The problem asks us to find the intersections of the given lines with the Cartesian axes (x-axis and y-axis). The lines are: 25) $x + 2y - 4 = 0$ 26) $3x - 4y - 8 = 0$ 27) $x - y + 1 = 0$

1. Problem Description

The problem asks us to find the intersections of the given lines with the Cartesian axes (x-axis and y-axis). The lines are:

25) $x + 2y - 4 = 0$

26) $3x - 4y - 8 = 0$

27) $x - y + 1 = 0$

2. Solution Steps

To find the intersection with the x-axis, we set $y = 0$ in the equation of the line and solve for $x$.

To find the intersection with the y-axis, we set $x = 0$ in the equation of the line and solve for $y$.

25) $x + 2y - 4 = 0$

x-axis intersection ($y=0$): $x + 2(0) - 4 = 0 \Rightarrow x = 4$. Intersection point: $(4, 0)$

y-axis intersection ($x=0$): $0 + 2y - 4 = 0 \Rightarrow 2y = 4 \Rightarrow y = 2$. Intersection point: $(0, 2)$

26) $3x - 4y - 8 = 0$

x-axis intersection ($y=0$): $3x - 4(0) - 8 = 0 \Rightarrow 3x = 8 \Rightarrow x = \frac{8}{3}$. Intersection point: $(\frac{8}{3}, 0)$

y-axis intersection ($x=0$): $3(0) - 4y - 8 = 0 \Rightarrow -4y = 8 \Rightarrow y = -2$. Intersection point: $(0, -2)$

27) $x - y + 1 = 0$

x-axis intersection ($y=0$): $x - 0 + 1 = 0 \Rightarrow x = -1$. Intersection point: $(-1, 0)$

y-axis intersection ($x=0$): $0 - y + 1 = 0 \Rightarrow y = 1$. Intersection point: $(0, 1)$

3. Final Answer

25) x-axis: $(4, 0)$, y-axis: $(0, 2)$

26) x-axis: $(\frac{8}{3}, 0)$, y-axis: $(0, -2)$

27) x-axis: $(-1, 0)$, y-axis: $(0, 1)$