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Systems of equationsAlgebraHigh schoolabout 1:11

Solve the system 2x - 3y = 4, -3x + 8y = -6: animated lesson

A narrated, step-by-step systems of equations video lesson with captions. Final answer: x = 2, y = 0.

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−3y=4−3x+8y=−6\begin{cases} 2x-3y = 4 \\ -3x+8y = -6 \end{cases}
  2. 2. Steps, part 1
    1. {2x−3y=4−3x+8y=−6\begin{cases} 2x - 3y = 4 \\ -3x + 8y = -6 \end{cases}
      Write every equation in standard form
    2. [2−34−38−6]\begin{bmatrix}2 & -3 & 4 \\ -3 & 8 & -6\end{bmatrix}
      Build the augmented matrix [A | b]
    3. Gauss–Jordan elimination
    4. R1→12R1: [1−322−38−6]R_{1} \to \tfrac{1}{2} R_{1}:\ \begin{bmatrix}1 & -\frac{3}{2} & 2 \\ -3 & 8 & -6\end{bmatrix}
      Divide R1 by 2
  3. 3. Steps, part 2
    1. R2→R2+3R1: [1−3220720]R_{2} \to R_{2} + 3R_{1}:\ \begin{bmatrix}1 & -\frac{3}{2} & 2 \\ 0 & \frac{7}{2} & 0\end{bmatrix}
      Clear column 1 using the pivot in R1
    2. R2→172R2: [1−322010]R_{2} \to \tfrac{1}{\frac{7}{2}} R_{2}:\ \begin{bmatrix}1 & -\frac{3}{2} & 2 \\ 0 & 1 & 0\end{bmatrix}
      Divide R2 by 7/2
    3. R1→R1+32R2: [102010]R_{1} \to R_{1} + \frac{3}{2} R_{2}:\ \begin{bmatrix}1 & 0 & 2 \\ 0 & 1 & 0\end{bmatrix}
      Clear column 2 using the pivot in R2
    4. [102010]\begin{bmatrix}1 & 0 & 2 \\ 0 & 1 & 0\end{bmatrix}
      Reduced row-echelon form
  4. 4. Steps, part 3
    1. x=2, y=0x = 2,\ y = 0
      Read off the unique solution
  5. 5. The answer
    x=2, y=0x = 2,\ y = 0

    2x - 3y = 4; -3x + 8y = -6 ⇒ x = 2, y = 0

  6. 6. Check
    • Equation 1: 2x - 3y = 4 ✓
    • Equation 2: -3x + 8y = -6 ✓

Transcript

The problem. Here is the problem: 2x minus 3y equals 4; negative 3x plus 8y equals negative 6. 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 7 over 2. 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 2, y equals 0. The system has exactly one solution; substituting it back satisfies every equation.

Check. Equation 1: 2x minus 3y equals 4, which checks out. Equation 2: negative 3x plus 8y equals negative 6, 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 - 3y = 4, -3x + 8y = -6: Animated Lesson