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Systems of equationsAlgebraHigh schoolabout 1:11

Solve the system 4x - y = -15 and 7x + y = -18: animated lesson

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

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−y=−157x+y=−18\begin{cases} 4x- y = -15 \\ 7x+ y = -18 \end{cases}
  2. 2. Steps, part 1
    1. {4x−y=−157x+y=−18\begin{cases} 4x - y = -15 \\ 7x + y = -18 \end{cases}
      Write every equation in standard form
    2. [4−1−1571−18]\begin{bmatrix}4 & -1 & -15 \\ 7 & 1 & -18\end{bmatrix}
      Build the augmented matrix [A | b]
    3. Gauss–Jordan elimination
    4. R1→14R1: [1−14−15471−18]R_{1} \to \tfrac{1}{4} R_{1}:\ \begin{bmatrix}1 & -\frac{1}{4} & -\frac{15}{4} \\ 7 & 1 & -18\end{bmatrix}
      Divide R1 by 4
  3. 3. Steps, part 2
    1. R2→R2−7R1: [1−14−1540114334]R_{2} \to R_{2} - 7R_{1}:\ \begin{bmatrix}1 & -\frac{1}{4} & -\frac{15}{4} \\ 0 & \frac{11}{4} & \frac{33}{4}\end{bmatrix}
      Clear column 1 using the pivot in R1
    2. R2→1114R2: [1−14−154013]R_{2} \to \tfrac{1}{\frac{11}{4}} R_{2}:\ \begin{bmatrix}1 & -\frac{1}{4} & -\frac{15}{4} \\ 0 & 1 & 3\end{bmatrix}
      Divide R2 by 11/4
    3. R1→R1+14R2: [10−3013]R_{1} \to R_{1} + \frac{1}{4} R_{2}:\ \begin{bmatrix}1 & 0 & -3 \\ 0 & 1 & 3\end{bmatrix}
      Clear column 2 using the pivot in R2
    4. [10−3013]\begin{bmatrix}1 & 0 & -3 \\ 0 & 1 & 3\end{bmatrix}
      Reduced row-echelon form
  4. 4. Steps, part 3
    1. x=−3, y=3x = -3,\ y = 3
      Read off the unique solution
  5. 5. The answer
    x=−3, y=3x = -3,\ y = 3

    4x - y = -15; 7x + y = -18 ⇒ x = -3, y = 3

  6. 6. Check
    • Equation 1: 4x - y = -15 ✓
    • Equation 2: 7x + y = -18 ✓

Transcript

The problem. Here is the problem: 4x minus y equals negative 15; 7x plus y equals negative 18. 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 4.

Steps, part 2. Clear column 1 using the pivot in R1. Divide R2 by 11 over 4. 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, y equals 3. The system has exactly one solution; substituting it back satisfies every equation.

Check. Equation 1: 4x minus y equals negative 15, which checks out. Equation 2: 7x plus y equals negative 18, 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 4x - y = -15 and 7x + y = -18 (Lesson)