Undo Multiplication with Division (UMD); also, Undo Addition with Subtraction (UAS)

The properties of equality that were mentioned earlier are, by some students, implemented incorrectly even when the situation calls for their use. For instance, when solving an equation like

\begin{displaymath}3x + 7 = 4, \end{displaymath}

two steps are called for:
\begin{displaymath}3x = -3 \qquad\mbox{({\bf {subtraction prop. of equality}}; $7$ subtracted from both sides)} \end{displaymath}

and
\begin{displaymath}x = -1 \qquad\mbox{({\bf {division prop. of equality}}; both sides divided by $3$)}. \end{displaymath}

Notice that, in the expression $(3x + 7)$, the order of operations dictates that the multiplication by 3 comes before the addition of 7, and the ``undoings" of these processes -- the subtraction of 7 and the division by 3 -- were carried out in reverse order. That is not to say that we could not have undone things in a different order, but students who do so often make the following error. Dividing by 3, they often neglect the fact that all terms on both sides are to be divided by 3. In other words, after dividing by 3 they write
\begin{displaymath}x + 7 = \frac{4}{3} \qquad\mbox{instead of}\qquad x + \frac{7}{3} = \frac{4}{3}. \end{displaymath}

They are too set on the idea that they will be subtracting 7 from both sides to realize that, having divided by 3 first, it is not 7, but rather $7/3$, which must be subtracted, giving the same answer $x=-1$ as before. One other note is in order here. If parentheses appear in an equation such as
\begin{displaymath}3(x - 1) = 5, \end{displaymath}

then the order of operations are preempted (the subtraction within the parentheses comes before the multiplication by 3). In solving for $x$, we may of course, distribute the 3, thereby eliminating the parentheses and making the problem appear like the last one discussed. Even fewer steps are required if one just ``undoes" the multiplication and subtraction in their opposite order:
\begin{displaymath}x - 1 = \frac{5}{3} \qquad (\mbox{{\bf {division prop. of equality}}; dividing both sides by 3}), \end{displaymath}

and then
\begin{displaymath}x = \frac{8}{3} \qquad (\mbox{{\bf {addition prop. of equality}}; adding 1 to both sides}). \end{displaymath}

Now let us return to the equation

\begin{displaymath}3x + 7 = 4, \end{displaymath}

and investigate the more telling errors that gave the titles UMD and UAS to this section. Some students recognize the need for two steps (like those carried out when this equation was being considered above) to isolate $x$, but have little feel for which operations will achieve this. For instance, realizing that, like the $4$ on the right-hand side of the equation, $7$ is a ``non-$x$" term, a student may write
\begin{displaymath}3x = \frac{4}{7}, \mbox{\hspace{0.6in} \epsfig{0.5in}{uas.eps}}\end{displaymath}

misunderstanding that she has subtracted 7 on the left side, but divided by 7 on the other side. The original equation and the new one no longer have the same solutions as a result. The same student may then recognize that she needs to move the $3$ over to the other side. Since the 3 is multiplied by the $x$, she should ``undo" this by dividing both sides by 3. But she may (wrongly) write
\begin{displaymath}x = \frac{4}{7} - 3, \mbox{\hspace{0.6in} \epsfig{0.5in}{umd.eps}}\end{displaymath}

having divided on the left but subtracted on the right. Again, the solution $x = -17/7$ is different from the one that solved the original equation $3x+7=4$, namely $x=-1$.

Worse still is when a student thinks he can solve in one step (that is, take care both of the multiplication by 3 and the addition of 7 via one operation). Such a student may write something like

\begin{displaymath}3x + 7 = 4 \quad\Rightarrow\quad x = \frac{4}{3+7}. \mbox{\hspace{0.6in} \epsfig{0.5in}{uas.eps}/\epsfig{0.5in}{umd.eps}}\end{displaymath}

Again, the answer this student gets, $x = \frac{2}{5}$, is different than the correct one $x=-1$.




Top Algebra Errors Made by Calculus Students (full document)
Full List of Grading Codes

Thomas L. Scofield 2003-09-04