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Systems of equationsAlgebraHigh schoolabout 1:11

Solve the system -x + 3y = 11, -4x + 7y = 19: 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
    {−x+3y=11−4x+7y=19\begin{cases} - x+3y = 11 \\ -4x+7y = 19 \end{cases}
  2. 2. Steps, part 1
    1. {−x+3y=11−4x+7y=19\begin{cases} -x + 3y = 11 \\ -4x + 7y = 19 \end{cases}
      Write every equation in standard form
    2. [−1311−4719]\begin{bmatrix}-1 & 3 & 11 \\ -4 & 7 & 19\end{bmatrix}
      Build the augmented matrix [A | b]
    3. Gauss–Jordan elimination
    4. R1→1−1R1: [1−3−11−4719]R_{1} \to \tfrac{1}{-1} R_{1}:\ \begin{bmatrix}1 & -3 & -11 \\ -4 & 7 & 19\end{bmatrix}
      Divide R1 by -1
  3. 3. Steps, part 2
    1. R2→R2+4R1: [1−3−110−5−25]R_{2} \to R_{2} + 4R_{1}:\ \begin{bmatrix}1 & -3 & -11 \\ 0 & -5 & -25\end{bmatrix}
      Clear column 1 using the pivot in R1
    2. R2→1−5R2: [1−3−11015]R_{2} \to \tfrac{1}{-5} R_{2}:\ \begin{bmatrix}1 & -3 & -11 \\ 0 & 1 & 5\end{bmatrix}
      Divide R2 by -5
    3. R1→R1+3R2: [104015]R_{1} \to R_{1} + 3R_{2}:\ \begin{bmatrix}1 & 0 & 4 \\ 0 & 1 & 5\end{bmatrix}
      Clear column 2 using the pivot in R2
    4. [104015]\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

    -x + 3y = 11; -4x + 7y = 19 ⇒ x = 4, y = 5

  6. 6. Check
    • Equation 1: -x + 3y = 11 ✓
    • Equation 2: -4x + 7y = 19 ✓

Transcript

The problem. Here is the problem: negative x plus 3y equals 11; negative 4x plus 7y equals 19. 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 1.

Steps, part 2. Clear column 1 using the pivot in R1. Divide R2 by negative 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 4, y equals 5. The system has exactly one solution; substituting it back satisfies every equation.

Check. Equation 1: negative x plus 3y equals 11, which checks out. Equation 2: negative 4x plus 7y equals 19, which checks out.

About systems of equations

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

Solve the system -x + 3y = 11, -4x + 7y = 19 (Lesson)