How do you solve the system \[y = 2x\] and \[x + 3y = - 14\] using substitution?

Hint: Here we are given with two equations with x and y as variables. Now we have to find the values of these variables using the method of substitution. In this we will replace one of the variables with the other given such that the variable so replaced is expressed in the form of the other. Here y is expressed in the form of x so we will replace y from other equations with x. On solving it we will get the values of the system.

Complete step-by-step solution:
Let \[y = 2x\]…………….equation I
And \[x + 3y = - 14\]………….equation II .
Now we will put the value of y in terms of x in the equation II
\[\Rightarrow x + 3\left( {2x} \right) = - 14\]
On multiplying we get,
\[\Rightarrow x + 6x = - 14\]
On adding like terms we get,
\[\Rightarrow 7x = - 14\]
Now we will divide 14 by 7,
\[\Rightarrow x = - \dfrac{{14}}{7}\]
And we get,
\[\Rightarrow x = - 2\]
This is the value of x .now putting this value of x in equation I we get,
\[\Rightarrow y = 2\left( { - 2} \right)\]
On multiplying we get,
\[\Rightarrow y = - 4\]
This is the value of y.

Thus using substitution we get the x and y values as \[x = - 2\] and \[y = - 4\].

Note: Note that here we can replace x from equation II in the form of y by taking the value from equation I. it is not necessary that we should replace as mentioned above only. Both methods are correct. But choose that method having lesser calculations.

Alternate method:
From equation I we get, \[x = \dfrac{y}{2}\]
Putting this value in equation II we get
\[\Rightarrow \dfrac{y}{2} + 3y = - 14\]
Taking LCM on LHS we get,
\[\Rightarrow \dfrac{{y + 6y}}{2} = - 14\]
On adding the terms in numerator and transposing 2 on other side,
\[\Rightarrow 7y = - 14 \times 2\]
On multiplying we get,
\[\Rightarrow 7y = - 28\]
Now on dividing by 7 we get,
\[\Rightarrow y = - \dfrac{{28}}{7} = - 4\]
This is just a cross verification that the method we used above is correct.