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Math  /  Algebra

QuestionSolving a 3×33 \times 3 system of line olve the system. 2x+2y+z=−2−x−2y+2z=−52x+4y+z=0\begin{aligned} 2 x+2 y+z & =-2 \\ -x-2 y+2 z & =-5 \\ 2 x+4 y+z & =0 \end{aligned}

Studdy Solution

STEP 1

1. We are given a system of three linear equations with three variables: x x , y y , and z z .
2. The system is consistent and has a unique solution.
3. We can use methods such as substitution, elimination, or matrix operations to solve the system.

STEP 2

1. Use elimination to simplify the system.
2. Solve for one variable.
3. Substitute back to find the remaining variables.
4. Verify the solution by substituting back into the original equations.

STEP 3

First, let's eliminate one of the variables from two pairs of equations. We will eliminate z z from the first two equations.
Multiply the first equation by 2 to match the coefficient of z z in the second equation:
4x+4y+2z=−4−x−2y+2z=−5\begin{aligned} 4x + 4y + 2z &= -4 \\ -x - 2y + 2z &= -5 \end{aligned}
Subtract the second equation from the modified first equation:
(4x+4y+2z)−(−x−2y+2z)=−4−(−5)5x+6y=1\begin{aligned} (4x + 4y + 2z) - (-x - 2y + 2z) &= -4 - (-5) \\ 5x + 6y &= 1 \end{aligned}

STEP 4

Now, eliminate z z from the first and third equations. Use the original first equation and multiply it by 1, so it remains the same:
2x+2y+z=−22x+4y+z=0\begin{aligned} 2x + 2y + z &= -2 \\ 2x + 4y + z &= 0 \end{aligned}
Subtract the first equation from the third equation:
(2x+4y+z)−(2x+2y+z)=0−(−2)2y=2\begin{aligned} (2x + 4y + z) - (2x + 2y + z) &= 0 - (-2) \\ 2y &= 2 \end{aligned}

STEP 5

Solve for y y from the equation 2y=2 2y = 2 :
y=1y = 1

STEP 6

Substitute y=1 y = 1 back into the equation 5x+6y=1 5x + 6y = 1 :
5x+6(1)=15x + 6(1) = 1
Simplify and solve for x x :
5x+6=15x=−5x=−15x + 6 = 1 \\ 5x = -5 \\ x = -1

STEP 7

Substitute x=−1 x = -1 and y=1 y = 1 back into the first original equation to solve for z z :
2(−1)+2(1)+z=−22(-1) + 2(1) + z = -2
Simplify and solve for z z :
−2+2+z=−2z=−2-2 + 2 + z = -2 \\ z = -2

STEP 8

Verify the solution by substituting x=−1 x = -1 , y=1 y = 1 , and z=−2 z = -2 back into the original equations:
1. 2(−1)+2(1)+(−2)=−2 2(-1) + 2(1) + (-2) = -2 (True)
2. −(−1)−2(1)+2(−2)=−5 -(-1) - 2(1) + 2(-2) = -5 (True)
3. 2(−1)+4(1)+(−2)=0 2(-1) + 4(1) + (-2) = 0 (True)
All equations are satisfied.
The solution is:
(x,y,z)=(−1,1,−2)(x, y, z) = (-1, 1, -2)

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