Rational expressions and rational equations

An expression that is a ratio of two polynomials is called rational expression.

i.e., a rational expression is of the form, $\dfrac{p(x)}{q(x)}$, where $p(x)$ and $q(x)$ are polynomials.

$\dfrac{2x^2+11}{4x+8}$ is a rational expression.

An equation that contains one or more fraction of polynomials (polynomials in the numerator and denominator) is called rational equation.

$\dfrac{x}{2x+5}+5=\dfrac{4x^2+5}{3x-5}$ is a rational equation.

Rational functions

A function that is a ratio of two polynomials is called rational function. A rational function is of the form,

$f(x)=\dfrac{p(x)}{q(x)}$, where $p(x)$ and $q(x)$ are polynomials and $q(x)\ne0$.

Example: The rational function $f$ is defined as $f(x)=\dfrac{11x+2}{3x^2-48}$. Find $f(2)$ and $f(4)$.

Solution:

Substituting, $x=2$ in the function, we get,

$f(2)=\dfrac{11\cdot 2+2}{3\cdot 2^2-48}$

$\hphantom{00000}=\dfrac{24}{-36}$

$\hphantom{00000}=\boxed{-\dfrac{2}{3}}$

Substituting, $x=4$ in the function, we get,

$f(4)=\dfrac{11\cdot 4+2}{3\cdot 4^2-48}$

$\hphantom{00000}=\dfrac{46}{0}$

Since there is zero in the denominator, $\boxed{f(4)\: is\: undefined.}$

Domain of a rational function

Domain of a rational function is all the input values (i.e., the $x$ values if the function is, $f(x)$) that do not make the function undefined.

For a rational function $f(x)$, domain is all the real numbers minus the $x$ values that make the denominator of the function zero, as zero denominator will make the function undefined. So, to find the domain, first find the $x$ values if any that make the function undefined and exclude those values from the real numbers.

Example 1:Find all the $x$ values for which the expression is undefined, $\dfrac{3x+5}{8x-16}$

Solution:
To find the $x$ values that makes a rational expression undefined, set the denominator of the expression equal to zero and solve for the variable, $x$.

$8x-16=0$

Solving for $x$, we get $x=2$. So, the anwser is $\boxed{x=2}$

Example 2:Find the values of $x$, which are not in the domain of the function,

$g(x)=\dfrac{3x-2}{x^2+3x-10}$

Solution:
Set the denominator of the function, g(x) equal to zero and solve for $x$.

$x^2+3x-10=0$

This is a quadratic equation and solving for $x$, we get

$x=2$ and $x=-5$.

We need to exclude these $x$ values as they cannot be in the domain. So the answer is $\boxed{x=2,-5}$.

Simplifying rational expression

To simplify a rational expression, find the greatest common factor of the polynomial in the numerator and that of the denominator. Then cancel out the same factors between the numerator and the denominator if any.

Example 1:Simplify, $\dfrac{2u+4}{6u^2+12u}$

Solution:In the numerator, the greatest common factor is $2$, and in the denominator, the greatest common factor is $6u$. Factoring out these factors,

$\dfrac{2u+4}{6u^2+12u}=\dfrac{2(u+2)}{6u(u+2)}$

Canceling out factor $2$ and $u+2$,

$\dfrac{2u+4}{6u^2+12u}=\dfrac{1}{3u}.$

Example 2:Simplify, $\dfrac{5x^2-5x-60}{x^2+8x+15}$.

Solution: First find the greatest common factor (GCF) of the polynomial in the numerator and that of the polynomial in the denominator. The GCF of the polynomial in the numerator is $5$ and there is no common factor for the polynomial in the denominator. Factoring the $5$ out,

$\dfrac{5x^2-5x-60}{x^2+8x+15}$$=\dfrac{5(x^2-x-12)}{x^2+8x+15}$

There are two quadratic trinomials, one in the numerator and the other in the denominator. Factoring them,

$x^2-x-12=(x+3)(x-4)$ and

$x^2+8x+15=(x+3)(x+5)$

Putting these in the polynomial expression,

$\dfrac{5x^2-5x-60}{x^2+8x+15}$$=\dfrac{5(x+3)(x-4)}{(x+3)(x+5)}$

Canceling out $x+3$,

$\hphantom{00000}$$=\boxed{\dfrac{5(x-4)}{x+5}}$

Example 3: Simplify, $\dfrac{3y^2-48}{3y^2+7y-20}$

Solution: In the numerator, the GCF is 3. factoring this out,

$\dfrac{3y^2-48}{3y^2+7y-20}$$=\dfrac{3(y^2-16)}{3y^2+7y-20}$

Factoring the binomial in the numerator and the trinomial in the denominator:

