Question

How do you solve the system \[4x=-11+y\] and \[y=-6x-9\]?

Answer 1

Hint: Assume the given equations as equation (1) and (2) respectively. Now, substitute the value of ‘y’ from equation (1), given in terms of ‘x’, in equation (1) and solve the equation for the value of x. Once the value of x is found, substitute it in any of the two equations to get the value of y.

Complete step by step solution:
Here, we have been provided with two equations: \[4x=-11+y\] and \[y=-6x-9\] and we are asked to solve this system, that means we have to find the values of the variables x and y.
Let us assume the two given equations as equation (1) and equation (2) respectively, so we have,
\[\Rightarrow 4x=-11+y\] - (1)
\[\Rightarrow y=-6x-9\] - (2)
Now, here we are going to use the method of substitution to solve the system. It states that we have to select one of the two equations and write one of the variables in terms of the other. Now, we have to substitute this obtained value in the non – selected equation to convert it into an equation containing only one variable that can be solved easily.
Here, let us select equation (2) in which the value of y is already given in terms of x, so substituting the value of y from equation (2) in equation (1) we get,
$\begin{align}
  & \Rightarrow 4x=-11+\left( -6x-9 \right) \\
 & \Rightarrow 4x=-11-6x-9 \\
 & \Rightarrow 4x+6x=-20 \\
 & \Rightarrow 10x=-20 \\
\end{align}$
Dividing both the sides with 10 we get,
\[\therefore x=-2\]
So we have found the value of x therefore substituting this value in equation (2) we get,
\[\begin{align}
  & \Rightarrow y=-6\left( -2 \right)-9 \\
 & \Rightarrow y=12-9 \\
 & \therefore y=3 \\
\end{align}\]
Hence, the solution of the given system of equations can be given as (x, y) = (-2, 3).

Note: Here you can also select equation (1) in place of equation (2) at the initial step of the solution and then proceed. We will get the same answer. There are two other methods also that can be used to solve this system, they are: the elimination method and the cross multiplication method. The elimination method will also be an easy method here. However, we may not prefer the cross multiplication method unless stated to use because of its lengthy formula which must be remembered carefully.