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Systems of equationsAlgebraHigh schoolabout 1:14

Solve the system -2x - 3y = -5 and 5x - y = -30: 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
    {−2x−3y=−55x−y=−30\begin{cases} -2x-3y = -5 \\ 5x- y = -30 \end{cases}
  2. 2. Steps, part 1
    1. {−2x−3y=−55x−y=−30\begin{cases} -2x - 3y = -5 \\ 5x - y = -30 \end{cases}
      Write every equation in standard form
    2. [−2−3−55−1−30]\begin{bmatrix}-2 & -3 & -5 \\ 5 & -1 & -30\end{bmatrix}
      Build the augmented matrix [A | b]
    3. Gauss–Jordan elimination
    4. R1→1−2R1: [132525−1−30]R_{1} \to \tfrac{1}{-2} R_{1}:\ \begin{bmatrix}1 & \frac{3}{2} & \frac{5}{2} \\ 5 & -1 & -30\end{bmatrix}
      Divide R1 by -2
  3. 3. Steps, part 2
    1. R2→R2−5R1: [132520−172−852]R_{2} \to R_{2} - 5R_{1}:\ \begin{bmatrix}1 & \frac{3}{2} & \frac{5}{2} \\ 0 & -\frac{17}{2} & -\frac{85}{2}\end{bmatrix}
      Clear column 1 using the pivot in R1
    2. R2→1−172R2: [13252015]R_{2} \to \tfrac{1}{-\frac{17}{2}} R_{2}:\ \begin{bmatrix}1 & \frac{3}{2} & \frac{5}{2} \\ 0 & 1 & 5\end{bmatrix}
      Divide R2 by -17/2
    3. R1→R1−32R2: [10−5015]R_{1} \to R_{1} - \frac{3}{2} R_{2}:\ \begin{bmatrix}1 & 0 & -5 \\ 0 & 1 & 5\end{bmatrix}
      Clear column 2 using the pivot in R2
    4. [10−5015]\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

    -2x - 3y = -5; 5x - y = -30 ⇒ x = -5, y = 5

  6. 6. Check
    • Equation 1: -2x - 3y = -5 ✓
    • Equation 2: 5x - y = -30 ✓

Transcript

The problem. Here is the problem: negative 2x minus 3y equals negative 5; 5x minus y equals negative 30. 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 2.

Steps, part 2. Clear column 1 using the pivot in R1. Divide R2 by negative 17 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 negative 5, y equals 5. The system has exactly one solution; substituting it back satisfies every equation.

Check. Equation 1: negative 2x minus 3y equals negative 5, which checks out. Equation 2: 5x minus y equals negative 30, 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 = -5 and 5x - y = -30 (Lesson)