$\hphantom{00000}$$=\dfrac{3(y+4)(y-4)}{(3y-5)(y+4)}$

Canceling out $y+4$, we get

$\hphantom{00000}=\boxed{\dfrac{3(y-4)}{3y-5}}$

Multiplication and division of rational expressions

When multiplying rational expressions, first simplify each expressions if possible, then multiply the numerators and the denominators. The again simplify if possible.

i.e., $\dfrac{p}{q}\cdot \dfrac{r}{s}=\dfrac{pr}{qs}$

To divide rational expressions, take the reciprocal of the second rational expression (i.e., the divisor) and multiply with the first rational expression.

i.e., $\dfrac{p}{q}\div \dfrac{r}{s}=\dfrac{p}{q}\cdot \dfrac{s}{r}$

Example 1: Simplify, $\dfrac{4x+20}{3}\cdot \dfrac{2x}{x+5}$.

Solution:

First simplify each expression by factoring if possible.

$\dfrac{4x+20}{3}\cdot \dfrac{2x}{x+5}$$=\dfrac{4(x+5)}{3}\cdot \dfrac{2x}{x+5}$.

Multiply the numerators,

$\hphantom{00000}=\dfrac{4(x+5)\:2x}{3(x+5)}$.

Canceling out $x+5$ and simplifying,

$\hphantom{00000}=\boxed{\dfrac{8x}{3}}$

Example 2: Divide, $\dfrac{2x^2-5x-3}{3x^2+14x-5}\div \dfrac{4x-12}{x+5}$

Solution:

Take the reciprocal of the second rational expression (the divisor) and change the division into multiplication,

$\dfrac{2x^2-5x-3}{3x^2+14x-5}\div \dfrac{4x-12}{x+5}$$=\dfrac{2x^2-5x-3}{3x^2+14x-5}\cdot \dfrac{x+5}{4x-12}$.

Now, do the possible factoring,

$2x^2-5x-3=(2x+1)(x-3)$

$3x^2+14x-5=(3x-1)(x+5)$

$4x-12=4(x-3)$

Substituting these into the expression,

$\dfrac{2x^2-5x-3}{3x^2+14x-5}\div \dfrac{4x-12}{x+5}$$=\dfrac{(2x+1)(x-3)}{(3x-1)(x+5)}\cdot \dfrac{x+5}{4(x-3)}$

Canceling out the common factors,$x-3$ and $x+5$, we get

$\hphantom{00000}=\boxed{\dfrac{2x+1}{4(3x-1)}}$

Addition and subtraction of rational expressions

Addition and subtraction of rational expressions is same as the addition and subtraction of numbers. To add/subtract two or more rational expressions, the denominators of all the expressions should be the same. So, if the denominators are not the same, we need to find the common denominator, first. Once we have the common denominator, then we can add/subtract the numerators and put the common denominator as the denominator.

Addition/subtraction of rational expressions of same denominators.

If $p,q,r$ and $s$ are polynomials, then

$\dfrac{p}{q}+\dfrac{r}{q}-\dfrac{s}{q}=\dfrac{p+r-s}{q}$

Example:Simplify, $\dfrac{4x-5}{8x+7}+\dfrac{11x}{8x+7}$.

Solution:
The denominators of the two rational expressions are the same. Therefore,

$\dfrac{4x-5}{8x+7}+\dfrac{11x}{8x+7}$$=\dfrac{4x-5+11x}{8x+7}$.

Combining the $x$ terms of the numerator,

$\hphantom{00000}=\dfrac{15x-5}{8x+7}$.

Factoring the numerator,

$\hphantom{00000}=\boxed{\dfrac{5(3x-1)}{8x+7}}$.

Addition and subtraction of rational expressions of different denominators.

Least common denominator (LCD)

To add or subtract rational expressions of different denominators, first we need to find the least common denominator or LCD in short, of the rational expressions.

If there are fractions of numbers, then the LCD is the least common multiplier (LCM) of the integers in the denominators of the fractions. LCM of two or more integers is the smallest integer that can be divided evenly by all the integers.

For example, the LCM of the numbers, $12,18$ and $30$ is $180$ (see the example below).

For exponents, the LCM is the exponent with the greatest power.

For example, the LCM of the exponents, $x^2, x$ and $x^5$, is $x^5$.

Example 1: Find the least common multiplier of the following numbers, $12,18$ and $30$.

Solution: LCM is the smallest integer that can be divided evenly by the given integers $12,18$ and $30$. Find an integer that can divide all the three numbers evenly without remainder. Start with a smaller number. Then go from there to higher numbers and keep dividing until you cannot divide all the numbers further by an integer.

$\begin{align*} 2&\vert \underline{12,18,30}\\ 3&\vert \underline{\hphantom{0}6,\hphantom{0}9,15}\\ &\hphantom{00}2,\hphantom{0}3,\hphantom{0}5\\ \end{align*}$

Now, multiply all the numbers on the left of the vertical line and all the numbers below the last horizontal line. The product is the least common multiplier.

