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Systems of equationsAlgebraHigh schoolabout 1:12

Solve the system -2x - y = -12, 9x - 3y = 9: animated lesson

A narrated, step-by-step systems of equations video lesson with captions. Final answer: x = 3, y = 6.

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 (0.5 MB) · Captions (WebVTT)

Steps on screen

  1. 1. The problem
    Solving an equation
    {−2x−y=−129x−3y=9\begin{cases} -2x- y = -12 \\ 9x-3y = 9 \end{cases}
  2. 2. Steps, part 1
    1. {−2x−y=−129x−3y=9\begin{cases} -2x - y = -12 \\ 9x - 3y = 9 \end{cases}
      Write every equation in standard form
    2. [−2−1−129−39]\begin{bmatrix}-2 & -1 & -12 \\ 9 & -3 & 9\end{bmatrix}
      Build the augmented matrix [A | b]
    3. Gauss–Jordan elimination
    4. R1→1−2R1: [11269−39]R_{1} \to \tfrac{1}{-2} R_{1}:\ \begin{bmatrix}1 & \frac{1}{2} & 6 \\ 9 & -3 & 9\end{bmatrix}
      Divide R1 by -2
  3. 3. Steps, part 2
    1. R2→R2−9R1: [11260−152−45]R_{2} \to R_{2} - 9R_{1}:\ \begin{bmatrix}1 & \frac{1}{2} & 6 \\ 0 & -\frac{15}{2} & -45\end{bmatrix}
      Clear column 1 using the pivot in R1
    2. R2→1−152R2: [1126016]R_{2} \to \tfrac{1}{-\frac{15}{2}} R_{2}:\ \begin{bmatrix}1 & \frac{1}{2} & 6 \\ 0 & 1 & 6\end{bmatrix}
      Divide R2 by -15/2
    3. R1→R1−12R2: [103016]R_{1} \to R_{1} - \frac{1}{2} R_{2}:\ \begin{bmatrix}1 & 0 & 3 \\ 0 & 1 & 6\end{bmatrix}
      Clear column 2 using the pivot in R2
    4. [103016]\begin{bmatrix}1 & 0 & 3 \\ 0 & 1 & 6\end{bmatrix}
      Reduced row-echelon form
  4. 4. Steps, part 3
    1. x=3, y=6x = 3,\ y = 6
      Read off the unique solution
  5. 5. The answer
    x=3, y=6x = 3,\ y = 6

    -2x - y = -12; 9x - 3y = 9 ⇒ x = 3, y = 6

  6. 6. Check
    • Equation 1: -2x - y = -12 ✓
    • Equation 2: 9x - 3y = 9 ✓

Transcript

The problem. Here is the problem: negative 2x minus y equals negative 12; 9x minus 3y equals 9. 2 linear equations in 2 unknowns (x, y).

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 | constant. Gauss–Jordan elimination. Use row operations to make each pivot 1 and clear the rest of its column. Divide R1 by negative 2.

Steps, part 2. Clear column 1 using the pivot in R1. Divide R2 by negative 15 over 2. Clear column 2 using the pivot in R2. Reduced row-echelon form.

Steps, part 3. Read off the unique solution.

The answer. So the answer is x equals 3, y equals 6. The system has exactly one solution; substituting it back satisfies every equation.

Check. Equation 1: negative 2x minus y equals negative 12, which checks out. Equation 2: 9x minus 3y equals 9, which checks out.

About systems of equations

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

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Solve the system -2x - y = -12, 9x - 3y = 9: Animated Lesson