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\title{Solutions to Linear Equations}
\date{September 27, 2005}

\begin{document}
\maketitle
\section{Objective} To learn how to solve linear equations that do not appear to be   linear.\\
\textbf{Note} When an equation contains a fraction, and that fraction contains an $x$ in its denominator, then the equation is called a \emph{fractional equation}.  If a fractional equation has the same solution as a linear equation, we say that it reduces to a linear equation.
 
\begin{itemize}
%---------------------------------------------------
\item \textbf{Recall 1}: Watch how quickly the fractional equation 
$$\frac{3}{(x + 2)} = \frac{2}{(3x - 1)}$$
reduces to a linear equation. \\ 
Multiply both sides by $(x + 2)(3x - 1)$ to  get
$$(x + 2)(3x - 1)\frac{3}{(x + 2)} = (x + 2)(3x - 1)\frac{2}{(3x - 1)}$$
which simplifies to
$$3(3x - 1) = 2(x + 2).$$
Notice that this is a linear equation whose solution is $x = 1$. 
%---------------------------------------------------
\item \textbf{Recall 2}: Some rational equations that convert to linear ones have no solutions.  Here is an example.  In attempting to solve the equation
$$\frac{x}{(x - 2)}= 2 + \frac{2}{(x - 2)},$$
we see that the common denominator is $(x - 2)$ and that by multiplying the  2 by  $\frac{(x - 2)}{(x - 2)}$ we have 
$$\frac{x}{(x - 2)}= \frac{2(x - 2)}{(x - 2)} + \frac{2}{(x - 2)}$$
or
$$ x = 2(x - 2) + 2 $$
and so $x = 2$ is a candidate for a solution.  However, if we check this  solution, we find that it cannot satisfy the equation, since it would  require us to divide by zero.  We must conclude that this equation has  no solutions.
%---------------------------------------------------
\item
\textbf{Recall 3}: There are equations that contain powers of $x$ greater than  1, but which quickly convert to linear equations.  For example,  consider the equation
$$x^2 + 2x + 3 = x^2 - x - 3.$$
By subtracting $x^2$ from both sides, the equation becomes linear. 
%---------------------------------------------------
\item \textbf{Recall 4}:  Occasionally, the disguise is more subtle.  For example,  the equation
$$2x^2 + 7 = (x - 3)(2x + 5)$$
is immediately reduced to a linear equation once the product on the  right-hand side is multiplied out and $2x^2$ is subtracted from both sides.  
%---------------------------------------------------
\item \textbf{Recall 5}: Any linear equation is mathematically equivalent to an  equation of the form 
$$ax+b=0.$$
where a and b are fixed real numbers.  We call this the \emph{standard form}  of a linear equation of one variable $x$.

For example, the first  equation of Recall 4 is equivalent to the second; and the second is in  standard form, where $a = 7$ and $b = 14$. 
\end{itemize}
%---------------------------------------------------
\section{Examples}
\begin{itemize} 
\item \textbf{Example 1.} Show that the equation
$$\frac{6}{(x^2 - 4)} = \frac{5}{(x + 2)} + \frac{3}{(x - 2)}$$
can be reduced to a linear equation.\\
\emph{Solution}: The common denominator of the right-hand side is
$(x + 2)(x - 2)$, which is the same as $x^2 - 4$.  The equation therefore reduces to
$$\frac{6}{(x^2 - 4)} = \frac{5(x - 2) + 3(x + 2)}{(x^2 - 4)}$$
and hence to
$$6 = 5(x - 2) + 3(x + 2).$$
Using the distributive law, we see that this is equivalent to
$$6 = 5x - 10 + 3x + 6,$$
which is clearly linear.
%---------------------------------------------------
\item \textbf{Example 2.}Show that the equation $(x - 3)(x + 4) = (x + 5)(x + 2)$ is linear.
\emph{Solution}: Multiply out each side to get
$$x^2 + x - 12 = x^2 + 7x + 10.$$
Subtract $x^2$ from both sides to get
$$x - 12 = 7x + 10,$$
which is linear.
%---------------------------------------------------
\item \textbf{Example 3.} Show that the equation $\sqrt{4 - 3x} = 5$
is linear.\\
\emph{Solution}: You should convince yourself that squaring both sides of an  equation gives you a valid equation. 
In symbols, this means: 
if $A = B$, then $A^2 = B^2$.
By squaring both sides of the equation, we have $4 - 3x = 25$.  
\end{itemize}
%---------------------------------------------------
\textbf{Warning:} Taking square roots of both sides of an equation may introduce extra solutions to an equation, since $A^2 = B^2$, implies that $A = \pm B$.
%---------------------------------------------------
\begin{itemize}
\item \textbf{Example 4.} Find a solution to $(x - 1)^2 - 4(x - 2)^2 = 0$.\\
\emph{Solution}: We may add $4(x - 2)^2$ to both sides to get
$$(x - 1)^2 = 4(x - 2)^2.$$
\emph{Remember}: if $A^2 = B^2$, then $A = \pm B$.
So
$$(x - 1) = \pm 2(x - 2).$$
This means that we have reduced our equation to two different linear  equations, namely
$$(x - 1) = 2(x - 2)$$
and
$$(x - 1) = -2(x - 2).$$
The solution to the first is $x = 3$; the solution to the second is $x = \frac{5}{3}$. 

