Math Problem Solver

The system of linear equations has exactly one solution.

The substitutions are \( a = 2 \), \( b = \frac{11}{7} \), \( c = \frac{5}{7} \).

\[a = 2\]

\[b = \frac{11}{7}\]

\[c = \frac{5}{7}\]

Solve \(5 a - 4 b - c = 3\) for \(a\). \[a = \frac{4 b}{5} + \frac{c}{5} + \frac{3}{5}\]

Substitute \(\frac{4 b}{5} + \frac{c}{5} + \frac{3}{5}\) for \(a\) in each of the remaining equations and simplify. $$\begin{aligned}4 a - 6 b + 2 c &= 0 \\ 4 \left(\frac{4 b}{5} + \frac{c}{5} + \frac{3}{5}\right) - 6 b + 2 c &= 0 \\ b - c &= \frac{6}{7} \end{aligned}$$$$\begin{aligned}a + b - 5 c &= 0 \\ \left(\frac{4 b}{5} + \frac{c}{5} + \frac{3}{5}\right) + b - 5 c &= 0 \\ 3 b - 8 c &= -1 \end{aligned}$$

Solve \(b - c = \frac{6}{7}\) for \(b\). \[b = c + \frac{6}{7}\]

Substitute \(c + \frac{6}{7}\) for \(b\) in \(3 b - 8 c = -1\) and simplify. $$\begin{aligned}3 b - 8 c &= -1 \\ 3 \left(c + \frac{6}{7}\right) - 8 c &= -1 \\ c &= \frac{5}{7} \end{aligned}$$

Use substitution of the numerical value of \(c\) to get the values of \(a\) and \( b\). $$\begin{aligned}b &= c + \frac{6}{7} \\ b &= \frac{5}{7} + \fra\frac{5}{7}{6}{7} \\ b &= \frac{11}{7}\end{aligned}$$$$\begin{aligned}a &= \frac{4 b}{5} + \frac{c}{5} + \frac{3}{5} \\ a &= \fra\frac{5}{7}{4 \\frac{5}{7}dot \fra\frac{5}{7}{11}{7}}{5} + \fra\frac{5}{7}{\frac{5}{7}}{5} + \fra\frac{5}{7}{3}{5} \\ a &= 2\end{aligned}$$