Solve $\frac3x - \frac4y = 1$ and $\frac7x + \frac2y = \frac{11}{12}$ How can we solve the following simultaneous equations:
$$\frac3x - \frac4y = 1$$
$$\frac7x + \frac2y = \frac{11}{12}$$
 A: Take $\frac{1}{x}$ as $w$ and $\frac{1}{y}$ as $v$ . Now you will get two linear equations . Solve the two equations for $w$ and $v$ and the subsequently put $w=1/x$ and $v=1/y$ . 
A: $\frac{3}{x}-\frac{4}{y}=1$ ---(1)
$\frac{7}{x}+\frac{2}{y}=\frac{11}{12}$ ---(2)
2(2): $\frac{14}{x}+\frac{4}{y}=\frac{11}{6}$ ---(3)
(1)+(3): $\frac{17}{x}=\frac{17}{6}$
Hence $x=6$, and $y=-8$.
A: Hint:
Solve one for one of the variables, substitute back into the second and solve for other variable.
From the first equation, we have:
$$x = \dfrac{3y}{y+4}$$
A: One way to solve two equations like this is to add a multiple of one to the other.  For example, Take $2$ times the second equation and add it to the first.  Explicitly,
$$(\frac{3}{x}-\frac{4}{y})+2(\frac{7}{x}+\frac{2}{y})=1+2\left(\frac{11}{12}\right)  $$
This gives us that,
$$\frac{17}{x}=\frac{17}{6} $$
Multiply both sides, by $x$ and then 6 to get,
$$(6)(17)=17x $$
so,
$$x=6 $$
Once we have what $x$ is equal to we can plug that back into one of our equations.  Let's do that with the first equation.
$$ \frac{1}{2}-\frac{4}{y}=1$$
Solving this gets us that $y=-8$.  You can check your answer by making sure this $x,y$ pair satisfy both equations at once.  Hope that helps!
