Find the equation of the line through the intersection of the lines $2x+3y4=0$ and $x-5y=7$that has its x-intercept equal to $4$.

Solution: Equation of the required line can be given as
$2x+3y-4+k\left(x-5y-7\right)=0$
$\left(2+k\right)x+\left(3-5k\right)y-\left(4+7k\right)=0\to \left(1\right)$
As, x- intercept is given, we can put $y=0$
$\therefore \left(2+k\right)x=4+7k$
$\Rightarrow \left(2+k\right)\left(-4\right)=4+7k$
$\Rightarrow -8-4k=4+7k\Rightarrow k=-\frac{12}{11}$
So, putting value of $k$ in (1), we get the required equation.
$\left(2-\frac{12}{11}\right)x+\left(3+\frac{60}{11}\right)-\left(4-\frac{84}{11}\right)=0$
$\Rightarrow 10x+93y+40=0$, which is the required equation.