How do you solve the system $x-4y=2$ and $x-2y=4$ using substitution?

Hint: We are given two equations as $x-4y=2$ and $x-2y=4$. We will first learn the type of equation it is and then we will learn about the method that we can use to solve such problems. We are asked to use the substitution method, so we will learn the substitution method and then use it to solve our problem. We will substitute the value of $y$ and solve for $x$.

Complete step by step solution:
We are given the equations as $x-4y=2$ and $x-2y=4$. We can see from the equations that it has the highest power as 1 only. It means that they are linear equations. We are asked to find the solution and solution means those values which when inserted in the equation, satisfy the equation. So, to find that we have to look for those values of $x$ and $y$ that satisfies the given equations, $x-4y=2$ and $x-2y=4$. Now as we have linear equations in 2 variables, we can solve the problem in different ways. The various different methods are:
1. Substitution.
2. Elimination.
3. Cross multiplication.
4. Graphing.
Here we will use the substitution method to solve the problem. In this, we substitute the value of one variable in terms of the other variable using one equation and then solve for that variable. Now, we have $x-4y=2$ and $x-2y=4$. We will find the value of $y$ from the equation, $x-4y=2$. So, we have,
$\begin{align}
  & x-4y=2 \\
 & \Rightarrow x-2=4y \\
\end{align}$
Simplifying we get,
$\Rightarrow \dfrac{x-2}{4}=y$
Now, we will use this value of $y$ and substitute it in the other equation $x-2y=4$. So, putting $y=\dfrac{x-2}{4}$, we get,
$\Rightarrow x-2\left( \dfrac{x-2}{4} \right)=4$
Simplifying, we get,
$\Rightarrow x-\dfrac{\left( x-2 \right)}{2}=4$
Now, taking the LCM and simplifying, we get,
$\Rightarrow \dfrac{2x-x+2}{2}=4$
Cross multiplying, we get,
$\begin{align}
  & \Rightarrow x+2=8 \\
 & \Rightarrow x=8-2 \\
 & \Rightarrow x=6 \\
\end{align}$
So, we get the value of $x=6$. We will put the value of $x=6$ in $\dfrac{x-2}{4}=y$ to get the value of $y$. So, putting the value of $x=6$ in $\dfrac{x-2}{4}=y$, we get,
$\Rightarrow y=\dfrac{6-2}{4}=\dfrac{4}{4}=1$
So, $y=1$. So, we get the solution as $x=6$ and $y=1$.

Note: We can check the answer to be double sure about the solution. If $x=6$ and $y=1$ satisfies both the equation, then it is the solution of our problem. So, we will put $x=6$ and $y=1$ in the equation $x-4y=2$. So, we get,
$\begin{align}
  & 6-4\times 1=2 \\
 & \Rightarrow 6-4=2 \\
 & \Rightarrow 2=2 \\
\end{align}$
This is true, so it satisfies the solution. Now, we will similarly put $x=6$ and $y=1$ in the equation $x-2y=4$. So, we get,
$\begin{align}
  & 6-2\times 1=4 \\
 & \Rightarrow 6-2=4 \\
 & \Rightarrow 4=4 \\
\end{align}$
So, this is also true and it satisfies this equation. Hence our solution is correct.