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Rational Expressions

Expression is a finite combination of symbols that are well formed and rational expressions is that where the numerator and the denominator or both of them are polynomials.

$\text{Rational expression}$ = $\frac{\text{Algebraic Expressions(s)}}{\text{Algebraic Expressions(s)}}$

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Rational Expression Definition

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Rational expression is an expression where the numerator and the denominator or both of them are polynomials.

If p(x) and q(x) are two polynomials, q(x) $\neq$ 0, then the quotient $\frac{p(x)}{q(x)}$ is called a rational expression.

Where, p(x) is known as numerator and q(x) is known as the denominator of the rational expression.

Simplifying rational expression means reduced the expression into the lowest form.

Steps for simplifying the rational expressions:

Step 1: Factorize each of the two polynomials p(x) and q(x)
.

Step 2: Find highest common divisor of p(x) and q(x).

Step 3: If h.c.f. = 1 , then the given rational expression $\frac{p(x)}{q(x)}$ is in its lowest terms.

Step 4: If h.c.f $\neq$ 1, then divide the numerator p(x) and denominator q(x) by the h.c.f of p(x) and q(x).

Step 5: The rational expression obtained in Step 3 or Step 4 is in its lowest terms.

Let us reduce the rational expression $\frac{x^2-5x-6}{x^2+3x+2}$ to its lowest terms.

Let p(x) = x2 - 5x - 6 = (x - 6)(x + 1)

Let q(x) = x2 + 3x + 2 = (x + 2)(x + 1)

Clearly h.c.f of p(x), q(x) = (x + 1)

$\frac{x^2-5x-6}{x^2+3x+2}$ = $\frac{(x-6)(x +1)}{(x+2)(x+1)}$

= $\frac{x - 6}{x + 2}$ (cancel out h.c.f = (x + 1))

Rational Expressions

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Adding rational expressions:

If $\frac{p(x)}{q(x)}$ and $\frac{r(x)}{s(x)}$ are two rational expressions, then their addition is given as:

$\frac{p(x)}{q(x)}$ + $\frac{r(x)}{s(x)}$ = $\frac{p(x) s(x) + r(x) q(x)}{q(x)s(x)}$ where q(x) $\neq$ 0, s(x)$\neq$ 0.

Subtracting rational expressions:

If $\frac{p(x)}{q(x)}$ and $\frac{r(x)}{s(x)}$ are two rational expressions, then their subtraction is given as:

$\frac{p(x)}{q(x)}$ - $\frac{r(x)}{s(x)}$ = $\frac{p(x) s(x) - r(x) q(x)}{q(x) s(x)}$ where, q(x) $\neq$ 0, s(x)$\neq$ 0.

Adding Rational Expressions with Different Denominators

If P = $\frac{x + 1}{x - 1}$ and Q = $\frac{x - 1}{x + 1}$ then let us find P + Q

P + Q = $\frac{x + 1}{x - 1}$ + $\frac{x - 1}{x + 1}$

= $\frac{(x + 1)(x + 1) + (x - 1)(x - 1)}{(x + 1)(x - 1)}$

= $\frac{(x + 1)^2 + (x - 1)^2}{(x - 1)(x + 1)}$ = $\frac{x^2 + 2x + 1 + x^2 - 2x + 1}{(x + 1)(x - 1)}$

= $\frac{2x^2 + 2}{(x + 1)(x - 1)}$ = $\frac{2(x^2 + 1)}{(x + 1)(x - 1)}$

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Multiplying and Dividing Rational Expressions

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Multiplying rational expressions:

If $\frac{p(x)}{q(x)}$ and $\frac{r(x)}{s(x)}$ are rational expressions, then their multiplication is defined as,

$\frac{p(x)}{q(x)}$ $\times$ $\frac{r(x)}{s(x)}$ = $\frac{p(x)r(x)}{q(x)s(x)}$ where, q(x) $\neq$ 0, s(x) $\neq$ 0

Dividing rational expressions:

If $\frac{p(x)}{q(x)}$ and $\frac{r(x)}{s(x)}$ are rational expressions, then their division is defined as,

$\frac{\frac{p(x)}{q(x)}}{\frac{r(x)}{s(x)}}$ = $\frac{p(x)s(x)}{q(x)r(x)}$ where, q(x) $\neq$ 0, s(x) $\neq$ 0 and r(x) $\neq$ 0.

