Simultaneous equation with fractional solutions. How do you get to find $x$ when $y$ is a fraction ? Anyone mind to explain it step by step for the clearest explanation.=)
$$-7x +2y = 2$$
$$14x + 3y = -5$$
Answer: $x=?, y=-1/7$ 
 A: Inserting $y=-\frac{1}{7}$ in the first equation
$-7x-2\cdot \frac{1}{7}=2$
Transforming 2 into $\frac{14}{7}$ to have a common denominator with $2\cdot \frac{1}{7}$
$-7x-\frac{2}{7}=\frac{14}{7} \quad \quad |+\frac{2}{7}$ 
Adding $\frac{2}{7}$ on both sides.
$-7x\underbrace{-\frac{2}{7}+\color{blue}{\frac{2}{7}}}_{=0}=\frac{14}{7} +\color{blue}{\frac{2}{7}}$ 
$-7x+0=\frac{14}{7} +\color{blue}{\frac{2}{7}}$ 
$-7x=\frac{14}{7}+\frac{2}{7}$
$-7x=\frac{16}{7}$
$-x=\frac{16}{7\cdot 7}$
$x=-\frac{16}{49}$
A: The correct solution to your system of equations is $y = -\frac{1}{7}$ and you want to find $x$. To do so, substitute this value of $y$ into the first equation to get $$-7x - 2\cdot \frac{1}{7} = 2 \iff -7x = \frac{16}{7} \iff x = -\frac{16}{49}$$ by re-arranging and solving for $x$.
Alternatively, you could substitute your $y$ value into the second equation to get $$14x - 3\cdot \frac{1}{7} = -5 \iff 14x = -\frac{32}{7} \iff x = -\frac{16}{49}$$ by re-arranging and solving for $x$.
So the solution to your system of equations is $x = -\frac{16}{49}, y = -\frac{1}{7}$. 
A: The answers of @calculus and @ZainPatel are perfect, but I would like to offer a slicker method, if you don’t like working with fractions.
Perform the same first step as they do, namely substitute your known value of $y$ into the first equation, getting
$$
-7x-2\cdot\frac17=2\,,
$$
and multiply both sides of the equation by $7$, thus cleaering of fractions:
$$-49x-2=14\,.$$
This gives $49x=-16$, the rest is clear.
