1. **State the problem:** Solve the system of equations: $$x^{2}+y^{2}-5x+2y+1=0$$ $$3x-4y+1=0$$ 2. **Rewrite the linear equation to express one variable:** From $$3x-4y+1=0$$, solve for $$y$$: $$3x-4y+1=0 \implies 4y=3x+1 \implies y=\frac{3x+1}{4}$$ 3. **Substitute $$y$$ into the circle equation:** $$x^{2}+\left(\frac{3x+1}{4}\right)^{2}-5x+2\left(\frac{3x+1}{4}\right)+1=0$$ 4. **Simplify the equation:** $$x^{2}+\frac{(3x+1)^{2}}{16}-5x+\frac{2(3x+1)}{4}+1=0$$ Multiply through by 16 to clear denominators: $$16x^{2}+(3x+1)^{2}-80x+8(3x+1)+16=0$$ 5. **Expand and simplify:** $$(3x+1)^{2} = 9x^{2} + 6x + 1$$ So: $$16x^{2} + 9x^{2} + 6x + 1 - 80x + 24x + 8 + 16 = 0$$ Combine like terms: $$25x^{2} + (6x - 80x + 24x) + (1 + 8 + 16) = 0$$ $$25x^{2} - 50x + 25 = 0$$ 6. **Divide entire equation by 25:** $$x^{2} - 2x + 1 = 0$$ 7. **Factor the quadratic:** $$(x - 1)^{2} = 0$$ 8. **Solve for $$x$$:** $$x = 1$$ 9. **Find corresponding $$y$$:** $$y = \frac{3(1) + 1}{4} = \frac{4}{4} = 1$$ 10. **Final solution:** $$\boxed{(x,y) = (1,1)}$$