# Appendix A: Math review

## A.4: Solving systems of two equations

Another situation that comes up in physics is where something is described by a system of multiple equations and variables. As long as there are the same number of equations as unknowns, you can find a solution for all unknowns. For the physics covered in this book, we generally only encounter scenarios with two equations and two unknowns. As with quadratic equations, there are multiple methods of solving these systems, but I’ll just present the straightforward, brute force approach that will apply in all cases. I leave the more nuanced (though often more efficient) methods to a math class.

To solve a system of two equations and two unknowns, follow this process:

1. Solve one of the equations for one of the unknowns.
2. Substitute this into the other equation.
3. You will now have an equation with only one unknown to solve for. Solve for this unknown.
4. Now that you have a solution for one unknown, return to the first equation to find the remaining unknown.

#### Example A.4

Solve the system

$\begin{Bmatrix} T – xN &= & ma \\ N – mg &= & 0 \end{Bmatrix}$

for $$N$$ and $$T$$ in terms of all other variables.

There are two equations and two unknown variables you are tasked with solving for. Since $$N$$ shows up by itself in the second equation, we’ll start there:

\begin{align*} N – mg &= 0 \\ N &= mg \end{align*}

Now substitute this into the first equation:

\begin{align*} T – xN &= ma \\ T – x(mg) &= ma \\ T &= ma + xmg \\ &= m(a + xg) \end{align*}

#### Example A.5

Solve the system

$\begin{Bmatrix} \frac{1}{2}T – \frac{\sqrt{3}}{2}R & = & 0 \\ \frac{\sqrt{3}}{2}T + \frac{1}{2}R & = & mg \end{Bmatrix}$

for $$R$$ and $$T$$ in terms of all other variables.

This is not as simple as the previous example, but we follow the same steps. I’ll start by solving the first equation for $$T$$:

\begin{align*} \frac{1}{2}T – \frac{\sqrt{3}}{2}R &= 0 \\ \frac{1}{2}T &= \frac{\sqrt{3}}{2}R \\ T &= \sqrt{3}R \end{align*}

Now substitute this into the second equation:

\begin{align*} \frac{\sqrt{3}}{2}T + \frac{1}{2}R &= mg \\ \frac{\sqrt{3}}{2}\left(\sqrt{3}R\right) + \frac{1}{2}R &= mg \\ \frac{3}{2}R + \frac{1}{2}R &= mg \\ R\left(\frac{3}{2} + \frac{1}{2}\right) &= mg \\ 2R &= mg \\ R &= \frac{1}{2}mg \end{align*}

Finally, return to our previous solution for $$T$$:

\begin{align*} T &= \sqrt{3}R \\ &= \sqrt{3}\left(\frac{1}{2}mg\right) \\ &= \frac{\sqrt{3}}{2}mg \end{align*}