Let $a$,$b$ be the roots of $x^2-6cx-7d=0$, and let $c$, $d$ be the roots of $x^2-6ax-7b=0$. Find $b+d$. 
Let $a$,$b$ be the roots of the equation $x^{2}-6 c x-7 d=0$ and let $c$,$d$ be the roots of the equation $x^{2}-6ax-7 b=0$. $a,b,c,d$ are distinct. Find value of $b+d$.

Using the relation between roots and coefficients of an equation, I got $a+b+c+d=6(a+c)$, and hence getting $b+d= 5(a+c)$, but I don't know what to do next.
Also, $abcd= 49bd$, but I don't know how to use it here.
 A: Alternative approach:
From the original equations you have that:

*

*$[E_1:] ~ a + b = 6c \implies b = 6c - a.$

*$[E_2:] ~ c + d = 6a \implies d = 6a - c.$
Using the two results above, you also have, from the original equations that:

*

*$[E_3:] ~ 6ac - a^2 = ab = -7d = -7(6a - c) = 7c - 42a.$

*$[E_4:] ~ 6ac - c^2 = cd = -7b = -7(6c - a) = 7a - 42c.$
Subtracting $E_3 - E_4$ above gives
$(c-a)(c+a) = c^2 - a^2 = 7(c-a) + 42(c-a) = 49(c-a).$
This implies that $c+a = 49.$
Then, adding $E_1 + E_2$ gives
$(a + c) + (b + d) = 6(a + c) \implies (b + d) = 5(a + c) = 5 \times 49.$
A: I like to work with a bit of generality whenever possible, so take the equations to be
$$x^2+pcx+qd=0 \qquad x^2+pax+qb=0$$
By Vieta's formula for the linear coefficients, we have a solvable linear system in $a$ and $c$:
$$
\left.\begin{align}
a+b=-pc \\ c+d=-pa
\end{align}\right\}\quad\to\quad a=\frac{b-dp}{p^2-1}\qquad c = \frac{d-bp}{p^2-1}\tag1$$
Then, by Vieta for the constant terms,
$$qd = ab=\frac{b-dp}{p^2-1}\,b \qquad\qquad qb=cd=\frac{d-bp}{p^2-1}\,d \tag2$$
Subtracting, and recalling the requirement that $b\neq d$,
$$q(d-b) = \frac{b^2-d^2}{p^2-1}=-\frac{(d-b) (b + d)}{p^2-1} \quad\to\quad b+d=-q(p^2-1) \tag3$$
For $p=-6$ and $q=-7$, the sum becomes $245$, in agreement with @user2661923's answer.
