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Systems of equationsAlgebraHigh schoolabout 1:27

Solve the system - 2y - 2z = 2, 2x + 2y - 3z = -14, 2y + 3z = 0: animated lesson

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

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)

Steps on screen

  1. 1. The problem
    Solving an equation
    {−2y−2z=22x+2y−3z=−142y+3z=0\begin{cases} -2y-2z = 2 \\ 2x+2y-3z = -14 \\ 2y+3z = 0 \end{cases}
  2. 2. Steps, part 1
    1. {−2y−2z=22x+2y−3z=−142y+3z=0\begin{cases} -2y - 2z = 2 \\ 2x + 2y - 3z = -14 \\ 2y + 3z = 0 \end{cases}
      Write every equation in standard form
    2. [0−2−2222−3−140230]\begin{bmatrix}0 & -2 & -2 & 2 \\ 2 & 2 & -3 & -14 \\ 0 & 2 & 3 & 0\end{bmatrix}
      Build the augmented matrix [A | b]
    3. Gauss–Jordan elimination
    4. [22−3−140−2−220230]\begin{bmatrix}2 & 2 & -3 & -14 \\ 0 & -2 & -2 & 2 \\ 0 & 2 & 3 & 0\end{bmatrix}
      Swap R1 and R2
  3. 3. Steps, part 2
    1. R1→12R1: [11−32−70−2−220230]R_{1} \to \tfrac{1}{2} R_{1}:\ \begin{bmatrix}1 & 1 & -\frac{3}{2} & -7 \\ 0 & -2 & -2 & 2 \\ 0 & 2 & 3 & 0\end{bmatrix}
      Divide R1 by 2
    2. R2→1−2R2: [11−32−7011−10230]R_{2} \to \tfrac{1}{-2} R_{2}:\ \begin{bmatrix}1 & 1 & -\frac{3}{2} & -7 \\ 0 & 1 & 1 & -1 \\ 0 & 2 & 3 & 0\end{bmatrix}
      Divide R2 by -2
    3. R1→R1−1R2, R3→R3−2R2: [10−52−6011−10012]R_{1} \to R_{1} - 1R_{2},\ R_{3} \to R_{3} - 2R_{2}:\ \begin{bmatrix}1 & 0 & -\frac{5}{2} & -6 \\ 0 & 1 & 1 & -1 \\ 0 & 0 & 1 & 2\end{bmatrix}
      Clear column 2 using the pivot in R2
    4. R1→R1+52R3, R2→R2−1R3: [100−1010−30012]R_{1} \to R_{1} + \frac{5}{2} R_{3},\ R_{2} \to R_{2} - 1R_{3}:\ \begin{bmatrix}1 & 0 & 0 & -1 \\ 0 & 1 & 0 & -3 \\ 0 & 0 & 1 & 2\end{bmatrix}
      Clear column 3 using the pivot in R3
  4. 4. Steps, part 3
    1. [100−1010−30012]\begin{bmatrix}1 & 0 & 0 & -1 \\ 0 & 1 & 0 & -3 \\ 0 & 0 & 1 & 2\end{bmatrix}
      Reduced row-echelon form
    2. x=−1, y=−3, z=2x = -1,\ y = -3,\ z = 2
      Read off the unique solution
  5. 5. The answer
    x=−1, y=−3, z=2x = -1,\ y = -3,\ z = 2

    -2y - 2z = 2; 2x + 2y - 3z = -14; 2y + 3z = 0 ⇒ x = -1, y = -3, z = 2

  6. 6. Check
    • Equation 1: -2y - 2z = 2 ✓
    • Equation 2: 2x + 2y - 3z = -14 ✓
    • Equation 3: 2y + 3z = 0 ✓

Transcript

The problem. Here is the problem: - 2y minus 2z equals 2; 2x plus 2y minus 3z equals negative 14; 2y plus 3z equals 0. 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. Swap R1 and R2.

Steps, part 2. Divide R1 by 2. Divide R2 by negative 2. Clear column 2 using the pivot in R2. 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 negative 1, y equals negative 3, z equals 2. The system has exactly one solution; substituting it back satisfies every equation.

Check. Equation 1: negative 2y minus 2z equals 2, which checks out. Equation 2: 2x plus 2y minus 3z equals negative 14, which checks out. Equation 3: 2y plus 3z equals 0, which checks out.

About systems of equations

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

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