A rational expression is one that has the form
$\frac{p\left(x\right)}{q\left(x\right)}$p(x)q(x),
where $p\left(x\right)$p(x) and $q\left(x\right)$q(x) are polynomials and $q\left(x\right)\ne0$q(x)≠0.
A rational equation is then an equation containing polynomials and at least one rational expression.
So, for example, the following are all rational equations:
$\frac{x}{x+1}=0$xx+1=0 | $4=\frac{p^2-9}{2p-1}$4=p2−92p−1 | $5x^3-1=\frac{2x^3-3}{2x}$5x3−1=2x3−32x |
Remember that a rational expression is undefined whenever the denominator of the expression is zero.
So an expression such as $\frac{1}{x}$1x is undefined for $x=0$x=0 and defined for every other value of $x$x.
More complicated expression such as $\frac{1}{x^2-3x+2}$1x2−3x+2 are also undefined for certain values of $x$x, but these values are less obvious. To find the values of $x$x for which the expression is undefined, we can construct an equation by setting the denominator to equal zero.
$x^2-3x+2$x2−3x+2 | $=$= | $0$0 | (Setting the denominator to zero) |
$\left(x-2\right)\left(x-1\right)$(x−2)(x−1) | $=$= | $0$0 | (Factoring the expression) |
$x$x | $=$= | $1,2$1,2 | (Solving for each factor) |
So we can conclude that the expression $\frac{1}{x^2-3x+2}$1x2−3x+2 is undefined for $x=1$x=1 and $x=2$x=2.
A rational equation is then undefined whenever any of the expressions in the equation are undefined.
A rational expression is undefined whenever the denominator of the expression is zero.
A rational equation is then undefined whenever any expression in the equation is undefined.
For what values of $x$x is the following rational equation undefined?
$\frac{x-3}{\left(x-3\right)\left(x+7\right)}=0$x−3(x−3)(x+7)=0
Think: A rational equation is undefined when the expressions of the equation are undefined. In this case, that happens when the denominator $\left(x-3\right)\left(x+7\right)$(x−3)(x+7) is zero.
Do: To determine when the denominator is zero, we set it equal to zero and solve for $x$x:
$\left(x-3\right)\left(x+7\right)$(x−3)(x+7) | $=$= | $0$0 | (Setting the denominator to zero) |
$x-3$x−3 | $=$= | $0$0 | (Solving when one factor is zero) |
$x$x | $=$= | $3$3 | (Adding $3$3 to both sides) |
$x+7$x+7 | $=$= | $0$0 | (Solving for the other factor) |
$x$x | $=$= | $-7$−7 | (Subtracting $7$7 from both sides) |
$x$x | $=$= | $-7,3$−7,3 | (Combining both solutions) |
So we can conclude that the equation $\frac{x-3}{\left(x-3\right)\left(x+7\right)}=0$x−3(x−3)(x+7)=0 is undefined for $x=-7$x=−7 and $x=3$x=3.
Reflect: The expression $\frac{x-3}{\left(x-3\right)\left(x+7\right)}$x−3(x−3)(x+7) has a common factor of $x-3$x−3 between the numerator and denominator.
If we were to cancel this common factor, we would end up with the expression $\frac{1}{x+7}$1x+7. Notice that this expression is only undefined when $x=-7$x=−7, and is defined when $x=3$x=3.
So the expressions $\frac{x-3}{\left(x-3\right)\left(x+7\right)}$x−3(x−3)(x+7) and $\frac{1}{x+7}$1x+7 are different, since the second one is defined for $x=3$x=3.
Now we can simplify a rational expression by canceling common factors, but only over values for which it is defined.
That is, $\frac{x-3}{\left(x-3\right)\left(x+7\right)}=\frac{1}{x+7}$x−3(x−3)(x+7)=1x+7 for all real values of $x$x except for $x=-7$x=−7 and $x=3$x=3.
An algebraic expression is undefined whenever the denominator is zero before any factors have been canceled.
For what value of $w$w is the following rational equation undefined?
$\frac{4}{w}=0$4w=0
For what values of $b$b is the following rational equation undefined?
$\frac{6}{b-2}-\frac{7b}{\left(b-5\right)\left(b-8\right)}=0$6b−2−7b(b−5)(b−8)=0
Write each answer on the same line, separated by commas.
For what values of $j$j is the following rational equation undefined?
$-\frac{4}{j}+\frac{7}{j+5}=\frac{24}{j^2+j}$−4j+7j+5=24j2+j
Write each answer on the same line, separated by commas.
We've already looked at how to solve equations, whether that be in one, two or three steps. As you know, equations can involve a number of different operations and there are different methods for solving equations which you may want to review.
In this chapter, we are going to look at examples of equations that involve addition and subtraction of algebraic terms, including ones with fractions.
Solve $\frac{-7}{100}+\frac{x}{100}=\frac{7x}{100}+\frac{7}{100}$−7100+x100=7x100+7100 for $x$x.
Think: How do we move these terms around to get $x$x by itself.
Do:
$\frac{-7}{100}+\frac{x}{100}$−7100+x100 | $=$= | $\frac{7x}{100}+\frac{7}{100}$7x100+7100 | Multiply all the terms by the common denominator, $100$100 |
$-7+x$−7+x | $=$= | $7x+7$7x+7 | Rearrange the expression so all the $x$x terms are on one side |
$-7-7$−7−7 | $=$= | $7x-x$7x−x | Now let's simplify |
$-14$−14 | $=$= | $6x$6x | |
$\frac{-14}{6}$−146 | $=$= | $x$x | Solve for $x$x and simplify the fraction |
$x$x | $=$= | $\frac{-7}{3}$−73 |
Solve the following equation: $5x-\frac{104}{5}=x$5x−1045=x
Solve the following equation: $\frac{5x}{3}-3=\frac{3x}{8}$5x3−3=3x8
Solve for the unknown.
$\frac{1}{x}-\frac{10x}{3}=-\frac{7}{3}$1x−10x3=−73
Write all solutions on the same line, separated by commas.
Solve $\frac{7}{n+1}-\frac{4}{n}=\frac{1}{n+1}$7n+1−4n=1n+1.
Solve simple rational and radical equations in one variable, and give examples showing how extraneous solutions may arise.