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Systems of equationsAlgebraHigh schoolabout 1:12

Solve the system 5x - 4y = -5, 4x - 8y = 20: animated lesson

A narrated, step-by-step systems of equations video lesson with captions. Final answer: x = -5, 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
    {5x−4y=−54x−8y=20\begin{cases} 5x-4y = -5 \\ 4x-8y = 20 \end{cases}
  2. 2. Steps, part 1
    1. {5x−4y=−54x−8y=20\begin{cases} 5x - 4y = -5 \\ 4x - 8y = 20 \end{cases}
      Write every equation in standard form
    2. [5−4−54−820]\begin{bmatrix}5 & -4 & -5 \\ 4 & -8 & 20\end{bmatrix}
      Build the augmented matrix [A | b]
    3. Gauss–Jordan elimination
    4. R1→15R1: [1−45−14−820]R_{1} \to \tfrac{1}{5} R_{1}:\ \begin{bmatrix}1 & -\frac{4}{5} & -1 \\ 4 & -8 & 20\end{bmatrix}
      Divide R1 by 5
  3. 3. Steps, part 2
    1. R2→R2−4R1: [1−45−10−24524]R_{2} \to R_{2} - 4R_{1}:\ \begin{bmatrix}1 & -\frac{4}{5} & -1 \\ 0 & -\frac{24}{5} & 24\end{bmatrix}
      Clear column 1 using the pivot in R1
    2. R2→1−245R2: [1−45−101−5]R_{2} \to \tfrac{1}{-\frac{24}{5}} R_{2}:\ \begin{bmatrix}1 & -\frac{4}{5} & -1 \\ 0 & 1 & -5\end{bmatrix}
      Divide R2 by -24/5
    3. R1→R1+45R2: [10−501−5]R_{1} \to R_{1} + \frac{4}{5} R_{2}:\ \begin{bmatrix}1 & 0 & -5 \\ 0 & 1 & -5\end{bmatrix}
      Clear column 2 using the pivot in R2
    4. [10−501−5]\begin{bmatrix}1 & 0 & -5 \\ 0 & 1 & -5\end{bmatrix}
      Reduced row-echelon form
  4. 4. Steps, part 3
    1. x=−5, y=−5x = -5,\ y = -5
      Read off the unique solution
  5. 5. The answer
    x=−5, y=−5x = -5,\ y = -5

    5x - 4y = -5; 4x - 8y = 20 ⇒ x = -5, y = -5

  6. 6. Check
    • Equation 1: 5x - 4y = -5 ✓
    • Equation 2: 4x - 8y = 20 ✓

Transcript

The problem. Here is the problem: 5x minus 4y equals negative 5; 4x minus 8y equals 20. 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 5.

Steps, part 2. Clear column 1 using the pivot in R1. Divide R2 by negative 24 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 5, y equals negative 5. The system has exactly one solution; substituting it back satisfies every equation.

Check. Equation 1: 5x minus 4y equals negative 5, which checks out. Equation 2: 4x minus 8y equals 20, 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 - 4y = -5, 4x - 8y = 20: Animated Lesson