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Systems of equationsAlgebraHigh schoolabout 1:31

Solve the system -x - 2y = 8, 2y - z = -4, 2x - 3y - 2z = -2: animated lesson

A narrated, step-by-step systems of equations video lesson with captions. Final answer: x = -4, y = -2, z = 0.

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Video with a synthetic English voice and captions. Pick another language for subtitles in that language, or open the interactive version.

Video narrated by a synthetic voice (Piper TTS, en_US-ljspeech-high voice (LJ Speech dataset, public domain)); the steps come from our maths engine and the answer was checked independently. Download MP4 (0.6 MB) · Captions (WebVTT)

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1. The problem
Solving an equation
$\begin{cases} - x-2y = 8 \\ 2y- z = -4 \\ 2x-3y-2z = -2 \end{cases}$
2. Steps, part 1
1. $\begin{cases} -x - 2y = 8 \\ 2y - z = -4 \\ 2x - 3y - 2z = -2 \end{cases}$
  Write every equation in standard form
2. $\begin{bmatrix}-1 & -2 & 0 & 8 \\ 0 & 2 & -1 & -4 \\ 2 & -3 & -2 & -2\end{bmatrix}$
  Build the augmented matrix [A | b]
3. Gauss–Jordan elimination
4. $R_{1} \to \tfrac{1}{-1} R_{1}:\ \begin{bmatrix}1 & 2 & 0 & -8 \\ 0 & 2 & -1 & -4 \\ 2 & -3 & -2 & -2\end{bmatrix}$
  Divide R1 by -1
3. Steps, part 2
1. $R_{3} \to R_{3} - 2R_{1}:\ \begin{bmatrix}1 & 2 & 0 & -8 \\ 0 & 2 & -1 & -4 \\ 0 & -7 & -2 & 14\end{bmatrix}$
  Clear column 1 using the pivot in R1
2. $R_{2} \to \tfrac{1}{2} R_{2}:\ \begin{bmatrix}1 & 2 & 0 & -8 \\ 0 & 1 & -\frac{1}{2} & -2 \\ 0 & -7 & -2 & 14\end{bmatrix}$
  Divide R2 by 2
3. $R_{1} \to R_{1} - 2R_{2},\ R_{3} \to R_{3} + 7R_{2}:\ \begin{bmatrix}1 & 0 & 1 & -4 \\ 0 & 1 & -\frac{1}{2} & -2 \\ 0 & 0 & -\frac{11}{2} & 0\end{bmatrix}$
  Clear column 2 using the pivot in R2
4. $R_{3} \to \tfrac{1}{-\frac{11}{2}} R_{3}:\ \begin{bmatrix}1 & 0 & 1 & -4 \\ 0 & 1 & -\frac{1}{2} & -2 \\ 0 & 0 & 1 & 0\end{bmatrix}$
  Divide R3 by -11/2
4. Steps, part 3
1. $R_{1} \to R_{1} - 1R_{3},\ R_{2} \to R_{2} + \frac{1}{2} R_{3}:\ \begin{bmatrix}1 & 0 & 0 & -4 \\ 0 & 1 & 0 & -2 \\ 0 & 0 & 1 & 0\end{bmatrix}$
  Clear column 3 using the pivot in R3
2. $\begin{bmatrix}1 & 0 & 0 & -4 \\ 0 & 1 & 0 & -2 \\ 0 & 0 & 1 & 0\end{bmatrix}$
  Reduced row-echelon form
3. $x = -4,\ y = -2,\ z = 0$
  Read off the unique solution
5. The answer
$x = -4,\ y = -2,\ z = 0$
-x - 2y = 8; 2y - z = -4; 2x - 3y - 2z = -2 ⇒ x = -4, y = -2, z = 0
6. Check
- Equation 1: -x - 2y = 8 ✓
- Equation 2: 2y - z = -4 ✓
- Equation 3: 2x - 3y - 2z = -2 ✓

Transcript

The problem. Here is the problem: negative x minus 2y equals 8; 2y minus z equals negative 4; 2x minus 3y minus 2z equals negative 2. 3 linear equations in 3 unknowns (x, y, z).

Steps, part 1. Write every equation in standard form. Move all variable terms to the left and constants to the right. Build the augmented matrix [A | b]. Columns: x, y, z | constant. Gauss–Jordan elimination. Use row operations to make each pivot 1 and clear the rest of its column. Divide R1 by negative 1.

Steps, part 2. Clear column 1 using the pivot in R1. Divide R2 by 2. Clear column 2 using the pivot in R2. Divide R3 by negative 11 over 2.

Steps, part 3. Clear column 3 using the pivot in R3. Reduced row-echelon form. Read off the unique solution.

The answer. So the answer is x equals negative 4, y equals negative 2, z equals 0. The system has exactly one solution; substituting it back satisfies every equation.

Check. Equation 1: negative x minus 2y equals 8, which checks out. Equation 2: 2y minus z equals negative 4, which checks out. Equation 3: 2x minus 3y minus 2z equals negative 2, which checks out.

About systems of equations

Solve systems of two or three linear equations by elimination and substitution.

Solve the system -x - 2y = 8, 2y - z = -4, 2x - 3y - 2z = -2