Anybody know how to solve this problem?

Find the equation of the straight line which is parallel to the line whose equation is 3x + 4y = 0, and which passes through the point of intersection of the lines x - 2y = a and x + 3y = 2a. Express the answer in the form Ax + By = C

I. The slope of the new straight line:

$$\small{\text{$ 3x + 4y = 0 \rm{~~or~~}y = -\frac34 \cdot x \rm{~~so~the ~ slope ~is ~~} -\frac34 $.}} \small{\text{$ \rm{~The ~ slope ~of~a ~parallel ~line ~is ~ also ~} -\frac34 $}}$$

II. intersection of the lines x - 2y = a and x + 3y = 2a:

$$\small{\text{$ \begin{array}{lrcl} (1): & x - 2y &=& a \\ (2): & x+3y &=& 2a\\ \hline (2)-(1): & x-x + 3y-(-2y) &=& 2a-a \\ & 3y+2y &=& a \\ & 5y &=& a \\ & \textcolor[rgb]{1,0,0}{y} & \textcolor[rgb]{1,0,0}{=}& \textcolor[rgb]{1,0,0}{\frac15 a}\\ \hline (1): & x - 2y &=& a \\ & x - 2y &=& a \\ & x &=& a + 2y \quad | \quad y = \frac15 a \\ & x &=& a + 2(\frac15 a) \\ & \textcolor[rgb]{1,0,0}{x} & \textcolor[rgb]{1,0,0}{=}& \textcolor[rgb]{1,0,0}{\frac75 a} \\ \end{array} $}}$$

The intersection is $$(~\frac75 a ~|~ \frac15 a~)$$

III. Find the equation of the straight line:

$$\small{\text{$ \begin{array}{rcl} y &=&mx+b \qquad | \qquad m = -\frac34 \\ y&=&-\frac34 \cdot x+b \end{array} $}}$$

find b:

$$\small{\text{$ \begin{array}{rcl} y_s&=&-\frac34 \cdot x_s+b \qquad | \qquad x_s = \frac75 a \rm{~~and~~} y_s = \frac15 a\\\\ \frac15 a &=&-\frac34 \cdot \frac75 a+b \\\\ b&=& \frac15 a+\frac34 \cdot \frac75 a \\\\ b&=& \frac{25}{20} a \\\\ \textcolor[rgb]{1,0,0}{b}&\textcolor[rgb]{1,0,0}{=}& \textcolor[rgb]{1,0,0}{\frac{5}{4} a} \\\\ \end{array} $}}$$

The equation of the straight line is:

$$\small{\text{$ \begin{array}{lrcl} &y &=& mx+b \qquad | \qquad m = -\frac34 \rm{~~and~~} b = \frac54 a\\\\ &\mathbf{y} &\mathbf{=}& \mathbf{-\frac34 \cdot x + \frac{5}{4} a}\\\\ \rm{or~~} \\\\ &\mathbf{\frac34 \cdot x } + 1\cdot \mathbf{y} &\mathbf{=}& \mathbf{ \frac{5}{4} a } \end{array} $}}$$

Parallel lines have the same slope, so let's find the slope of the line 3x + 4y  = 0:

     3x + 4y  =  0    --->     4y  =  -3x     --->     y  =  (-3/4)x

The slope is  -3/4.

The equation of any line with a slope of  -3/4  is  y - k  =  (-3/4)(x - h)  where (h, k) is some point.

Now, let's find the point of intersection:

     x - 2y  =  a     --->     x  =  2y + a

     x + 3y  =  2a  --->     x  =  -3y + 2a

Combining these equations:  2y + a  =  -3y + 2a     --->     5y  =  a     --->     y  =  (1/5)a

Since  x - 2y  =  a, substituting:  x -2(1/5)a  =  a     --->     x  =  (7/5)a

So, an equation is:  y - (1/5)a  =  (-3/4)[x - (7/5)a]

Some algebra will make the equation look prettier!