%---------------------------------------------------

\item \textbf{Example 5.} Solve the equation $$\frac{a}{(x^2 - 4)} = \frac{b}{(x + 2)} + \frac{c}{(x - 2)}$$ for $x$ in terms of $a$, $b$ and $c$.\\
\emph{Solution}: Treat the $a$, $b$ and $c$ as if they are numbers and proceed as  illustrated in the solution to Example 1.  \\
The common denominator of  the right-hand side is $x^2 - 4$.  Reduce the right hand side to a single  fraction so the equation looks like
$$\frac{a}{(x^2 - 4)} = \frac{b(x - 2) + c(x + 2)}{(x^2 - 4)}.$$
Here is a special fact that comes in handy:
If  $\frac{A}{B} = \frac{C}{B}$ and $B \ne 0$, then $A = C$.
In other words, whenever two fractions with the same non-zero denominators are equal, their numerators are  equal.  Applying this to our equation, we see that
$$a = b(x - 2) + c(x + 2).$$
The distributive law then tells us that 
$$a = bx - 2b + cx + 2c.$$  
Collecting  the terms involving $x$ on the left side and the terms not involving $x$ on  the right side gives
$$bx + cx = a + 2b - 2c.$$
We then factor out $x$ from the left-hand side, giving
$$x(b + c) =  a + 2b - 2c.$$
As long as $b \ne -c$, we have a solution: $x = \frac{(a + 2b - 2c)}{(b+c)}$.\\
When $b = -c$ we have two cases:\\
i) If $a + 2b - 2c = 0$ (that is, $a = 4c$), then our equation reduces to $0 = 0$, so any real number, except for 2 and -2, is a solution.

ii) If $a \ne 4c$, then the equation has no solutions.
\end{itemize}
%---------------------------------------------------
\textbf{Reminder:} Always check your solution to be sure that it makes sense.   If your solution causes a denominator in the equation to become zero,  then you must disregard that solution.
%---------------------------------------------------
\section{Exercises}
Solve for the variable ($x$, $s$, $y$, or $t$) in the following equations:\\

1. $x^2 + 3x - 5 = x^2 + 5x + 3$\\

 
2. $2x^2 - 3x + 7 = (2x + 3)(x - 2)$\\
 

 
3. $(s + 4)(2s - 1) = (2s + 3)(s - 2)$\\

 
4. $\frac{2}{(x + 5)}  = \frac{3}{(x - 2)}$\\

 
5. $\frac{1}{(x - 1)} + \frac{3}{(x - 1)}  = 2$\\

 
6.  $\frac{1}{(x - 2)} + \frac{5}{(2 - x)} = 1$\\

 
7. $\frac{2}{(2y - 1)} = \frac{7}{(y + 4)}$ [Hint:  $(2y - 1)(y + 4)$ is a common divisor.]\\

 
8.  $\frac{3x}{(2x - 6)}= 3 + \frac{3}{(x - 3)}$ [Hint:  Factor out the 2 from the denominator of the left-hand side.]\\

 
9.  $\frac{3}{(t - 3)}+ \frac{2}{(t - 4)} =  \frac{5}{(t - 3)(t - 4)}$ [Hint:  $(t - 3)(t - 4)$ is a  common divisor.]\\

 
10.  $\frac{7}{(2x + 3)(x - 2)} = \frac{4}{(2x + 3)} + \frac{2}{(x - 2)}$  [Hint:  $(2x + 3)(x - 2)$ is a common divisor.]\\

 
11.  $\frac{(4 - 3x)}{(6x - 8)} = \frac{5}{(3x - 4)}$  [Hint:  Factor out a 2 from the denominator of the first term.]\\

 
12.  $\frac{2}{(x - 2)} + \frac{7}{(x+2)} = \frac{8}{(x^2 - 4)}$   [Hint: $x^2 - 4 = (x - 2)(x + 2)$.]\\

 
13. $\sqrt (7 - 3y) = 6$\\

 
14. $\sqrt[3]{5x-9}=-2$\\

 
15.  Solve the following equation for $x$ in terms of $a$ and $b$, where $a \ne b$:
   $\frac{(x + a)}{(x - b)}= \frac{a}{b}$.\\

 
16. $2 + \frac{2}{(2 + \frac{2}{x})} = \frac{5}{(x + 1)}$    [Hint: Simplify the compound fraction on the left-hand side.] 

%---------------------------------------------------
\section{Solutions} \mbox{}

1.  $x = -4$\\

 
2.  $x = \frac{13}{2}$\\

 
3.  $s = \frac{-1}{4}$\\

 
4.  $x = -19$\\

 
5.  $x = 3$\\

 
6.  $x = -2$\\

 
7.  $y = \frac{15}{12}$\\

 
8.  $x = 4$\\
 

10. $x = \frac{9}{8}$\\
 

11.  $x = -2$\\
 

12. $x = 2$\\
 

13.  $y = \frac{-29}{3}$\\
 

14. $x = \frac{1}{5}$\\
 

15.  $x = \frac{2ab}{(a - b)}$\\
 

 16. $x = 1$


\end{document}
