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Systems of equationsAlgebraHigh schoolabout 1:12

Solve the system -9x + 5y = -61 and 8x + y = 27: animated lesson

A narrated, step-by-step systems of equations video lesson with captions. Final answer: x = 4, y = -5.

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+5y=−618x+y=27\begin{cases} -9x+5y = -61 \\ 8x+ y = 27 \end{cases}
  2. 2. Steps, part 1
    1. {−9x+5y=−618x+y=27\begin{cases} -9x + 5y = -61 \\ 8x + y = 27 \end{cases}
      Write every equation in standard form
    2. [−95−618127]\begin{bmatrix}-9 & 5 & -61 \\ 8 & 1 & 27\end{bmatrix}
      Build the augmented matrix [A | b]
    3. Gauss–Jordan elimination
    4. R1→1−9R1: [1−596198127]R_{1} \to \tfrac{1}{-9} R_{1}:\ \begin{bmatrix}1 & -\frac{5}{9} & \frac{61}{9} \\ 8 & 1 & 27\end{bmatrix}
      Divide R1 by -9
  3. 3. Steps, part 2
    1. R2→R2−8R1: [1−596190499−2459]R_{2} \to R_{2} - 8R_{1}:\ \begin{bmatrix}1 & -\frac{5}{9} & \frac{61}{9} \\ 0 & \frac{49}{9} & -\frac{245}{9}\end{bmatrix}
      Clear column 1 using the pivot in R1
    2. R2→1499R2: [1−5961901−5]R_{2} \to \tfrac{1}{\frac{49}{9}} R_{2}:\ \begin{bmatrix}1 & -\frac{5}{9} & \frac{61}{9} \\ 0 & 1 & -5\end{bmatrix}
      Divide R2 by 49/9
    3. R1→R1+59R2: [10401−5]R_{1} \to R_{1} + \frac{5}{9} R_{2}:\ \begin{bmatrix}1 & 0 & 4 \\ 0 & 1 & -5\end{bmatrix}
      Clear column 2 using the pivot in R2
    4. [10401−5]\begin{bmatrix}1 & 0 & 4 \\ 0 & 1 & -5\end{bmatrix}
      Reduced row-echelon form
  4. 4. Steps, part 3
    1. x=4, y=−5x = 4,\ y = -5
      Read off the unique solution
  5. 5. The answer
    x=4, y=−5x = 4,\ y = -5

    -9x + 5y = -61; 8x + y = 27 ⇒ x = 4, y = -5

  6. 6. Check
    • Equation 1: -9x + 5y = -61 ✓
    • Equation 2: 8x + y = 27 ✓

Transcript

The problem. Here is the problem: negative 9x plus 5y equals negative 61; 8x plus y equals 27. 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 49 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 4, y equals negative 5. The system has exactly one solution; substituting it back satisfies every equation.

Check. Equation 1: negative 9x plus 5y equals negative 61, which checks out. Equation 2: 8x plus y equals 27, which checks out.

About systems of equations

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

Solve the system -9x + 5y = -61 and 8x + y = 27 (Lesson)