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Math  /  Algebra

QuestionFind the complete solution of the linear system, or show that it is incor there is no solution, enter NO SOLUTION.) {x+2y−z=−192x+3y−4z=−483x+6y−3z=−56(x,y,z)=(□)\begin{array}{l} \left\{\begin{array}{r} x+2 y-z=-19 \\ 2 x+3 y-4 z=-48 \\ 3 x+6 y-3 z=-56 \end{array}\right. \\ (x, y, z)=(\square) \end{array}

Studdy Solution

STEP 1

1. We are given a system of three linear equations with three variables xx, yy, and zz.
2. We need to determine if there is a unique solution, infinitely many solutions, or no solution.
3. We will use the method of elimination or substitution to solve the system.

STEP 2

1. Simplify the system of equations.
2. Use elimination to reduce the system to two equations with two variables.
3. Solve the reduced system for two variables.
4. Back-substitute to find the third variable.
5. Verify the solution by substituting back into the original equations.

STEP 3

First, observe the equations:
\begin{align*}
1. & \quad x + 2y - z = -19 \\
2. & \quad 2x + 3y - 4z = -48 \\
3. & \quad 3x + 6y - 3z = -56 \end{align*}
Notice that equation 3 is a multiple of equation 1. Simplify equation 3 by dividing by 3:
x+2y−z=−563 x + 2y - z = -\frac{56}{3}
Compare this with equation 1:
x+2y−z=−19 x + 2y - z = -19
Since −563≠−19-\frac{56}{3} \neq -19, equation 3 is inconsistent with equation 1.

STEP 4

Since equation 3 is inconsistent with equation 1, the system of equations has no solution.

STEP 5

Given the inconsistency found in Step 1, the system has no solution.
The solution is:
NO SOLUTION

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