Math Problem Solver

The system of linear equations has exactly one solution.

The substitutions are \( x = \frac{15}{28} \), \( y = \frac{25}{28} \), \( z = \frac{11}{4} \).

\[x = \frac{15}{28}\]

\[y = \frac{25}{28}\]

\[z = \frac{11}{4}\]

Solve \(2 x + 3 y - z = 1\) for \(x\). \[x = - \frac{3 y}{2} + \frac{z}{2} + \frac{1}{2}\]

Substitute \(- \frac{3 y}{2} + \frac{z}{2} + \frac{1}{2}\) for \(x\) in each of the remaining equations and simplify. $$\begin{aligned}x - 2 y + 3 z &= 7 \\ \left(- \frac{3 y}{2} + \frac{z}{2} + \frac{1}{2}\right) - 2 y + 3 z &= 7 \\ y - z &= - \frac{13}{7} \end{aligned}$$$$\begin{aligned}3 x + y - 2 z &= -3 \\ 3 \left(- \frac{3 y}{2} + \frac{z}{2} + \frac{1}{2}\right) + y - 2 z &= -3 \\ 7 y + z &= 9 \end{aligned}$$

Solve \(y - z = - \frac{13}{7}\) for \(y\). \[y = z - \frac{13}{7}\]

Substitute \(z - \frac{13}{7}\) for \(y\) in \(7 y + z = 9\) and simplify. $$\begin{aligned}7 y + z &= 9 \\ 7 \left(z - \frac{13}{7}\right) + z &= 9 \\ z &= \frac{11}{4} \end{aligned}$$

Use substitution of the numerical value of \(z\) to get the values of \(x\) and \( y\). $$\begin{aligned}y &= z - \frac{13}{7} \\ y &= \frac{11}{4} - \frac{13}{7} \\ y &= \frac{25}{28}\end{aligned}$$$$\begin{aligned}x &= - \frac{3 y}{2} + \frac{z}{2} + \frac{1}{2} \\ x &= - \frac{3 \cdot \frac{25}{28}}{2} + \frac{\frac{11}{4}}{2} + \frac{1}{2} \\ x &= \frac{15}{28}\end{aligned}$$