- Find all pairs of values of \(x\) and \(y\) for which \[\begin{equation*} \frac{3x+y+1}{8} = \frac{x-y}{5} = \frac{x^2 - y^2}{5}. \end{equation*}\]

So either \(x-y=0\), or we can divide by \(x-y\) to give \(x+y=1\).

#### Case 1: \(x - y = 0\)

In this case, the first equation holds if and only if \(3x+y+1=0\).

But as \(x=y\), this gives \(4x+1=0\), so \(x=y=-\frac14\) is a solution.

#### Case 2: \(x+y = 1\)

In this case, we can write \(y = 1 - x\) and thus the first equation becomes \[\begin{align*} &&\frac{3x + (1-x) + 1}{8} &= \frac{x - (1-x)}{5} \\ \iff&& \frac{2x + 2}{8} &= \frac{2x - 1}{5} \\ \iff&& 5(x+1) &= 4(2x-1) \\ \iff&& 9 &= 3x \\ \iff&& x &= 3. \end{align*}\]So \(x=3\), \(y=-2\) is a second solution.

- Solve the equation
\[\begin{equation*}
\sqrt{3-3x} - \sqrt{2-x} = 1.
\end{equation*}\]
*The expression \(\sqrt{u}\) denotes the positive square root of \(u\).*

so that \(x = 1\) or \(x = -2\).

These are only *candidate solutions*; the original equation implies these are the only solutions possible, but they are only possibilities. Our squaring may have introduced extra ‘solutions’. (For example, if \(x=1\), then \(x^2=1\), but \(x^2=1\) has two solutions, \(x=1\) and \(x=-1\).) So we need to check whether these candidate solutions are actually solutions of the original equation.

so this is *not* a solution to the original problem.

So \(x = -2\) is the unique solution to the original problem.