Algebra - Linear Systems with Two Variables

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Section 7.1 : Linear Systems with Two Variables

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2. Use the Method of Substitution to find the solution to the following system or to determine if the system is inconsistent or dependent.

\[\begin{align*}7x - 8y & = - 12\\ - 4x + 2y & = 3\end{align*}\]

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Start Solution

Before 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 Substitution tells us that we first need to solve one of the equations for one of the variables. The equation we solve and the variable we solve for technically doesn’t matter as noted above.

However, there is often one equation/variable combination that is “easier” than the others. In this case we can solve the second equation for \(y\) without a lot of extra work so let’s do that.

\[\begin{align*} - 4x + 2y & = 3\\ 2y & = 4x + 3\hspace{0.25in} \Rightarrow \hspace{0.25in}y = 2x + \frac{3}{2}\end{align*}\]

Note that you will often get fractions showing up at this step and there isn’t going to be a whole lot that you can do about it so don’t worry when they show up!

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We now take the equation for \(y\) we found above and substitute this into the other equation (the first equation in this case). Doing this gives,

\[\begin{align*}7x - 8y & = - 12\\ 7x - 8\left( {2x + \frac{3}{2}} \right) & = - 12\end{align*}\] Show Step 3

We can now solve the equation we found in the previous step for \(x\). Doing this gives,

\[\begin{align*}7x - 8\left( {2x + \frac{3}{2}} \right) & = - 12\\ 7x - 16x - 12 & = - 12\\ - 9x & = 0\hspace{0.25in} \to \hspace{0.25in}x = 0\end{align*}\]

Do not get excited about the zero here! They will be answers occasionally.

Show Step 4

Finally, we can plug the value of \(x\) we found in the previous step into the equation for \(y\) we found in the first step. This gives,

\[y = 2\left( 0 \right) + \frac{3}{2} = \frac{3}{2}\]

The solution to the system is then : \(\require{bbox} \bbox[2pt,border:1px solid black]{{x = 0,\,\,y = \frac{3}{2}}}\).