Solve the system by elimination.⎧⎩⎨⎪⎪−2x+2y+3z=0−2x−y+z=−32x+3y+3z=5
First, we can multiply the second equation by -1 to simplify the coefficients of the x terms:
-2x + 2y + 3z = 0
2x + y - z = 3
2x + 3y + 3z = 5
Now, we can add the first equation to the second equation to eliminate the x terms:
-2x + 2y + 3z = 0
2x + y - z = 3
--------------
3y + 2z = 3
Next, we multiply the first equation by 2 and add it to the third equation to eliminate the x terms:
-4x + 4y + 6z = 0
2x + 3y + 3z = 5
--------------
7y + 9z = 5
Now we have the following two equations:
3y + 2z = 3
7y + 9z = 5
We can multiply the first equation by 7 and the second equation by 3 in order to make the coefficient of y equal:
21y + 14z = 21
21y + 27z = 15
Now, we can subtract the first equation from the second equation to eliminate the y terms:
21y - 21y + 27z - 14z = 15 - 21
13z = -6
z = -6/13
Substituting this value of z back into the first equation, we can solve for y:
3y + 2z = 3
3y + 2(-6/13) = 3
3y - 12/13 = 3
3y = 3 + 12/13
3y = 39/13 + 12/13
3y = 51/13
y = 17/13
Finally, we substitute the values of y and z back into any of the original equations to solve for x:
-2x + 2y + 3z = 0
-2x + 2(17/13) + 3(-6/13) = 0
-2x + 34/13 - 18/13 = 0
-2x = 18/13 - 34/13
-2x = -16/13
x = (-16/13)(-1/2)
x = 8/13
Therefore, the solution to the system of equations is x = 8/13, y = 17/13, and z = -6/13.
To solve the system of equations by elimination, we need to eliminate one variable at a time by adding or subtracting the equations.
First, let's eliminate the variable x. We will use the first two equations. The goal is to get coefficients of x that cancel each other when we add the equations together.
Start by multiplying the second equation by -1 to make the coefficients of x in both equations the same. We get:
-2x + y - z = 3
Now, add this new equation to the first equation:
(-2x + 2y + 3z) + (-2x + y - z) = 0 + 3
Simplifying the equation:
-4x + 3y + 2z = 3
Now, we have one equation with only y and z.
Next, let's eliminate the variable x again. This time, we'll use the second and third equations. Multiply the second equation by 2 to match the coefficients of x in the third equation. We get:
-4x - 2y + 2z = -6
Now, add this equation to the third equation:
(-4x + 3y + 2z) + (-4x - 2y + 2z) = 3 + 5
Simplifying the equation:
-8x + y + 4z = 8
Now, we have a new equation with only y and z.
Finally, we have two equations that we can solve simultaneously:
-4x + 3y + 2z = 3
-8x + y + 4z = 8
Now, you can use any method like substitution or elimination to solve this 2x3 system of equations.
To solve the system of equations by elimination, we will eliminate one variable at a time.
Step 1: Eliminate the variable x
We will eliminate x by adding the first equation to the second equation:
(-2x + 2y + 3z) + (-2x - y + z) = (0 - 3)
Combine like terms:
-4x + y + 4z = -3 (Equation 3)
Step 2: Eliminate the variable x (Again)
Next, we eliminate x from Equation 2 and Equation 3:
-4x + y + 4z = -3 (Equation 3)
2x - 3y - 3z = 5 (Equation 2)
Multiply Equation 2 by 2 to make the coefficients of x match in both equations:
4x - 6y - 6z = 10 (Equation 4)
Now, adding Equation 3 and Equation 4:
(-4x + y + 4z) + (4x - 6y - 6z) = (-3 + 10)
Combine like terms:
-5y - 2z = 7 (Equation 5)
We have eliminated the variable x, and Equation 5 represents a simplified equation with y and z.
Step 3: Solve the remaining equations
Now, we have the following two equations:
-5y - 2z = 7 (Equation 5)
2x + 3y + 3z = 5 (Equation 1)
To continue solving, you can solve Equation 5 for y or z and substitute the result back into one of the original equations. However, the equations in this system are not easily solved by elimination alone. Another method such as substitution or matrices may be more efficient for finding the solution to this system.