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# Exploring Rational Expressions

Top
 Sub Topics If two or more rational numbers are joined together with the mathematical operator, it forms the rational expression. We solve the rational expression in order to get the number in the form of the rational number. When we proceed to solve the rational numbers, with the addition or subtraction operation, we first need to convert the two rational numbers into their equivalent form, such that the denominators of the two rational numbers becomes same and now either the addition of the subtraction operation is performed on the numerators and we get the sum or the difference of the two rational numbers. Now let us say that the operation of multiplication or division is to be performed between the two rational numbers, and then a different method is used.In case the two numbers are to be multiplied, then we will simply multiply the numerator with the numerator and the denominator is to be multiplied with the denominator. The resultant rational number is then converted into the standard form. To convert the product into the standard form, we will find the HCF of the numerator and the denominator and then divide both the numerator and the denominator by the HCF.In case of division, we will change the divisor to its reciprocal and then along with it the sign of division will change to the sign of multiplication. Thus it is now the ordinary expression of the multiplication of two rational numbers, which can be solved using the method of multiplication.

## Graphing Rational Function

Rational function can be defined as Ratio of two Polynomials. In Rational Function numerator and denominator values are present, denominator value of rational function cannot be zero. In other words rational function can be defined as ratio of two polynomials.

Example:

$\frac{p^{2} + 5}{p + 2}$, as it is the ratio of two polynomials. Now we will see Graphing rational function.

Suppose we have a rational function $f (p)$ = $\frac{p^{2} - 4}{p^{2} - 4p}$, then we can plot graph of rational function as shown below.

Solution:

Here we need to follow some steps to plot graph of rational function.

Step 1:

First we check whether function is rational function or not. As we know that ratio of two polynomials is known as rational function. So rational function is:

$\Rightarrow\ f (p)$ = $\frac{p^{2} - 4}{p^{2} - 4p}$,

Here in this function we cannot put value of denominator as zero because if we put value of denominator as zero then whole function changes to infinity. There is no $y$-intercept present for graph.

Step 2:

Put numerator value equals to zero to get value of $x$-coordinate.

So we can write numerator value as:

$\Rightarrow\ p^{2} - 4$ = $0$, so here we get two values of $p$ that is $p = + 2$.

Above function can also be written as:

= $p^{2} - 4$ = $p (p - 4)$ = $0$, so here we get value of '$p$' as $0$ and $4$.

Step 3:

Now put different values of '$p$' to get other values.

If we put value of '$p$' as $1, 3$ and $5$ then we get $1$, $\frac{-5}{3}$ and $\frac{21}{5}$ respectively. Now we can easily plot graph of rational function using these coordinates. Graph of rational function is shown below:

## Add and Subtract Rational Expression

Rational expression can be defined as a function which is given as Ratio of two Polynomials. In other words, rational expression can be expressed as a fraction which is comprised of numerator and denominator, where as numerator is the upper part of ratio and denominator is lower part.

Let’s consider a function $f\ (y)$ which is represented as:

$f\ (y)$ = $\frac{P\ (x)}{Q\ (x)}$ where $P\ (x)$ and $Q\ (x)$ are two polynomials. Function $f\ (x)$ is a Rational Expression. $f\ (y)$ = sin $y$ is not a Rational Function. Here polynomial in denominator cannot be zero that is $B\ (y) \neq 0$.

Let’s see how to add subtract rational expressions. Following steps are used:

1)
First of all, factor all the values and get the Least Common Denominator.

2)
Then multiply the numerators and denominators with such a denominator that makes the denominator common.

3)
Add the numerators together by taking L.H.S.

4)
Factor the numerator.

5)
Finally, cancel all common factors.

Let’s take an example to understand the concept of addition and subtraction of rational expressions.

$\frac{3y + 1}{(y^{2} – 1)}$ + $\frac{y}{(y + 1)}$ = $\frac{3y + 1}{(y + 1)}$ $\frac{(y– 1) + y}{(y + 1)}$, least common denominator is $(y + 1) (y – 1)$.

