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Systems of equationsAlgebraHigh schoolabout 1:12

Solve the system 3x - 4y = -2, 2x - 6y = -18: animated lesson

A narrated, step-by-step systems of equations video lesson with captions. Final answer: x = 6, 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
    {3x−4y=−22x−6y=−18\begin{cases} 3x-4y = -2 \\ 2x-6y = -18 \end{cases}
  2. 2. Steps, part 1
    1. {3x−4y=−22x−6y=−18\begin{cases} 3x - 4y = -2 \\ 2x - 6y = -18 \end{cases}
      Write every equation in standard form
    2. [3−4−22−6−18]\begin{bmatrix}3 & -4 & -2 \\ 2 & -6 & -18\end{bmatrix}
      Build the augmented matrix [A | b]
    3. Gauss–Jordan elimination
    4. R1→13R1: [1−43−232−6−18]R_{1} \to \tfrac{1}{3} R_{1}:\ \begin{bmatrix}1 & -\frac{4}{3} & -\frac{2}{3} \\ 2 & -6 & -18\end{bmatrix}
      Divide R1 by 3
  3. 3. Steps, part 2
    1. R2→R2−2R1: [1−43−230−103−503]R_{2} \to R_{2} - 2R_{1}:\ \begin{bmatrix}1 & -\frac{4}{3} & -\frac{2}{3} \\ 0 & -\frac{10}{3} & -\frac{50}{3}\end{bmatrix}
      Clear column 1 using the pivot in R1
    2. R2→1−103R2: [1−43−23015]R_{2} \to \tfrac{1}{-\frac{10}{3}} R_{2}:\ \begin{bmatrix}1 & -\frac{4}{3} & -\frac{2}{3} \\ 0 & 1 & 5\end{bmatrix}
      Divide R2 by -10/3
    3. R1→R1+43R2: [106015]R_{1} \to R_{1} + \frac{4}{3} R_{2}:\ \begin{bmatrix}1 & 0 & 6 \\ 0 & 1 & 5\end{bmatrix}
      Clear column 2 using the pivot in R2
    4. [106015]\begin{bmatrix}1 & 0 & 6 \\ 0 & 1 & 5\end{bmatrix}
      Reduced row-echelon form
  4. 4. Steps, part 3
    1. x=6, y=5x = 6,\ y = 5
      Read off the unique solution
  5. 5. The answer
    x=6, y=5x = 6,\ y = 5

    3x - 4y = -2; 2x - 6y = -18 ⇒ x = 6, y = 5

  6. 6. Check
    • Equation 1: 3x - 4y = -2 ✓
    • Equation 2: 2x - 6y = -18 ✓

Transcript

The problem. Here is the problem: 3x minus 4y equals negative 2; 2x minus 6y equals negative 18. 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 3.

Steps, part 2. Clear column 1 using the pivot in R1. Divide R2 by negative 10 over 3. 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 6, y equals 5. The system has exactly one solution; substituting it back satisfies every equation.

Check. Equation 1: 3x minus 4y equals negative 2, which checks out. Equation 2: 2x minus 6y equals negative 18, 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 3x - 4y = -2, 2x - 6y = -18 (Lesson)