$2\cdot3\cdot2\cdot3\cdot5=\boxed{180}$

Example 2: Find the least common multiple of $7u^4$ and $5u^6$

Solution:
First find the least common multiplier of the constants, $7$ and $5$. There is no common factor between these two numbers. So, just multiply the two numbers, which is the LCM. i.e., Therefore, just multiply the two numbers, $7\cdot 5=35$. This is the LCM for the constants.

Next, find the LCM for the variables. We have $u^4$ and $u^6$, take the one with the largest exponent. i.e., the LCM is $u^6$.

Multiply the two LCMs together, $\boxed{35u^6}$. This is the LCM of the two polynomials.

Example 3: Find the least common denominator of $\dfrac{7x}{2x-6}$ and $\dfrac{5x+1}{8x-24}$

Solution
We have the denominators, $2x-6$ and $8x-24$. We can factorize these denominators and we will get,

$2(x-3)$ and $8(x-3)$.

For the constants,$2$ and $8$, LCM is $8$. And considering the variables factors, we have the factor, $x-3$ in both of the denominators, their LCM is $x-3$.

Multiplying the two LCMs we will get the LCM of the two denominators.

So the answer is LCD $=8(x-3)$.

Example 4: Subtract and simplify:$\dfrac{8x+2}{11x-3}-\dfrac{2}{3-11x}$

Solution:
We can make the denominators of both the radical expressions by factoring out a $-1$ in the denominator of the first or the second radical expression. Factoring out a $-1$ in the second one,

$\dfrac{8x+2}{11x-3}-\dfrac{2}{3-11x}$$=\dfrac{8x+2}{11x-3}-\dfrac{2}{-(11x-3)}$

$\hphantom{00000}=\dfrac{8x+2}{11x-3}+\dfrac{2}{11x-3}$

Both have the same denominators, put that as the common denominator and add the polynomials in the numerators,

$\hphantom{00000}=\dfrac{8x+2+2}{11x-3}$

$\hphantom{00000}=\boxed{\dfrac{8x+4}{11x-3}}$

Example 5:Add and simplify as much as possible, $\dfrac{2}{2x^2-x-21}+\dfrac{3}{3x+9}$

Solution:
Factoring the denominators, $2x^2-x-21$, and $3x+9$ we get,

$2x^2-x-21=(x+3)(2x-7)$

and

$3x+9=3(x+3)$

Now,

$\dfrac{2}{2x^2-x-21}+\dfrac{3}{3x+9}$$=\dfrac{2}{(x+3)(2x-7)}+\dfrac{3}{3(x+3)}$

$\hphantom{00000}=\dfrac{2}{(x+3)(2x-7)}+\dfrac{1}{x+3}$

LCD of the two rational expressions is $(x+3)(2x-7)$. With this LCD, we can write

$\hphantom{00000}=\dfrac{2}{(x+3)(2x-7)}+\dfrac{2x-7}{(x+3)(2x-7)}$

$\hphantom{00000}=\dfrac{2+2x-7}{(x+3)(2x-7)}$

$\hphantom{00000}=\boxed{\dfrac{2x-5}{(x+3)(2x-7)}}$

Complex fractions

In a complex fraction there is one or more fraction in the numerator, or the denominator or both.

Example 1: Simplify $\dfrac{\dfrac{4x^2y^5}{5uv^3}}{\dfrac{5x^3y^4}{11u^3v^2}}.$

Solution:In this complex rational expression, there is one rational expression in the numerator and another in the denominator. First, we can replace the division by the $\div$ symbol,

$\dfrac{\dfrac{4x^2y^5}{5uv^3}}{\dfrac{5x^3y^4}{11u^3v^2}}$$=\dfrac{4x^2y^5}{5uv^3} \div \dfrac{5x^3y^4}{11u^3v^2}$

Now, we can replace the $\div$ symbol by the multiplication symbol if we take the reciprocal of the divisor rational expression.

$\hphantom{00000}=\dfrac{4x^2y^5}{5uv^3} \cdot \dfrac{11u^3v^2}{5x^3y^4}$

Simplifying,

$\hphantom{00000}=\dfrac{44u^2y}{25vx}$

Example 2: Simplify, $\dfrac{\dfrac{u+5}{7}}{\dfrac{u^2-25}{u+2}}$

Solution:

$\dfrac{\dfrac{u+5}{7}}{\dfrac{u^2-25}{u+2}}$$=\dfrac{u+5}{7}\div \dfrac{u^2-25}{u+2}$

Take the reciprocal of the divisor rational expression and change division into multiplication.

