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Systems of equationsAlgebraHigh schoolabout 1:14

Solve the system -9x - 8y = -22, -5x + 5y = -50: animated lesson

A narrated, step-by-step systems of equations video lesson with captions. Final answer: x = 6, y = -4.

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
    {−9x−8y=−22−5x+5y=−50\begin{cases} -9x-8y = -22 \\ -5x+5y = -50 \end{cases}
  2. 2. Steps, part 1
    1. {−9x−8y=−22−5x+5y=−50\begin{cases} -9x - 8y = -22 \\ -5x + 5y = -50 \end{cases}
      Write every equation in standard form
    2. [−9−8−22−55−50]\begin{bmatrix}-9 & -8 & -22 \\ -5 & 5 & -50\end{bmatrix}
      Build the augmented matrix [A | b]
    3. Gauss–Jordan elimination
    4. R1→1−9R1: [189229−55−50]R_{1} \to \tfrac{1}{-9} R_{1}:\ \begin{bmatrix}1 & \frac{8}{9} & \frac{22}{9} \\ -5 & 5 & -50\end{bmatrix}
      Divide R1 by -9
  3. 3. Steps, part 2
    1. R2→R2+5R1: [1892290859−3409]R_{2} \to R_{2} + 5R_{1}:\ \begin{bmatrix}1 & \frac{8}{9} & \frac{22}{9} \\ 0 & \frac{85}{9} & -\frac{340}{9}\end{bmatrix}
      Clear column 1 using the pivot in R1
    2. R2→1859R2: [18922901−4]R_{2} \to \tfrac{1}{\frac{85}{9}} R_{2}:\ \begin{bmatrix}1 & \frac{8}{9} & \frac{22}{9} \\ 0 & 1 & -4\end{bmatrix}
      Divide R2 by 85/9
    3. R1→R1−89R2: [10601−4]R_{1} \to R_{1} - \frac{8}{9} R_{2}:\ \begin{bmatrix}1 & 0 & 6 \\ 0 & 1 & -4\end{bmatrix}
      Clear column 2 using the pivot in R2
    4. [10601−4]\begin{bmatrix}1 & 0 & 6 \\ 0 & 1 & -4\end{bmatrix}
      Reduced row-echelon form
  4. 4. Steps, part 3
    1. x=6, y=−4x = 6,\ y = -4
      Read off the unique solution
  5. 5. The answer
    x=6, y=−4x = 6,\ y = -4

    -9x - 8y = -22; -5x + 5y = -50 ⇒ x = 6, y = -4

  6. 6. Check
    • Equation 1: -9x - 8y = -22 ✓
    • Equation 2: -5x + 5y = -50 ✓

Transcript

The problem. Here is the problem: negative 9x minus 8y equals negative 22; negative 5x plus 5y equals negative 50. 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 9.

Steps, part 2. Clear column 1 using the pivot in R1. Divide R2 by 85 over 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 6, y equals negative 4. The system has exactly one solution; substituting it back satisfies every equation.

Check. Equation 1: negative 9x minus 8y equals negative 22, which checks out. Equation 2: negative 5x plus 5y equals negative 50, 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 -9x - 8y = -22, -5x + 5y = -50 (Lesson)