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 1:15

Solve the system 7x - 8y = -45/2, 5x - 1y = -9: animated lesson

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

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−8y=−4525x−1y=−9\begin{cases} 7x-8y = \frac{-45}{2} \\ 5x-1y = -9 \end{cases}
  2. 2. Steps, part 1
    1. {7x−8y=−4525x−y=−9\begin{cases} 7x - 8y = -\frac{45}{2} \\ 5x - y = -9 \end{cases}
      Write every equation in standard form
    2. [7−8−4525−1−9]\begin{bmatrix}7 & -8 & -\frac{45}{2} \\ 5 & -1 & -9\end{bmatrix}
      Build the augmented matrix [A | b]
    3. Gauss–Jordan elimination
    4. R1→17R1: [1−87−45145−1−9]R_{1} \to \tfrac{1}{7} R_{1}:\ \begin{bmatrix}1 & -\frac{8}{7} & -\frac{45}{14} \\ 5 & -1 & -9\end{bmatrix}
      Divide R1 by 7
  3. 3. Steps, part 2
    1. R2→R2−5R1: [1−87−451403379914]R_{2} \to R_{2} - 5R_{1}:\ \begin{bmatrix}1 & -\frac{8}{7} & -\frac{45}{14} \\ 0 & \frac{33}{7} & \frac{99}{14}\end{bmatrix}
      Clear column 1 using the pivot in R1
    2. R2→1337R2: [1−87−45140132]R_{2} \to \tfrac{1}{\frac{33}{7}} R_{2}:\ \begin{bmatrix}1 & -\frac{8}{7} & -\frac{45}{14} \\ 0 & 1 & \frac{3}{2}\end{bmatrix}
      Divide R2 by 33/7
    3. R1→R1+87R2: [10−320132]R_{1} \to R_{1} + \frac{8}{7} R_{2}:\ \begin{bmatrix}1 & 0 & -\frac{3}{2} \\ 0 & 1 & \frac{3}{2}\end{bmatrix}
      Clear column 2 using the pivot in R2
    4. [10−320132]\begin{bmatrix}1 & 0 & -\frac{3}{2} \\ 0 & 1 & \frac{3}{2}\end{bmatrix}
      Reduced row-echelon form
  4. 4. Steps, part 3
    1. x=−32, y=32x = -\frac{3}{2},\ y = \frac{3}{2}
      Read off the unique solution
  5. 5. The answer
    x=−32, y=32x = -\frac{3}{2},\ y = \frac{3}{2}

    7x - 8y = -45/2; 5x - y = -9 ⇒ x = -3/2, y = 3/2

  6. 6. Check
    • Equation 1: 7x - 8y = -45/2 ✓
    • Equation 2: 5x - y = -9 ✓

Transcript

The problem. Here is the problem: 7x minus 8y equals negative 45 over 2; 5x minus 1y equals negative 9. 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 33 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 negative 3 over 2, y equals 3 over 2. The system has exactly one solution; substituting it back satisfies every equation.

Check. Equation 1: 7x minus 8y equals negative 45 over 2, which checks out. Equation 2: 5x minus y equals negative 9, 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 - 8y = -45/2, 5x - 1y = -9 (Lesson)