Skip to content
MathSolver Pro
  • Open the math solver
  • Ask the AI tutor
  • Find an AP course
  • Create a quiz
  • Open the calculator
  • Graphing Calculator
  • Browse all calculators
  • Read math guides
↑ ↓Enter to open · Esc to close
Solve a problem
Systems of equationsAlgebraHigh schoolabout 1:14

Solve the system 7x + 4y = 32/3, 6x - 4y = 24: animated lesson

A narrated, step-by-step systems of equations video lesson with captions. Final answer: x = 8/3, y = -2.

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
    {7x+4y=3236x−4y=24\begin{cases} 7x+4y = \frac{32}{3} \\ 6x-4y = 24 \end{cases}
  2. 2. Steps, part 1
    1. {7x+4y=3236x−4y=24\begin{cases} 7x + 4y = \frac{32}{3} \\ 6x - 4y = 24 \end{cases}
      Write every equation in standard form
    2. [743236−424]\begin{bmatrix}7 & 4 & \frac{32}{3} \\ 6 & -4 & 24\end{bmatrix}
      Build the augmented matrix [A | b]
    3. Gauss–Jordan elimination
    4. R1→17R1: [14732216−424]R_{1} \to \tfrac{1}{7} R_{1}:\ \begin{bmatrix}1 & \frac{4}{7} & \frac{32}{21} \\ 6 & -4 & 24\end{bmatrix}
      Divide R1 by 7
  3. 3. Steps, part 2
    1. R2→R2−6R1: [14732210−5271047]R_{2} \to R_{2} - 6R_{1}:\ \begin{bmatrix}1 & \frac{4}{7} & \frac{32}{21} \\ 0 & -\frac{52}{7} & \frac{104}{7}\end{bmatrix}
      Clear column 1 using the pivot in R1
    2. R2→1−527R2: [147322101−2]R_{2} \to \tfrac{1}{-\frac{52}{7}} R_{2}:\ \begin{bmatrix}1 & \frac{4}{7} & \frac{32}{21} \\ 0 & 1 & -2\end{bmatrix}
      Divide R2 by -52/7
    3. R1→R1−47R2: [108301−2]R_{1} \to R_{1} - \frac{4}{7} R_{2}:\ \begin{bmatrix}1 & 0 & \frac{8}{3} \\ 0 & 1 & -2\end{bmatrix}
      Clear column 2 using the pivot in R2
    4. [108301−2]\begin{bmatrix}1 & 0 & \frac{8}{3} \\ 0 & 1 & -2\end{bmatrix}
      Reduced row-echelon form
  4. 4. Steps, part 3
    1. x=83, y=−2x = \frac{8}{3},\ y = -2
      Read off the unique solution
  5. 5. The answer
    x=83, y=−2x = \frac{8}{3},\ y = -2

    7x + 4y = 32/3; 6x - 4y = 24 ⇒ x = 8/3, y = -2

  6. 6. Check
    • Equation 1: 7x + 4y = 32/3 ✓
    • Equation 2: 6x - 4y = 24 ✓

Transcript

The problem. Here is the problem: 7x plus 4y equals 32 over 3; 6x minus 4y equals 24. 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 7.

Steps, part 2. Clear column 1 using the pivot in R1. Divide R2 by negative 52 over 7. 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 8 over 3, y equals negative 2. The system has exactly one solution; substituting it back satisfies every equation.

Check. Equation 1: 7x plus 4y equals 32 over 3, which checks out. Equation 2: 6x minus 4y equals 24, 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 7x + 4y = 32/3, 6x - 4y = 24 (Lesson)