Math Problem Solver

The system of equations has one solution.\[x = \frac{140}{27}\] \[y = \frac{38}{27}\]

Solve by substitution:

Solve \(\frac{6 x}{5} - 3 y = 2\) for \(x\). \[x = \frac{5 y}{2} + \frac{5}{3}\]

Substitute \(\frac{5 y}{2} + \frac{5}{3}\) for \(x\) in \(x + 2 y = 8\) and simplify. $$\begin{aligned}x + 2 y &= 8 \\ \left(\frac{5 y}{2} + \frac{5}{3}\right) + 2 y &= 8 \\ y &= \frac{38}{27} \end{aligned}$$

Use substitution of the numerical value of \(y\) to get the values of \(x\). $$\begin{aligned}x &= \frac{5 y}{2} + \frac{5}{3} \\ x &= \frac{5 \cdot \frac{38}{27}}{2} + \frac{5}{3} \\ x &= \frac{140}{27}\end{aligned}$$

Solve by Gauss-Jordan Elimination:

Begin by writing the augmented matrix of the system of equations. $$\begin{bmatrix}\frac{6}{5} & -3 &\bigm |& 2\\1 & 2 &\bigm |& 8\end{bmatrix}$$

Then use a series of elementary row operations to convert the matrix into reduced-row echelon form. The three elementary row operations are:

  1. Swap the positions of any two rows.

  2. Multiply any row by a nonzero scalar.

  3. Multiply a row by a nonzero scalar and add it to any other row.

First, switch the rows in the matrix such that the row with the leftmost non-zero entry with the greatest magnitude is at the top.

$$\begin{bmatrix}\frac{6}{5} & -3 &\bigm |& 2\\1 & 2 &\bigm |& 8\end{bmatrix}$$

Multiply row \(1\) by scalar \(\frac{5}{6}\) to make the leading term \(1\).

$$\begin{bmatrix}1 & - \frac{5}{2} &\bigm |& \frac{5}{3}\\1 & 2 &\bigm |& 8\end{bmatrix}$$

Multiply row \(1\) by scalar \(-1\) and add it to row \(2\).

$$\begin{bmatrix}1 & - \frac{5}{2} &\bigm |& \frac{5}{3}\\0 & \frac{9}{2} &\bigm |& \frac{19}{3}\end{bmatrix}$$

Multiply row \(2\) by scalar \(\frac{2}{9}\) to make the leading term \(1\).

$$\begin{bmatrix}1 & - \frac{5}{2} &\bigm |& \frac{5}{3}\\0 & 1 &\bigm |& \frac{38}{27}\end{bmatrix}$$

Multiply row \(2\) by scalar \(\frac{5}{2}\) and add it to row \(1\).

$$\begin{bmatrix}1 & 0 &\bigm |& \frac{140}{27}\\0 & 1 &\bigm |& \frac{38}{27}\end{bmatrix}$$

Once the matrix is in reduced-row echelon form, convert the matrix back into linear equations to find the solution. $$\begin{aligned}1 \cdot x+ 0 \cdot y = \frac{140}{27} \\ x = \frac{140}{27}\end{aligned}$$$$\begin{aligned}0 \cdot x+ 1 \cdot y = \frac{38}{27} \\ y = \frac{38}{27}\end{aligned}$$

Solve by matrix inversion:

In cases where the coefficient matrix of the system of equations is invertible, we can use the inverse to solve the system. Use this method with care as matrix inversion can be numerically unstable for ill-conditioned matrices.

Express the linear equations in the form \(A \times X = B\) where \(A\) is the coefficient matrix, \(X\) is the matrix of unknowns, and \(B\) is the constant matrix.$$\left[\begin{matrix}\frac{6}{5} & -3\\1 & 2\end{matrix}\right] \times \left[\begin{matrix}x\\y\end{matrix}\right] = \left[\begin{matrix}2\\8\end{matrix}\right]$$

The product of \(A\) and its inverse \(A^{-1}\) is the identity matrix. Any matrix multiplied by the identity matrix remains unchanged, so this yields the matrix of unknowns on the left hand side of the equation, and the solution matrix on the right. $$\begin{aligned} A \times X &= B\\ A^{-1} \times A \times X &= A^{-1} \times B \\ I \times X &= A^{-1} \times B \\ X &= A^{-1} \times B \end{aligned}$$

Using a computer algebra system, calculate \(A^{-1}\). $$\left[\begin{matrix}\frac{10}{27} & \frac{5}{9}\\- \frac{5}{27} & \frac{2}{9}\end{matrix}\right]$$

Multiply both sides of the equation by the inverse. $$\left[\begin{matrix}\frac{10}{27} & \frac{5}{9}\\- \frac{5}{27} & \frac{2}{9}\end{matrix}\right] \times \left[\begin{matrix}\frac{6}{5} & -3\\1 & 2\end{matrix}\right] \times \left[\begin{matrix}x\\y\end{matrix}\right] = \left[\begin{matrix}\frac{10}{27} & \frac{5}{9}\\- \frac{5}{27} & \frac{2}{9}\end{matrix}\right] \times \left[\begin{matrix}2\\8\end{matrix}\right]$$ $$\left[\begin{matrix}x\\y\end{matrix}\right] = \left[\begin{matrix}\frac{140}{27}\\\frac{38}{27}\end{matrix}\right]$$

\[x = \frac{140}{27}\]\[y = \frac{38}{27}\]