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Systems of equationsAlgebraHigh schoolabout 1:14

Solve the system -8x - 5y = 13 and -7x + 6y = 84: animated lesson

A narrated, step-by-step systems of equations video lesson with captions. Final answer: x = -6, y = 7.

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
    {−8x−5y=13−7x+6y=84\begin{cases} -8x-5y = 13 \\ -7x+6y = 84 \end{cases}
  2. 2. Steps, part 1
    1. {−8x−5y=13−7x+6y=84\begin{cases} -8x - 5y = 13 \\ -7x + 6y = 84 \end{cases}
      Write every equation in standard form
    2. [−8−513−7684]\begin{bmatrix}-8 & -5 & 13 \\ -7 & 6 & 84\end{bmatrix}
      Build the augmented matrix [A | b]
    3. Gauss–Jordan elimination
    4. R1→1−8R1: [158−138−7684]R_{1} \to \tfrac{1}{-8} R_{1}:\ \begin{bmatrix}1 & \frac{5}{8} & -\frac{13}{8} \\ -7 & 6 & 84\end{bmatrix}
      Divide R1 by -8
  3. 3. Steps, part 2
    1. R2→R2+7R1: [158−13808385818]R_{2} \to R_{2} + 7R_{1}:\ \begin{bmatrix}1 & \frac{5}{8} & -\frac{13}{8} \\ 0 & \frac{83}{8} & \frac{581}{8}\end{bmatrix}
      Clear column 1 using the pivot in R1
    2. R2→1838R2: [158−138017]R_{2} \to \tfrac{1}{\frac{83}{8}} R_{2}:\ \begin{bmatrix}1 & \frac{5}{8} & -\frac{13}{8} \\ 0 & 1 & 7\end{bmatrix}
      Divide R2 by 83/8
    3. R1→R1−58R2: [10−6017]R_{1} \to R_{1} - \frac{5}{8} R_{2}:\ \begin{bmatrix}1 & 0 & -6 \\ 0 & 1 & 7\end{bmatrix}
      Clear column 2 using the pivot in R2
    4. [10−6017]\begin{bmatrix}1 & 0 & -6 \\ 0 & 1 & 7\end{bmatrix}
      Reduced row-echelon form
  4. 4. Steps, part 3
    1. x=−6, y=7x = -6,\ y = 7
      Read off the unique solution
  5. 5. The answer
    x=−6, y=7x = -6,\ y = 7

    -8x - 5y = 13; -7x + 6y = 84 ⇒ x = -6, y = 7

  6. 6. Check
    • Equation 1: -8x - 5y = 13 ✓
    • Equation 2: -7x + 6y = 84 ✓

Transcript

The problem. Here is the problem: negative 8x minus 5y equals 13; negative 7x plus 6y equals 84. 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 8.

Steps, part 2. Clear column 1 using the pivot in R1. Divide R2 by 83 over 8. 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 6, y equals 7. The system has exactly one solution; substituting it back satisfies every equation.

Check. Equation 1: negative 8x minus 5y equals 13, which checks out. Equation 2: negative 7x plus 6y equals 84, 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 -8x - 5y = 13 and -7x + 6y = 84 (Lesson)