Solve a problem

Systems of equationsAlgebraHigh schoolabout 1:09

Solve the system x - 5y = -2, -4x - 2y = -14: animated lesson

A narrated, step-by-step systems of equations video lesson with captions. Final answer: x = 3, y = 1.

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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. The problem
Solving an equation
$\begin{cases} x-5y = -2 \\ -4x-2y = -14 \end{cases}$
2. Steps, part 1
1. $\begin{cases} x - 5y = -2 \\ -4x - 2y = -14 \end{cases}$
  Write every equation in standard form
2. $\begin{bmatrix}1 & -5 & -2 \\ -4 & -2 & -14\end{bmatrix}$
  Build the augmented matrix [A | b]
3. Gauss–Jordan elimination
4. $R_{2} \to R_{2} + 4R_{1}:\ \begin{bmatrix}1 & -5 & -2 \\ 0 & -22 & -22\end{bmatrix}$
  Clear column 1 using the pivot in R1
3. Steps, part 2
1. $R_{2} \to \tfrac{1}{-22} R_{2}:\ \begin{bmatrix}1 & -5 & -2 \\ 0 & 1 & 1\end{bmatrix}$
  Divide R2 by -22
2. $R_{1} \to R_{1} + 5R_{2}:\ \begin{bmatrix}1 & 0 & 3 \\ 0 & 1 & 1\end{bmatrix}$
  Clear column 2 using the pivot in R2
3. $\begin{bmatrix}1 & 0 & 3 \\ 0 & 1 & 1\end{bmatrix}$
  Reduced row-echelon form
4. $x = 3,\ y = 1$
  Read off the unique solution
4. The answer
$x = 3,\ y = 1$
x - 5y = -2; -4x - 2y = -14 ⇒ x = 3, y = 1
5. Check
- Equation 1: x - 5y = -2 ✓
- Equation 2: -4x - 2y = -14 ✓

Transcript

The problem. Here is the problem: x minus 5y equals negative 2; negative 4x minus 2y equals negative 14. 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. Clear column 1 using the pivot in R1.

Steps, part 2. Divide R2 by negative 22. Clear column 2 using the pivot in R2. Reduced row-echelon form. Read off the unique solution.

The answer. So the answer is x equals 3, y equals 1. The system has exactly one solution; substituting it back satisfies every equation.

Check. Equation 1: x minus 5y equals negative 2, which checks out. Equation 2: negative 4x minus 2y equals negative 14, which checks out.

About systems of equations

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

Solve the system x - 5y = -2, -4x - 2y = -14 (Lesson)