find the equation of the line which is perpendicular to the line 3x+2y-5=0 and which passes through the point of intersection of the lines 4

We can do this as follows: (Note: In copying the LaTeX from Tekmaker and pasting it here the third equation has been renumbered to (1)!!!. The third equation below should be numbered (3), of course.)

$$Let's first rewrite the two given equations as: \begin{equation} y1 = -\frac{3}{2}x+\frac{5}{2} \end{equation} \begin{equation} y2 = -4x+5 \end{equation}$$

$$and the line perpendicular to the first of the above as: \begin{equation} y3 = mx+c \end{equation} where $m$ is its gradient and $c$ is its intersection with the y-axis.\\\\$$

$$The gradient, $m$, is given by -1/the gradient of the line to which it is perpendicular, so $m=\frac{2}{3}$\\\\ To find the point of intersection of lines (1) and (2) we need to equate them:$$ -\frac{3}{2}x+\frac{5}{2}=-4x+5 $$and solve for $x$. This is straightforward and results in $x=1$. Plugging this back in either equation (1) or equation (2) we find that the value of y at the intersection point is also $y=1$\\\\ Now we put the known values of $x$ and $y$ at the intersection point, together with our value for $m$ into equation (3):$$ 1=\frac{2}{3}*1+c$$

$$from which we get $c=\frac{1}{3}$; so, using equation (3), we get the equation of the perpendicular line as:$$ y3=\frac{2}{3}x+\frac{1}{3}

Let's plot a graph to see if the result looks sensible: