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Systems of equationsAlgebraHigh schoolabout 1:24

Solve the system x + y + z = 6, x - y + z = 2, 2x + y - z = 1: animated lesson

A narrated, step-by-step systems of equations video lesson with captions. Final answer: x = 1, y = 2, z = 3.

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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
$\begin{cases} x+ y+ z = 6 \\ x- y+ z = 2 \\ 2x+ y- z = 1 \end{cases}$
2. Steps, part 1
1. $\begin{cases} x + y + z = 6 \\ x - y + z = 2 \\ 2x + y - z = 1 \end{cases}$
  Write every equation in standard form
2. $\begin{bmatrix}1 & 1 & 1 & 6 \\ 1 & -1 & 1 & 2 \\ 2 & 1 & -1 & 1\end{bmatrix}$
  Build the augmented matrix [A | b]
3. Gauss–Jordan elimination
4. $R_{2} \to R_{2} - 1R_{1},\ R_{3} \to R_{3} - 2R_{1}:\ \begin{bmatrix}1 & 1 & 1 & 6 \\ 0 & -2 & 0 & -4 \\ 0 & -1 & -3 & -11\end{bmatrix}$
  Clear column 1 using the pivot in R1
3. Steps, part 2
1. $R_{2} \to \tfrac{1}{-2} R_{2}:\ \begin{bmatrix}1 & 1 & 1 & 6 \\ 0 & 1 & 0 & 2 \\ 0 & -1 & -3 & -11\end{bmatrix}$
  Divide R2 by -2
2. $R_{1} \to R_{1} - 1R_{2},\ R_{3} \to R_{3} + 1R_{2}:\ \begin{bmatrix}1 & 0 & 1 & 4 \\ 0 & 1 & 0 & 2 \\ 0 & 0 & -3 & -9\end{bmatrix}$
  Clear column 2 using the pivot in R2
3. $R_{3} \to \tfrac{1}{-3} R_{3}:\ \begin{bmatrix}1 & 0 & 1 & 4 \\ 0 & 1 & 0 & 2 \\ 0 & 0 & 1 & 3\end{bmatrix}$
  Divide R3 by -3
4. $R_{1} \to R_{1} - 1R_{3}:\ \begin{bmatrix}1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 2 \\ 0 & 0 & 1 & 3\end{bmatrix}$
  Clear column 3 using the pivot in R3
4. Steps, part 3
1. $\begin{bmatrix}1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 2 \\ 0 & 0 & 1 & 3\end{bmatrix}$
  Reduced row-echelon form
2. $x = 1,\ y = 2,\ z = 3$
  Read off the unique solution
5. The answer
$x = 1,\ y = 2,\ z = 3$
x + y + z = 6; x - y + z = 2; 2x + y - z = 1 ⇒ x = 1, y = 2, z = 3
6. Check
- Equation 1: x + y + z = 6 ✓
- Equation 2: x - y + z = 2 ✓
- Equation 3: 2x + y - z = 1 ✓

Transcript

The problem. Here is the problem: x plus y plus z equals 6; x minus y plus z equals 2; 2x plus y minus z equals 1. 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. Clear column 1 using the pivot in R1.

Steps, part 2. Divide R2 by negative 2. Clear column 2 using the pivot in R2. Divide R3 by negative 3. Clear column 3 using the pivot in R3.

Steps, part 3. Reduced row-echelon form. Read off the unique solution.

The answer. So the answer is x equals 1, y equals 2, z equals 3. The system has exactly one solution; substituting it back satisfies every equation.

Check. Equation 1: x plus y plus z equals 6, which checks out. Equation 2: x minus y plus z equals 2, which checks out. Equation 3: 2x plus y minus z equals 1, which checks out.

About systems of equations

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

Solve the system x + y + z = 6, x - y + z = 2, 2x + y - z…