Solve the following system of linear equations: 3x − 4y = 5 and 9x + y = 4.

$$3x - 4y = 5 \quad \text{(1)}$$ $$9x + y = 4 \quad \text{(2)}$$

Multiply equation (2) by 4 to make the coefficients of $y$ opposites.

$$4 \times (9x + y) = 4 \times 4$$ $$36x + 4y = 16 \quad \text{(3)}$$

$$(3x - 4y) + (36x + 4y) = 5 + 16$$ $$3x + 36x - 4y + 4y = 21$$ $$39x = 21$$

$$x = \frac{21}{39}$$ $$x = \frac{7}{13}$$

$$9x + y = 4$$ $$9\left(\frac{7}{13}\right) + y = 4$$ $$\frac{63}{13} + y = 4$$

$$y = 4 - \frac{63}{13}$$ $$y = \frac{4 \times 13}{13} - \frac{63}{13}$$ $$y = \frac{52}{13} - \frac{63}{13}$$ $$y = \frac{52 - 63}{13}$$ $$y = -\frac{11}{13}$$

The solution to the system of equations is $x = \frac{7}{13}$ and $y = -\frac{11}{13}$.

The solution set is $\left\{\left(\frac{7}{13}, -\frac{11}{13}\right)\right\}$.