An equation of a straight line passing through the inter-section of the straight lines $3x – 4y + 1 = 0$ and $5x + y – 1 = 0$ and making non-zero, equal intercepts on the axes is

Equation of any line through the point of intersection of the given lines is 

$(3x-4y+1)+k(5x+y-1)=0$

or $(3+5k)x+(k-4)y+1-k=0$

or $\frac{x}{(k-1)/(3+5k)}+\frac{y}{(k-1)/(k-4)}=1$

Since $x$-intercept = $y$-intercept

$\Rightarrow \frac{k-1}{3+5k}=\frac{k-1}{k-4}\Rightarrow (k-1)(3+5k-k+4)=0$

$\Rightarrow k=1\text{ or }k=-7/4$

For $k$ = 1, (1) becomes 8x – 3y = 0 which makes zero intercepts on the axes.

$\therefore k=-7/4\Rightarrow$ The required equation is

$\begin{array}{l}4(3x-4y+1)-7(5x+y-1)=0\\ \Rightarrow 23x+23y=11.\end{array}$