Can someone show me how to solve the system by elimination with?
-2x+2y+3z=0
-2x-y+z=-3
2x+3y+3z=5
add the 3rd equation to the 1st and then the 2nd to eliminate x
1 & 3 ... 5y + 6z = 5 ... (A)
2 & 3 ... 2y + 4z = 2 ... (B)
... multiply by 3/2 ... 3y + 6z = 3 ... (C)
subtract (C) from (A) to eliminate z
... 2y = 2
substitute the y-value back to find x and z
So it would go like
-2x+2y+3z=0
2x+3y+3z=5
Equal
0x+5y+6z=5?
-2x-y+z=-3
2x+3y+3z=5
Equal
0x+3y+3z=5
But if I multiply 3/2 with what you got (aka) 2 & 3 ... 2y + 4z = 2 ... (B)
3/2*2=3
3/2*4=6
3/2*2=3
So I’d plug in the variables and get
3y+6z=3
And then subtract
5y + 6z = 5
From
3y + 6z = 3
And get
2y+0z=2
So basically if our variable has 0 it drops (aka) it’s dead? As in we don’t use it?
So I’d get 2y=2... but you said substitute y back to get x and z but how do I do that????
(So basically I have 2 problems with how I just solved this problem #1 I have no clue how you got “B” because I guess I solved it wrong??? And I have no clue on how to substitute y back to find x and z).... I know all I did is somewhat retyped what you said do, but I was showing you why i needed help instead of just being given the answers I needed someone to help explain and guide me step by step so I’d be able to look back and try doing it on my own.
you have found y=1
You also have
5y+6z=5
so, 5+6z = 5, so z=0
Now you have any of the original equations, such as
2x+3y+3z=5
2x+3+0=5
x=1
So does that mean for “B” I was right? Or did you just go with mine?...and I’m still confused on how you found x, for everything else I understand now. Thank you
So it would go like
-2x+2y+3z=0
2x+3y+3z=5
Equal
0x+5y+6z=5?
-2x-y+z=-3
2x+3y+3z=5
Equal
0x+3y+3z=5
But if I multiply 3/2 with what you got (aka) 2 & 3 ... 2y + 4z = 2 ... (B)
3/2*2=3
3/2*4=6
3/2*2=3
So I’d plug in the variables and get
3y+6z=3
And then subtract
5y + 6z = 5
From
3y + 6z = 3
And get
2y+0z=2
So basically if our variable has 0 it drops (aka) it’s dead? As in we don’t use it?
So I’d get 2y=2... but you said substitute y back to get x and z but how do I do that????
(So basically I have 2 problems with how I just solved this problem #1 I have no clue how you got “B” because I guess I solved it wrong??? And I have no clue on how to substitute y back to find x and z).... I know all I did is somewhat retyped what you said do, but I was showing you why i needed help instead of just being given the answers I needed someone to help explain and guide me step by step so I’d be able to look back and try doing it on my own.
Sure, I can help you solve the system of equations by the method of elimination.
To begin, let's write down the system of equations:
Equation 1: -2x + 2y + 3z = 0
Equation 2: -2x - y + z = -3
Equation 3: 2x + 3y + 3z = 5
To eliminate one of the variables, we need to manipulate the equations so that when we add or subtract them, one of the variables will cancel out.
In this case, let's eliminate x. To do that, we'll need to multiply Equation 2 by -1 so that when we add it to Equation 1, the x terms will cancel out.
Multiply Equation 2 by -1:
-(-2x - y + z) = -1 * (-3)
2x + y - z = 3
Now, we have:
Equation 1: -2x + 2y + 3z = 0
Equation 2 (modified): 2x + y - z = 3
Equation 3: 2x + 3y + 3z = 5
Now, let's add Equation 1 and Equation 2:
(-2x + 2y + 3z) + (2x + y - z) = 0 + 3
2y + 2z = 3
Our system of equations becomes:
Equation 1: -2x + 2y + 3z = 0
Equation 2 (modified): 2y + 2z = 3
Equation 3: 2x + 3y + 3z = 5
Now, we can eliminate y. We'll need to multiply Equation 2 by -1, just like before, so that we can cancel out the y terms.
Multiply Equation 2 by -1:
-(2y + 2z) = -1 * 3
-2y - 2z = -3
Now, we have:
Equation 1: -2x + 2y + 3z = 0
Equation 2 (modified): -2y - 2z = -3
Equation 3: 2x + 3y + 3z = 5
Finally, let's add Equation 2 and Equation 3:
(-2y - 2z) + (2x + 3y + 3z) = -3 + 5
2x + y + z = 2
Our system of equations becomes:
Equation 1: -2x + 2y + 3z = 0
Equation 2 (modified): -2y - 2z = -3
Equation 3 (modified): 2x + y + z = 2
Now we have a simpler system of equations with only two variables x and y. Notice that z did not cancel out, which means we have three possible values for z.
To solve for x and y, we can now use any method such as substitution or by solving two equations at a time.