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Systems of equationsAlgebraHigh schoolabout 1:11

Solve the system 3x - 4y = 15, x + 4y = -27: animated lesson

A narrated, step-by-step systems of equations video lesson with captions. Final answer: x = -3, y = -6.

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=15x+4y=−27\begin{cases} 3x-4y = 15 \\ x+4y = -27 \end{cases}
  2. 2. Steps, part 1
    1. {3x−4y=15x+4y=−27\begin{cases} 3x - 4y = 15 \\ x + 4y = -27 \end{cases}
      Write every equation in standard form
    2. [3−41514−27]\begin{bmatrix}3 & -4 & 15 \\ 1 & 4 & -27\end{bmatrix}
      Build the augmented matrix [A | b]
    3. Gauss–Jordan elimination
    4. R1→13R1: [1−43514−27]R_{1} \to \tfrac{1}{3} R_{1}:\ \begin{bmatrix}1 & -\frac{4}{3} & 5 \\ 1 & 4 & -27\end{bmatrix}
      Divide R1 by 3
  3. 3. Steps, part 2
    1. R2→R2−1R1: [1−4350163−32]R_{2} \to R_{2} - 1R_{1}:\ \begin{bmatrix}1 & -\frac{4}{3} & 5 \\ 0 & \frac{16}{3} & -32\end{bmatrix}
      Clear column 1 using the pivot in R1
    2. R2→1163R2: [1−43501−6]R_{2} \to \tfrac{1}{\frac{16}{3}} R_{2}:\ \begin{bmatrix}1 & -\frac{4}{3} & 5 \\ 0 & 1 & -6\end{bmatrix}
      Divide R2 by 16/3
    3. R1→R1+43R2: [10−301−6]R_{1} \to R_{1} + \frac{4}{3} R_{2}:\ \begin{bmatrix}1 & 0 & -3 \\ 0 & 1 & -6\end{bmatrix}
      Clear column 2 using the pivot in R2
    4. [10−301−6]\begin{bmatrix}1 & 0 & -3 \\ 0 & 1 & -6\end{bmatrix}
      Reduced row-echelon form
  4. 4. Steps, part 3
    1. x=−3, y=−6x = -3,\ y = -6
      Read off the unique solution
  5. 5. The answer
    x=−3, y=−6x = -3,\ y = -6

    3x - 4y = 15; x + 4y = -27 ⇒ x = -3, y = -6

  6. 6. Check
    • Equation 1: 3x - 4y = 15 ✓
    • Equation 2: x + 4y = -27 ✓

Transcript

The problem. Here is the problem: 3x minus 4y equals 15; x plus 4y equals negative 27. 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 16 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 negative 3, y equals negative 6. The system has exactly one solution; substituting it back satisfies every equation.

Check. Equation 1: 3x minus 4y equals 15, which checks out. Equation 2: x plus 4y equals negative 27, 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 = 15, x + 4y = -27: Animated Lesson