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Systems of equationsAlgebraHigh schoolabout 1:13

Solve the system -3x - 6y = 15, -5x - 1y = 16: animated lesson

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

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
    {−3x−6y=15−5x−1y=16\begin{cases} -3x-6y = 15 \\ -5x-1y = 16 \end{cases}
  2. 2. Steps, part 1
    1. {−3x−6y=15−5x−y=16\begin{cases} -3x - 6y = 15 \\ -5x - y = 16 \end{cases}
      Write every equation in standard form
    2. [−3−615−5−116]\begin{bmatrix}-3 & -6 & 15 \\ -5 & -1 & 16\end{bmatrix}
      Build the augmented matrix [A | b]
    3. Gauss–Jordan elimination
    4. R1→1−3R1: [12−5−5−116]R_{1} \to \tfrac{1}{-3} R_{1}:\ \begin{bmatrix}1 & 2 & -5 \\ -5 & -1 & 16\end{bmatrix}
      Divide R1 by -3
  3. 3. Steps, part 2
    1. R2→R2+5R1: [12−509−9]R_{2} \to R_{2} + 5R_{1}:\ \begin{bmatrix}1 & 2 & -5 \\ 0 & 9 & -9\end{bmatrix}
      Clear column 1 using the pivot in R1
    2. R2→19R2: [12−501−1]R_{2} \to \tfrac{1}{9} R_{2}:\ \begin{bmatrix}1 & 2 & -5 \\ 0 & 1 & -1\end{bmatrix}
      Divide R2 by 9
    3. R1→R1−2R2: [10−301−1]R_{1} \to R_{1} - 2R_{2}:\ \begin{bmatrix}1 & 0 & -3 \\ 0 & 1 & -1\end{bmatrix}
      Clear column 2 using the pivot in R2
    4. [10−301−1]\begin{bmatrix}1 & 0 & -3 \\ 0 & 1 & -1\end{bmatrix}
      Reduced row-echelon form
  4. 4. Steps, part 3
    1. x=−3, y=−1x = -3,\ y = -1
      Read off the unique solution
  5. 5. The answer
    x=−3, y=−1x = -3,\ y = -1

    -3x - 6y = 15; -5x - y = 16 ⇒ x = -3, y = -1

  6. 6. Check
    • Equation 1: -3x - 6y = 15 ✓
    • Equation 2: -5x - y = 16 ✓

Transcript

The problem. Here is the problem: negative 3x minus 6y equals 15; negative 5x minus 1y equals 16. 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 3.

Steps, part 2. Clear column 1 using the pivot in R1. Divide R2 by 9. 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 negative 3, y equals negative 1. The system has exactly one solution; substituting it back satisfies every equation.

Check. Equation 1: negative 3x minus 6y equals 15, which checks out. Equation 2: negative 5x minus y equals 16, 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 -3x - 6y = 15, -5x - 1y = 16 (Lesson)