1. We are given the system of equations: $$4x + 19y = 25$$ $$7x + 14y = 30$$ We need to find the value of $x$. 2. To solve for $x$, we can use the method of elimination or substitution. Here, we use elimination. 3. Multiply the first equation by 7 and the second equation by 4 to align the coefficients of $x$: $$7(4x + 19y) = 7(25) \Rightarrow 28x + 133y = 175$$ $$4(7x + 14y) = 4(30) \Rightarrow 28x + 56y = 120$$ 4. Subtract the second new equation from the first new equation to eliminate $x$: $$\cancel{28x} + 133y - (\cancel{28x} + 56y) = 175 - 120$$ $$133y - 56y = 55$$ $$77y = 55$$ 5. Solve for $y$: $$y = \frac{55}{77} = \frac{5}{7}$$ 6. Substitute $y = \frac{5}{7}$ back into the first original equation to solve for $x$: $$4x + 19\left(\frac{5}{7}\right) = 25$$ $$4x + \frac{95}{7} = 25$$ 7. Multiply both sides by 7 to clear the denominator: $$7 \times 4x + 7 \times \frac{95}{7} = 7 \times 25$$ $$28x + 95 = 175$$ 8. Subtract 95 from both sides: $$28x = 175 - 95$$ $$28x = 80$$ 9. Solve for $x$: $$x = \frac{80}{28} = \frac{20}{7} \approx 2.86$$ **Final answer:** $x = \frac{20}{7} \approx 2.86$