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Systems of equationsAlgebraHigh schoolabout 1:15

Solve the system -5x - 6y = 26, -3x - 5y = 17: animated lesson

A narrated, step-by-step systems of equations video lesson with captions. Final answer: x = -4, 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
    {−5x−6y=26−3x−5y=17\begin{cases} -5x-6y = 26 \\ -3x-5y = 17 \end{cases}
  2. 2. Steps, part 1
    1. {−5x−6y=26−3x−5y=17\begin{cases} -5x - 6y = 26 \\ -3x - 5y = 17 \end{cases}
      Write every equation in standard form
    2. [−5−626−3−517]\begin{bmatrix}-5 & -6 & 26 \\ -3 & -5 & 17\end{bmatrix}
      Build the augmented matrix [A | b]
    3. Gauss–Jordan elimination
    4. R1→1−5R1: [165−265−3−517]R_{1} \to \tfrac{1}{-5} R_{1}:\ \begin{bmatrix}1 & \frac{6}{5} & -\frac{26}{5} \\ -3 & -5 & 17\end{bmatrix}
      Divide R1 by -5
  3. 3. Steps, part 2
    1. R2→R2+3R1: [165−2650−7575]R_{2} \to R_{2} + 3R_{1}:\ \begin{bmatrix}1 & \frac{6}{5} & -\frac{26}{5} \\ 0 & -\frac{7}{5} & \frac{7}{5}\end{bmatrix}
      Clear column 1 using the pivot in R1
    2. R2→1−75R2: [165−26501−1]R_{2} \to \tfrac{1}{-\frac{7}{5}} R_{2}:\ \begin{bmatrix}1 & \frac{6}{5} & -\frac{26}{5} \\ 0 & 1 & -1\end{bmatrix}
      Divide R2 by -7/5
    3. R1→R1−65R2: [10−401−1]R_{1} \to R_{1} - \frac{6}{5} R_{2}:\ \begin{bmatrix}1 & 0 & -4 \\ 0 & 1 & -1\end{bmatrix}
      Clear column 2 using the pivot in R2
    4. [10−401−1]\begin{bmatrix}1 & 0 & -4 \\ 0 & 1 & -1\end{bmatrix}
      Reduced row-echelon form
  4. 4. Steps, part 3
    1. x=−4, y=−1x = -4,\ y = -1
      Read off the unique solution
  5. 5. The answer
    x=−4, y=−1x = -4,\ y = -1

    -5x - 6y = 26; -3x - 5y = 17 ⇒ x = -4, y = -1

  6. 6. Check
    • Equation 1: -5x - 6y = 26 ✓
    • Equation 2: -3x - 5y = 17 ✓

Transcript

The problem. Here is the problem: negative 5x minus 6y equals 26; negative 3x minus 5y equals 17. 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 5.

Steps, part 2. Clear column 1 using the pivot in R1. Divide R2 by negative 7 over 5. 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 4, y equals negative 1. The system has exactly one solution; substituting it back satisfies every equation.

Check. Equation 1: negative 5x minus 6y equals 26, which checks out. Equation 2: negative 3x minus 5y equals 17, 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 -5x - 6y = 26, -3x - 5y = 17 (Lesson)