Help to solve and understand Simultaneous equations. Could anyone help me solve this simultaneous equation:
$$6x + 7y = 12.5\tag1$$
$$7x + 5y = 14\tag2$$
I have watched loads of videos on YouTube, and still don't understand how to do it. I am trying to learn it. And practice with different equations. But don't understand how to do it. So any help will be much appreciated. 
 A: A method that always works for such (linear) equations is to isolate one of the variables in one of the equation, say $y$ in $(1)$, and then insert that expression for $y,$ which is now written in terms of $x,$ into the other equation. Then you will have an equation with only one variable, $x,$ which you can then solve. 
Having done this, you can insert that value for $x$ into the expression for $y,$ where you had isolated $y.$ Now you have both the $x$ and $y$.
Try it out, and if it is still causing you trouble, leave me a comment below and I'll show the full solution.
Good luck!
A side-note: Often, however, there are simpler ways of doing things, as @ParclyTaxel points out in the comments above. What is being referred to is multiplying equation $(2)$ with seven and then subtracting five times equation $(1).$ This is done to achieve a common factor, allowing the cancellation of a variable, as such:
\begin{align}
7(2):\quad & 49x+35y=7\cdot 14\\
5(1):\quad & 30x+35y=5\cdot 12.5
\end{align}
Subtracting then gives 
$$49x+35y-30x-35y=7\cdot 14-5\cdot 12.5$$
simplifying to 
$$19x=35.5.$$
Shortcuts like this comes with experience. The way described above, however, is straight-forward application of an algorithm, and doesn't require any "ideas", so it always works. 
A: we can also write
$$x+\frac{7}{6}=\frac{25}{12}$$
$$x+\frac{5}{7}y=2$$
multiplying by $$-1$$ we have
$$-x-\frac{7}{3}y=-\frac{25}{12}$$
$$x+\frac{5}{7}y=2$$
adding both we obatain:
$$\frac{5}{7}y-\frac{7}{3}y=2-\frac{25}{12}$$
can you finish?
simplifying the last equation we obtain:
$$-\frac{34}{21}y=-\frac{1}{12}$$
$$y=\frac{7}{38}$$
and then you can compute $$x$$
A: Here is an another example, then you can apply it to your problem:
(i) $x+y=4$ 
(ii) $2x+y=7$
Solving (i) for $x$ gives us $x=4-y$. Now put this into (ii) and you get $2(4-y)+y=7$ and you can now easily solve it for $y$. You should get $y=1$ and finally $x=4-1=3$.
A: We have \begin{align}6x + 7y &= 12.5\tag{1}\\
7x + 5y &= 14\tag{2}\end{align}
One way to solve this, would be to rearrange equation $(1)$ for $x$:
\begin{align}6x + 7y &= 12.5\\
6x&=12.5-7y\\
x&=\frac{12.5-7y}{6}\end{align}
We can then put this into equation $(2)$ and solve for $y$:
\begin{align}7x + 5y &= 14\\
7\left(\frac{12.5-7y}{6}\right)+5y&=14\\
\frac{87.5-49y}{6}+5y&=14\\
87.5-49y+30y&=84\\
87.5-19y&=84\\
19y&=3.5\\
y&=\frac{3.5}{19}\\
&=\frac{7}{38}\end{align}
Finally we can put this into equation $(1)$ to find that 
\begin{align}6x + 7\left(\frac{7}{38}\right) &= 12.5\\
6x+\frac{49}{38}&=12.5\\
6x&=12.5-\frac{49}{38}\\
6x&=\frac{213}{19}\\
x&=\frac{71}{38}\end{align}
We can then check this using equation $(2)$:
\begin{align}7x + 5y &=7\left(\frac{71}{38}\right)+5\left(\frac{7}{38}\right)\\
&=\frac{497}{38}+\frac{35}{38}\\
&=\frac{532}{38}\\
&=14\end{align}
Therefore we can be confident that the answer is $x=\dfrac{71}{38}$ and $y=\dfrac{7}{38}$
