Math  /  Algebra

QuestionUse any convenient method to solve the following system of equations. If the system is dependent, express the solution set in terms of one of the variables. Leave all fractional answers in fraction form. {3x12y+6z=292x+y2z=5x4y+2z=10\left\{\begin{array}{r} 3 x-12 y+6 z=-29 \\ 2 x+y-2 z=-5 \\ x-4 y+2 z=-10 \end{array}\right.

Studdy Solution

STEP 1

1. We are given a system of three linear equations with three variables: xx, yy, and zz.
2. The goal is to find the values of xx, yy, and zz that satisfy all three equations simultaneously.
3. The system may have a unique solution, infinitely many solutions (dependent), or no solution (inconsistent).

STEP 2

1. Choose a method to solve the system of equations.
2. Simplify the system using the chosen method.
3. Solve for one variable and substitute back to find others.
4. Check the solution or express the solution set if dependent.

STEP 3

We will use the elimination method to solve the system. This involves eliminating one variable at a time to simplify the system.

STEP 4

First, let's eliminate xx from the second and third equations. We can do this by manipulating the equations:
From the third equation: x4y+2z=10 x - 4y + 2z = -10
Multiply the third equation by 2: 2(x4y+2z)=2(10) 2(x - 4y + 2z) = 2(-10) 2x8y+4z=20 2x - 8y + 4z = -20
Now subtract the second equation from this result: (2x8y+4z)(2x+y2z)=20(5) (2x - 8y + 4z) - (2x + y - 2z) = -20 - (-5) 9y+6z=15 -9y + 6z = -15

STEP 5

Now, eliminate xx from the first and third equations:
From the third equation: x4y+2z=10 x - 4y + 2z = -10
Multiply the third equation by 3: 3(x4y+2z)=3(10) 3(x - 4y + 2z) = 3(-10) 3x12y+6z=30 3x - 12y + 6z = -30
Subtract the first equation from this result: (3x12y+6z)(3x12y+6z)=30(29) (3x - 12y + 6z) - (3x - 12y + 6z) = -30 - (-29) 0=1 0 = -1
This indicates that the system is inconsistent, as we have reached a contradiction.

STEP 6

Since we have a contradiction, the system of equations has no solution.

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