Then $[\frac{3y + 1}{(y + 1) (y– 1)}]$ + $[\frac{y}{(y + 1)}$ $\frac{(y – 1)}{(y – 1)}]$,

= $[\frac{(3y + 1) + y (y – 1)}{[(y + 1) (y– 1)]}]$,

= $[\frac{(3y + 1 + y^{2} – y)]}{[(y + 1) (y– 1)}]$,

= $[\frac{(y^{2} + 2y +1)]}{[(y + 1) (y– 1)}]$,

Fraction of numerator part will be $(y + 1)^{2}$,

Then final solution will be,

$\frac{3y + 1}{(y^{2} – 1)}$ + $\frac{y}{(y + 1)}$ = $\frac{(y + 1)}{(y – 1)}$.

## Inverse Variation

Statements of the inverse variations can be stated as follows:

Suppose we have two variables: “$a$” and “$b$” such that,

1) Value of “$b$” differs inversely as that of “$a$”,

2) “$b$” is inversely proportional to “$a$”.

Actual relation can be written as:

$b$ = $f\ a$, where '$f$' is some fixed real number and not equals to zero. “$f$” is also known as variation constant or proportionality constant.

Let us understand the formula for Inverse Variation:

Connection that we just saw between two variables, will always give multiplication as a constant. There are certain steps which have to be followed in order to retrieve the inverse variation relation among the variables. We have to interpret the statement in form of an inverse variation relation.

Step 1:

According to 2nd Point as shown above, “$b$” is inversely proportional to “$a$” values which means: $b$ = $f x$,

Step 2:

Next put the known values i.e. “$a$” and “$b$” in the equation, we wrote in step 1 to get the value of non - zero real number '$f$'.

Step 3:

Lastly, put the value of '$k$' in the original equation to get the desired equation.
Example:

If $b$ = $20$ when $a$ = $4$, then write the formula for relation between '$a$' and '$b$' given '$b$' is inversely proportional to '$a$'?

Step 1:

Convert the given statement to an inverse variation formula. In our example we will write it as follows: “$b$” is inversely proportional to “$a$” means

$b$ = $f \times a$,

Step 2:

Now substitute the known values i.e. at $a$ = $4$ we have $b$ = $20$. To get the corresponding value of '$f$' as follows:

$20$ = $f \times 4$,

Or $f$ = $5$,

So, general relation that we can write using value of '$f$' as:

$b$ = $5a$.

## Indirect Variations

Indirect variations are exactly opposite of Direct Variations. In direct variations we study that two variables are directly proportional to each other but here in indirect variations, variables are inversely proportional to each other. For example, relation between two variables values like speed and time have an inverse relation as when speed increases the time decreases.

So, if we increase our speed by twice its value, then time decrease by two times. We can say that product of two variables in indirect variation remains constant as compared to direct variations, where ration of two quantities remains constant.

General expression for indirect variation or indirect proportionality can be given by considering two variables '$n$' and '$m$'. '$n$' is inversely proportional to '$m$' and which is given as follows:

$N \times m$ = $k$ or $n$ = $\frac{k}{m}$,

Let us consider an example of indirect variation to understand it better.
Example:

If a car has an average speed of $80$ miles per hour and it takes $7$ hours to complete its journey, then how long it would take to complete the same journey with a speed of $100$ miles per hour?

Solution:

Let speed be denoted by variable '$s$' and time by '$t$'. Distance can be calculated as follows:

$D$ = $s \times t$ = $80 \times 7$ = $560$ miles,

To calculate the time taken to complete the same distance at a speed of $100$ miles per hour

$t$ = $\frac{560}{100}$ = $5.6$ hours.

This shows that speed and time are in indirect relation and thus are inversely proportional to each other. We saw that at speed $80$ miles / hour, time taken was $7$ hours and at speed $100$ miles / hour, time taken is $5.6$ hours.

## Direct Variations

Two variable quantities in Math are said to be in direct variation if, proportion of their values does not change.

For instance, suppose we have two variables “$a$” and “$b$” such that they are in a relation $a$ = $4b$ i.e. value of “$a$” will always be four times the corresponding value of “$b$” or we can say that the values of “$a$” and “$b$” are in a fixed Ratio $\frac{1}{4}$. That is, if value of '$b$' is $4$ then $a$’s value will be $16$ and similarly if $a$’s value is $20$, $b$’s value will be $5$.

