Question

How do you solve this system of linear equations using the elimination method: \[3x-y=-1,3x+3y=27\] ?

Answer 1

Hint: For such types of problems, we need to have a clear-cut idea about linear equations and simultaneous equations. In problems like these we need to eliminate one of the two variables by rearranging the equations or multiplying a number in such a manner that either adding or subtracting them directly helps us to get rid of one variable. We first multiply both sides of the first equation by $3$ and then add this equation with the second equation. This gets rid of y. We find a value of x, which if we put in one of the equations gives the value of y.

Complete step by step solution:
Now we start off with the solution to the problem by writing the two equations as,
\[3x-y=-1\]---- Equation \[1\]
\[3x+3y=27\]----- Equation \[2\]
We now multiply the first equation with \[3\]on both the sides and it transforms to,
\[9x-3y=-3\] ---- Equation \[3\]
We now add the third equation and the second equation after which, we get,
\[\Rightarrow \left( 9x-3y \right)+\left( 3x+3y \right)=-3+27\]
On evaluating we get,
\[\begin{align}
  & \Rightarrow \left( 9x-3y \right)+\left( 3x+3y \right)=-3+27 \\
 & \Rightarrow 12x=24 \\
 & \Rightarrow x=\dfrac{24}{12} \\
 & \Rightarrow x=2 \\
\end{align}\]
Now, putting this value of \[x\] in equation \[1\] we find the value of \[y\] as,
\[\begin{align}
  & 3x-y=-1 \\
 & \Rightarrow 3\times 2-y=-1 \\
 & \Rightarrow -y=-1-6 \\
 & \Rightarrow -y=-7 \\
 & \Rightarrow y=7 \\
\end{align}\]

Thus, the solution to our problem is \[x=2\] and \[y=7\].

Note: For these types of problems, we need to remember the theory of linear and simultaneous equations chapters very well. The problem can also be solved using the graphical method, the substitution method or even the method of matrices. All of the three methods would have given the same result. Even in the elimination method, we can try another variant. We can eliminate x instead of y by directly subtracting the first equation from the second.