find the equation of the line through the intersection of the lines 2x+3y-4 = 0 and x-5y = 7 that has its intercept equal to -4.

2x + 3y-4 = 0 and x-5y = 7
Here first we have to find the intersection point of these two lines.
So on solving both equation simultaneously we have x = 41/13 and y = -10/13

Let the equation of straight line in intercept form is x/a + y/b = 1
Here a is x-intercept and b = y-intercept and the point x = 41/13 and y = -10/13 will satisfy this line
As x-intercept is a = -4

$$\begin{gathered} \frac{x}{-4}+\frac{y}{b} = 1 \\ So puting the point is this equation: \\ \frac{41}{13\times -4}+\frac{-10}{13b} =1 \\ So b = \frac{-40}{93} \\ So the general equation will be \\ \frac{x}{-4}+\frac{y}{{\displaystyle \frac{-40}{93}}}=1 \\ Or \frac{x}{-4}+\frac{-93y}{40}=1 \left(ans\right) \end{gathered}$$