Section 7.1 : Linear Systems with Two Variables
4. Use the Method of Elimination to find the solution to the following system or to determine if the system is inconsistent or dependent.
\[\begin{align*}6x - 5y & = 8\\ - 12x + 2y & = 0\end{align*}\]Show All Steps Hide All Steps
Start SolutionBefore we get started with the solution process for this system we need to make it clear that there is no “one correct solution path”. There are lots of solution paths that we can take to find the solution to this system. All are correct and all will end up with the same solution to the system (provided the work has been done correctly of course…).
Okay, let’s get started on the solution to this system.
The Method of Elimination tells us that we first need to multiply one or both of the equations by constants so that one of the variables has the same coefficient but with opposite signs and then add the two equations.
For this system if we multiply the first equation by 2 then the first equation will have an \(x\) coefficient of 12 while the second equation will have an \(x\) coefficient of -12. This is exactly what we need so we’ll do that and then add the resulting equations.
\[\begin{align*} 6x-5y & =8 & \underrightarrow{\times \,\,2} \hspace{0.5in} & \,12x-10y=16 \\ -12x+2y & =0 & \underrightarrow{\mbox{same}} \hspace{0.5in} & \underline{-12x+2y=0} \\ & & & \hspace{0.45in} -8y=16 \end{align*}\] Show Step 2We can now easily solve the result from the above step to see that \(y = - 2\).
Show Step 3Finally, we can plug the value of \(y\) we found in the previous step in either of the original equations and solve for \(x\). We’ll use the first equation for this.
\[\begin{align*}6x - 5\left( { - 2} \right) & = 8\\ 6x + 10 & = 8\\ 6x & = - 2\hspace{0.25in} \to \hspace{0.25in}x = - \frac{1}{3}\end{align*}\]The solution to the system is then : \(\require{bbox} \bbox[2pt,border:1px solid black]{{x = - \frac{1}{3},\,\,y = - 2}}\).