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Systems of equationsAlgebraHigh schoolabout 1:13

Solve the system -4x - 3y = 35, -8x + 5y = 15: 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
    {−4x−3y=35−8x+5y=15\begin{cases} -4x-3y = 35 \\ -8x+5y = 15 \end{cases}
  2. 2. Steps, part 1
    1. {−4x−3y=35−8x+5y=15\begin{cases} -4x - 3y = 35 \\ -8x + 5y = 15 \end{cases}
      Write every equation in standard form
    2. [−4−335−8515]\begin{bmatrix}-4 & -3 & 35 \\ -8 & 5 & 15\end{bmatrix}
      Build the augmented matrix [A | b]
    3. Gauss–Jordan elimination
    4. R1→1−4R1: [134−354−8515]R_{1} \to \tfrac{1}{-4} R_{1}:\ \begin{bmatrix}1 & \frac{3}{4} & -\frac{35}{4} \\ -8 & 5 & 15\end{bmatrix}
      Divide R1 by -4
  3. 3. Steps, part 2
    1. R2→R2+8R1: [134−354011−55]R_{2} \to R_{2} + 8R_{1}:\ \begin{bmatrix}1 & \frac{3}{4} & -\frac{35}{4} \\ 0 & 11 & -55\end{bmatrix}
      Clear column 1 using the pivot in R1
    2. R2→111R2: [134−35401−5]R_{2} \to \tfrac{1}{11} R_{2}:\ \begin{bmatrix}1 & \frac{3}{4} & -\frac{35}{4} \\ 0 & 1 & -5\end{bmatrix}
      Divide R2 by 11
    3. R1→R1−34R2: [10−501−5]R_{1} \to R_{1} - \frac{3}{4} R_{2}:\ \begin{bmatrix}1 & 0 & -5 \\ 0 & 1 & -5\end{bmatrix}
      Clear column 2 using the pivot in R2
    4. [10−501−5]\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

    -4x - 3y = 35; -8x + 5y = 15 ⇒ x = -5, y = -5

  6. 6. Check
    • Equation 1: -4x - 3y = 35 ✓
    • Equation 2: -8x + 5y = 15 ✓

Transcript

The problem. Here is the problem: negative 4x minus 3y equals 35; negative 8x plus 5y equals 15. 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 4.

Steps, part 2. Clear column 1 using the pivot in R1. Divide R2 by 11. 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 negative 5. The system has exactly one solution; substituting it back satisfies every equation.

Check. Equation 1: negative 4x minus 3y equals 35, which checks out. Equation 2: negative 8x plus 5y equals 15, which checks out.

About systems of equations

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

More lessons like this

Solve the system -4x - 3y = 35, -8x + 5y = 15 (Lesson)