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5.04 Solving equations with rational expressions

Lesson

Identify allowable values of rational equations

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=p292p1 $5x^3-1=\frac{2x^3-3}{2x}$5x31=2x332x

 

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}$1x23x+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.

$x^2-3x+2$x23x+2 $=$= $0$0 (Setting the denominator to zero)
$\left(x-2\right)\left(x-1\right)$(x2)(x1) $=$= $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}$1x23x+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.

 

Remember!

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.

 

Worked example

Question 1

For what values of $x$x is the following rational equation undefined?

$\frac{x-3}{\left(x-3\right)\left(x+7\right)}=0$x3(x3)(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 $\left(x-3\right)\left(x+7\right)$(x3)(x+7) is zero.

Do: To determine when the denominator is zero, we set it equal to zero and solve for $x$x:

$\left(x-3\right)\left(x+7\right)$(x3)(x+7) $=$= $0$0 (Setting the denominator to zero)
$x-3$x3 $=$= $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 $\frac{x-3}{\left(x-3\right)\left(x+7\right)}=0$x3(x3)(x+7)=0 is undefined for $x=-7$x=7 and $x=3$x=3.

Reflect: The expression $\frac{x-3}{\left(x-3\right)\left(x+7\right)}$x3(x3)(x+7) has a common factor of $x-3$x3 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 $\frac{x-3}{\left(x-3\right)\left(x+7\right)}$x3(x3)(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, $\frac{x-3}{\left(x-3\right)\left(x+7\right)}=\frac{1}{x+7}$x3(x3)(x+7)=1x+7 for all real values of $x$x except for $x=-7$x=7 and $x=3$x=3.

 

Careful!

An algebraic expression is undefined whenever the denominator is zero before any factors have been canceled.

 

Practice questions

Question 2

For what value of $w$w is the following rational equation undefined?

$\frac{4}{w}=0$4w=0

Question 3

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$6b27b(b5)(b8)=0

  1. Write each answer on the same line, separated by commas.

question 4

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

  1. Write each answer on the same line, separated by commas.

 

Rational adding and subtracting

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.

 

Worked examples 

Question 5

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$77 $=$= $7x-x$7xx 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  

 

Practice questions

Question 6

Solve the following equation: $5x-\frac{104}{5}=x$5x1045=x

Question 7

Solve the following equation: $\frac{5x}{3}-3=\frac{3x}{8}$5x33=3x8

Question 8

Question 9

Solve for the unknown.

$\frac{1}{x}-\frac{10x}{3}=-\frac{7}{3}$1x10x3=73

  1. Write all solutions on the same line, separated by commas.

question 10

Solve $\frac{7}{n+1}-\frac{4}{n}=\frac{1}{n+1}$7n+14n=1n+1.

Outcomes

A.CED.1c

Create equations and inequalities in one variable and use them to solve problems. Include equations and inequalities arising from linear, quadratic, simple rational, and exponential functions.Extend to include more complicated function situations with the option to solve with technology.

A.CED.3a

Represent constraints by equations or inequalities, and by systems of equations and/or inequalities, and interpret solutions as viable or non-viable options in a modeling context. For example, represent inequalities describing nutritional and cost constraints on combinations of different foods.. While functions will often be linear, exponential, or quadratic, the types of problems should draw from more complicated situations.

A.CED.4d

Rearrange formulas to highlight a quantity of interest, using the same reasoning as in solving equations. While functions will often be linear, exponential, or quadratic, the types of problems should draw from more complicated situations.

A.REI.2

Solve simple rational and radical equations in one variable, and give examples showing how extraneous solutions may arise.

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