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Systems of equationsAlgebraHigh schoolabout 1:09

Solve the system y = 3x - 5, 2x + y = 10: animated lesson

A narrated, step-by-step systems of equations video lesson with captions. Final answer: x = 3, y = 4.

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
    {y=3x−52x+y=10\begin{cases} y = 3x-5 \\ 2x+ y = 10 \end{cases}
  2. 2. Steps, part 1
    1. {−3x+y=−52x+y=10\begin{cases} -3x + y = -5 \\ 2x + y = 10 \end{cases}
      Write every equation in standard form
    2. [−31−52110]\begin{bmatrix}-3 & 1 & -5 \\ 2 & 1 & 10\end{bmatrix}
      Build the augmented matrix [A | b]
    3. Gauss–Jordan elimination
    4. R1→1−3R1: [1−13532110]R_{1} \to \tfrac{1}{-3} R_{1}:\ \begin{bmatrix}1 & -\frac{1}{3} & \frac{5}{3} \\ 2 & 1 & 10\end{bmatrix}
      Divide R1 by -3
  3. 3. Steps, part 2
    1. R2→R2−2R1: [1−1353053203]R_{2} \to R_{2} - 2R_{1}:\ \begin{bmatrix}1 & -\frac{1}{3} & \frac{5}{3} \\ 0 & \frac{5}{3} & \frac{20}{3}\end{bmatrix}
      Clear column 1 using the pivot in R1
    2. R2→153R2: [1−1353014]R_{2} \to \tfrac{1}{\frac{5}{3}} R_{2}:\ \begin{bmatrix}1 & -\frac{1}{3} & \frac{5}{3} \\ 0 & 1 & 4\end{bmatrix}
      Divide R2 by 5/3
    3. R1→R1+13R2: [103014]R_{1} \to R_{1} + \frac{1}{3} R_{2}:\ \begin{bmatrix}1 & 0 & 3 \\ 0 & 1 & 4\end{bmatrix}
      Clear column 2 using the pivot in R2
    4. [103014]\begin{bmatrix}1 & 0 & 3 \\ 0 & 1 & 4\end{bmatrix}
      Reduced row-echelon form
  4. 4. Steps, part 3
    1. x=3, y=4x = 3,\ y = 4
      Read off the unique solution
  5. 5. The answer
    x=3, y=4x = 3,\ y = 4

    -3x + y = -5; 2x + y = 10 ⇒ x = 3, y = 4

  6. 6. Check
    • Equation 1: -3x + y = -5 ✓
    • Equation 2: 2x + y = 10 ✓

Transcript

The problem. Here is the problem: y equals 3x minus 5; 2x plus y equals 10. 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 3.

Steps, part 2. Clear column 1 using the pivot in R1. Divide R2 by 5 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 3, y equals 4. The system has exactly one solution; substituting it back satisfies every equation.

Check. Equation 1: negative 3x plus y equals negative 5, which checks out. Equation 2: 2x plus y equals 10, 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 y = 3x - 5, 2x + y = 10: Animated Lesson