To define the direct variations more specifically, let's say there are two variables $A$ and $B$ such that $A$ varies directly as $B$. When we plot the value of $A$ and $B$, then we get a straight line with all pairs of coordinates lying on this line. Also this line would pass through the origin i.e. $(0,\ 0)$. We can summarize as: Each variable is a constant multiple of every other number.

Direct variations can be expressed in form of equations as follows:

Suppose we have an equation $\frac{m}{n}$ = $10$. It states a direct variation between two variables $m$ and $n$. Constant ratio that is maintained among the random values of $m$ and $n$ is $10$. Direct variations can also be represented as algebraic expressions.

For example, we have an expression given as: $m$ = $x\ n$. Where, '$x$' is a constant non zero factor that can be termed as a constant ratio maintained between the two variables '$m$' and '$n$'. For any random value of '$m$' and '$n$' we can calculate various values of '$x$'. This can lead to many direct variations among different Sets of values of '$m$' and '$n$'.

## Solving Rational Equations

Solving a rational equation is very similar to solving normal mathematical equation. Rational equation is in the form of $\frac{p}{q}$ where one of them can be unknown variable. Let us see how to solve rational equations.

As we studied above that rational equations contain terms in Ratio form.

For Example: $\frac{4}{x + 7}$ = $\frac{2}{3x}$ is a rational equation because it has $x$ in denominator. Let us see steps for         solving rational equations:

Step 1: Assume that we have a rational equation $\frac{4}{x + 7}$ = $\frac{2}{3x}$.

Step 2: Now find the Least Common Denominator of all the Fractions present in the rational equation. In above rational equation least common denominator is $3x$.

Step 3: Now we will multiply all the terms by least common denominator.

$\frac{(3x)\ 4}{x}$ + $(3x)\ 7$ = $\frac{(3x)\ 2}{3x}$.

Step 4: Now just simplify all the terms of rational equation.

Step 5: We get $12 + 21x$ = $2$

Step 6: Now we have a simplified equation. From this equation we can easily find the value of variable '$x$'.

Step 7: Separate the variables and constant terms by moving them on either sides of equal sign.

$21x$ = - $12 + 2$,

$21x$ = - $10$,

$X$ = $\frac{-10}{21}$

## Multiply and Divide Rational Expression

Rational expression is in the form of $\frac{p}{q}$. Here $p$ and $q$ are two real expressions. We can perform arithmetic operations on rational expressions such as addition, subtraction, multiplication and division. Let us discuss multiplying and dividing rational expressions.

For multiplication we need two rational expressions. Rational expression has numerator and denominator. '$p$' is numerator and '$q$' is denominator.
Let us discuss steps for multiplication and division of rational expressions:

Step 1: Assume that we have two Rational Numbers $\frac{2x}{3}$ and $\frac{4y}{5}$.

Step 2: For multiplication we multiply the numerator with numerator and denominator with denominator.

Step 3: When we multiply these rational numbers $\frac{2x}{3}$ and $\frac{4y}{5}$ we will get $\frac{2x}{3}$ $\times$ $\frac{4y}{5}$ = $\frac{8xy}{15}$.

Step 4: Simplify the rational number.
Steps to find the division of rational expressions are:

Step 1: Division of rational expression is very similar to multiplication of rational expressions. In division of rational expressions we just reverse the second rational expression. Assume that we have two rational expressions $\frac{(2x + 2)}{(x + 1)}$ and $\frac{(2x + 1)}{(x + 1)}$.

Step 2: Now reverse the second rational expression as $\frac{(x + 1)}{(2x + 1)}$.

Step 3: Follow the same procedure of multiplication.

Step 4: Multiply $\frac{(2x + 2)}{(x + 1)}$ and $\frac{(x + 1)}{(2x + 1)}$.

Step 5: Division of $[\frac{(2x + 2)}{(x + 1)}]$ $\times$ $[\frac{(x + 1)}{(2x + 1)}]$.

Step 6: Result of division will be $\frac{2x+2}{2x+1}$.

These are the steps to multiply and divide rational expressions.