Solve the system by elimination.⎧⎩⎨⎪⎪−2x+2y+3z=0−2x−y+z=−32x+3y+3z=5

Answer 1

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
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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
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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.