$\hphantom{00000}=\dfrac{u+5}{7}\cdot \dfrac{u+2}{u^2-25}$

We need to factor, $u^2-25$. i.e., $u^2-25=(u+5)(u-5)$. Substituting these,

$\hphantom{00000}=\dfrac{u+5}{7}\cdot \dfrac{u+2}{(u+5)(u-5)}$

Canceling $u+5$,

$\hphantom{00000}=\dfrac{u+2}{7(u-5)}$

Example 3: Simplify, $\dfrac{6+\dfrac{15}{x}}{\dfrac{2}{x}+\dfrac{5}{x^2}}$

Solution:

LCD of the fractions in the numerator is $x$. And the LCD of the fractions in the denominator is $x^2$. With these LCDs, we can write,

$\dfrac{6+\dfrac{15}{x}}{\dfrac{2}{x}+\dfrac{5}{x^2}}$$=\dfrac{\dfrac{6x+15}{x}}{\dfrac{2x+5}{x^2}}$

Changing the division into multiplication by taking the reciprocal,

$\hphantom{00000}=\dfrac{6x+15}{x}\cdot \dfrac{x^2}{2x+5}$

Factoring the $6x+15$,

$\phantom{00000}=\dfrac{3(2x+5)}{x}\cdot \dfrac{x^2}{2x+5}$

Canceling, an $x$ and $2x+5$, we get

$\hphantom{00000}=\boxed{3x}$

Example 4: Simplify, $\dfrac{7v^{-1}y^{-1}}{2v^{-2}+5y^{-1}}$

Solution:

Making the exponents in the denominator positive,

$\dfrac{7v^{-1}y^{-1}}{2v^{-2}+5y^{-1}}$$=\dfrac{7v^{-1}y^{-1}}{\dfrac{2}{v^2}+\dfrac{5}{y}}$

LCD of the fractions in the denominator is $v^2y$. So we can write,

$\hphantom{00000}=\dfrac{7v^{-1}y^{-1}}{\dfrac{2y+5v^2}{v^2y}}$

$\hphantom{00000}=7v^{-1}y^{-1}\cdot \dfrac{v^2y}{2y+5v^2}$

$\hphantom{00000}=\boxed{\dfrac{7v}{2y+5v^2}}$

Solving rational equations

Example 1: Solve the equation, $\dfrac{2}{x}-\dfrac{1}{3}=5$

Solution: First find the LCD of all the terms of the equation. The LCD is $3x$,

Now multiply the LCD with the equation,

$3x\left(\dfrac{2}{x}-\dfrac{1}{3}\right)=3x\cdot5$

Simplifying,

$6-x=15x$

Solving for $x$,

$x=\dfrac{3}{8}$

Example 2: Solve, $\dfrac{4}{y-3}=2-\dfrac{7}{5y-15}$

Solution:
First you need to factor out completely. We can factor out $5y-15$, i.e., $5y-15y=5(y-3)$. So, we have

$\dfrac{4}{y-3}=2-\dfrac{7}{5(y-3)}$

Now find the LCD of all the terms of the equation. The LCD is $5(y-3)$.

Multiply the LCD,

$20=10(y-3)-7$

$20=10y-30-7$

Solving,

$y=\dfrac{57}{10}$

Example 3: Solve $\dfrac{15}{x}=x+2$.

Solution: The LCD of the equation is $x$.

Multiplying the equation with the LCD,

$15=x^2+2x$

Rewriting the equation with zero on one side,

$x^2+2x-15=0$

This is a quadratic equation. Solving this equation, we get

$x=-5$ and $x=3$.

Example 4: Solve for $m$, $c=\dfrac{a}{b(m+n)}$.

Solution:

First do the cross multiplication,

$c\cdot b(m+n)=a$

Now isolate, $m+n$, by dividing out $cb$,

$m+n=\dfrac{a}{cb}$

Subtract, $n$,

$\boxed{m=\dfrac{a}{cb}-n}$

Solving proportion

Example 1: Solve for $x$, $\dfrac{3}{x}=\dfrac{8}{7}$

Solution: Doing the cross multiplication,

$3\cdot7=8\cdot x$

Dividing out, $8$,

$\dfrac{21}{8}=x$

i.e., $\boxed{x=\dfrac{21}{8}}$

Example 2: Solve for $y$, $\dfrac{4}{5}=\dfrac{7}{y+2}$

Solution: Cross multiplying,

$4\cdot (y+2)=5\cdot 7$

Dividing $4$,

$y+2=\dfrac{35}{4}$

$y=\dfrac{35}{4}-2$

$y=\dfrac{27}{4}$

Example 3: Solve, $\dfrac{3}{x}=\dfrac{-4}{x-7}$

Solution: Cross multiplying,

$3(x-7)=-4x$

$3x-21=-4x$

$x=3$.