Solve the following system of linear equations. - 2w + x - y + 2z = 6 3x + 3y - 2z = -12 y - 2z = -5 z = 3 (w, x, y, z) = ( , , , )
Added by Julio P.
Close
Step 1
For the first equation: -2w + x - y + 2z = 0 Substituting the values, we get: -2(5) + 3 - 6 + 2(12) = -10 + 3 - 6 + 24 = 11 + 18 = 29 For the second equation: 3x + 3y - 2z = 6 Substituting the values, we get: 3(3) + 3(6) - 2(12) = 9 + 18 - 24 = 27 - 24 = 3 For Show more…
Show all steps
Your feedback will help us improve your experience
Oswaldo Jiménez and 86 other Calculus 1 / AB educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
Solve the system of linear equations: x + y - z = 0 x + 2y - 3z = -3 2x + 3y - 4z = -3
Dayna K.
Solve the system of linear equations using Gauss-Jordan elimination. $$\begin{array}{rr} x-2 y+3 z= & 5 \\ 3 x+6 y-4 z= & -12 \\ -x-4 y+6 z= & 16 \end{array}$$
Systems of Linear Equations and Inequalities
Systems of Linear Equations and Matrices
Recommended Textbooks
Calculus: Early Transcendentals
Thomas Calculus
Transcript
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD