9s - 4s = 55
5s = 55
Now we can solve for "s" by dividing both sides of the equation by 5:
5s/5 = 55/5
s = 11
Now that we have the value of "s", we can substitute it back into one of the original equations to solve for "r". Let's use the first equation:
2r + 3(11) = 29
2r + 33 = 29
2r = 29 - 33
2r = -4
Finally, dividing both sides by 2, we find:
2r/2 = -4/2
r = -2
Therefore, the solution to the system of equations is r = -2 and s = 11.
Correction
6r+9s=87...(*)
6r+4s=33...(**)