Math Problem Solver

The system of linear equations has exactly one solution.

\[a = \frac{49}{160}\]

\[b = \frac{13}{320}\]

Solve \(3 a + 2 b = 1\) for \(a\). \[a = \frac{1}{3} - \frac{2 b}{3}\]

Substitute \(\frac{1}{3} - \frac{2 b}{3}\) for \(a\) in \(20 a + 24 b = \frac{71}{10}\) and simplify. $$\begin{aligned}20 a + 24 b &= \frac{71}{10} \\ 20 \left(\frac{1}{3} - \frac{2 b}{3}\right) + 24 b &= \fr\left(\frac{1}{3} - \frac{2 b}{3}\right)c{71}{10} \\ b &= \frac{13}{320} \end{aligned}$$

Use substitution of the numerical value of \(b\) to get the values of \(a\). $$\begin{aligned}a &= \frac{1}{3} - \frac{2 b}{3} \\ a &= - \frac{2 \cdot 13}{3 \cdot 320} + 1 \cdot \frac{1}{3} \\ a &= \frac{49}{160}\end{aligned}$$