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Systems of equationsAlgebraHigh schoolabout 1:10

Solve the system 2x - y = -11, 8x + 8y = -32: animated lesson

A narrated, step-by-step systems of equations video lesson with captions. Final answer: x = -5, 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
    {2x−y=−118x+8y=−32\begin{cases} 2x- y = -11 \\ 8x+8y = -32 \end{cases}
  2. 2. Steps, part 1
    1. {2x−y=−118x+8y=−32\begin{cases} 2x - y = -11 \\ 8x + 8y = -32 \end{cases}
      Write every equation in standard form
    2. [2−1−1188−32]\begin{bmatrix}2 & -1 & -11 \\ 8 & 8 & -32\end{bmatrix}
      Build the augmented matrix [A | b]
    3. Gauss–Jordan elimination
    4. R1→12R1: [1−12−11288−32]R_{1} \to \tfrac{1}{2} R_{1}:\ \begin{bmatrix}1 & -\frac{1}{2} & -\frac{11}{2} \\ 8 & 8 & -32\end{bmatrix}
      Divide R1 by 2
  3. 3. Steps, part 2
    1. R2→R2−8R1: [1−12−11201212]R_{2} \to R_{2} - 8R_{1}:\ \begin{bmatrix}1 & -\frac{1}{2} & -\frac{11}{2} \\ 0 & 12 & 12\end{bmatrix}
      Clear column 1 using the pivot in R1
    2. R2→112R2: [1−12−112011]R_{2} \to \tfrac{1}{12} R_{2}:\ \begin{bmatrix}1 & -\frac{1}{2} & -\frac{11}{2} \\ 0 & 1 & 1\end{bmatrix}
      Divide R2 by 12
    3. R1→R1+12R2: [10−5011]R_{1} \to R_{1} + \frac{1}{2} R_{2}:\ \begin{bmatrix}1 & 0 & -5 \\ 0 & 1 & 1\end{bmatrix}
      Clear column 2 using the pivot in R2
    4. [10−5011]\begin{bmatrix}1 & 0 & -5 \\ 0 & 1 & 1\end{bmatrix}
      Reduced row-echelon form
  4. 4. Steps, part 3
    1. x=−5, y=1x = -5,\ y = 1
      Read off the unique solution
  5. 5. The answer
    x=−5, y=1x = -5,\ y = 1

    2x - y = -11; 8x + 8y = -32 ⇒ x = -5, y = 1

  6. 6. Check
    • Equation 1: 2x - y = -11 ✓
    • Equation 2: 8x + 8y = -32 ✓

Transcript

The problem. Here is the problem: 2x minus y equals negative 11; 8x plus 8y equals negative 32. 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 2.

Steps, part 2. Clear column 1 using the pivot in R1. Divide R2 by 12. 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 5, y equals 1. The system has exactly one solution; substituting it back satisfies every equation.

Check. Equation 1: 2x minus y equals negative 11, which checks out. Equation 2: 8x plus 8y equals negative 32, 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 2x - y = -11, 8x + 8y = -32 (Lesson)