Undefined Rational Expressions

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Rational expression is an expression having numerator and the denominator or both of them are polynomials. A rational expression undefined when the denominator is 0. To find where a rational expression is undefined, set the denominator equal to 0.

Let us show that $\frac{x^2 + 1}{x - 2}$ is undefined at x = 2.

Given expression is $\frac{x^2 + 1}{x - 2}$

We know that, rational expression is undefined when its denominator is 0

Denominator = x - 2

Put x - 2 = 0

=> x = 2

Given expression is undefined at x = 2. Hence Proved

Complex Rational Expressions

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A complex fraction is a rational expression that has a fraction in its numerator, denominator or both.

Simplifying Complex Rational Expressions:


To solve complex expressions follow the steps given below:

Step 1: Rewrite the numerator and denominator in a single fraction.

Step 2: Adding and subtracting rational expressions.

Step 3: Divide the numerator by the denominator.

Step 4: Simplify the rational expression.

Let us simplify $\frac{3 + \frac{2}{x+3}}{x + \frac{1}{x + 3}}$

$\frac{3 + \frac{2}{x+3}}{x + \frac{1}{x + 3}}$

$\frac{\frac{3(x + 3) + 2}{x+3}}{\frac{x(x+3)+ 1}{x+3}}$

$\frac{\frac{3x + 9 + 2}{x + 3}}{\frac{x^2 + 3x + 1}{x + 3}}$

$\frac{3x+11}{x+3}$ * $\frac{x + 3}{x^2 + 3x + 1}$


Canceling common terms, we get

$\frac{3x + 11}{x^2 + 3x + 1}$

Solving Rational Expressions

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Given below are the some of the examples in solving rational expressions.

Solved Examples

Question 1: If P = $\frac{4x}{x^2 - 1}$ and Q = $\frac{x + 1}{x - 1}$, then find P . Q
Solution:
P . Q = $\frac{4x}{x^2 - 1}$ $\times$ $\frac{x + 1}{x - 1}$
= $\frac{4x(x + 1)}{(x^2 - 1)(x - 1)}$ = $\frac{4x(x + 1)}{(x - 1)(x + 1)(x - 1)}$ [Using identity, $x^2 - 1 = (x - 1)(x + 1)$]
= $\frac{4x}{(x - 1)(x - 1)}$ (Cancel (x + 1) which is common in both numerator and denominator)

Question 2: If A = $\frac{x}{x + 1}$ and B = $\frac{x^2}{x^2 - 1}$, find $\frac{A}{B}$
Solution:
$\frac{A}{B}$ = $\frac{\frac{x}{x + 1}}{\frac{x^2}{x^2 - 1}}$
= $\frac{x(x^2 - 1)}{x^2(x + 1)}$ = $\frac{x(x + 1)(x - 1)}{x^2(x + 1)}$
= $\frac{x - 1}{x}$ (Cancel x(x + 1) common factor from both numerator and denominator)

Rational Expressions Word Problems

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Given below are some of the word problems on rational expressions.

Solved Examples

Question 1: Find a fraction such that if 2 is added to the numerator and 1 to the denominator it reduces to $\frac{1}{2}$, and becomes $\frac{3}{5}$, if added 3 to the denominator.
Solution:
Let fraction be $\frac{x}{y}$

Case 1: When 2 is added to the numerator and 1 to the denominator fraction reduces to $\frac{1}{2}$.

$\frac{x + 2}{y + 1} = \frac{1}{2}$


2(x + 2) = y + 1

2x + 4 = y + 1

2x - y + 3 = 0 ......................(i)

Case 2: When 3 is added to the denominator fraction reduces to $\frac{3}{5}$.

$\frac{x}{y + 3} = \frac{3}{5}$

5x = 3y + 9

5x - 3y - 9 = 0 ...................(ii)

Solving (i) and (ii), we have

x = -18 and y = -33

Therefore, required fraction be $\frac{18}{33}$.

Question 2: The sum of a number and its reciprocal is $\frac{13}{6}$. Find the number.
Solution:
Let the number be x.

x + $\frac{1}{x}$ = $\frac{13}{6}$  (According to statement)

$\frac{x^2 + 1}{x}$ = $\frac{13}{6}$

6($x^2$ + 1) = 13$x$ (Cross multiplication)

6$x^2$ + 6 = 13$x$

6$x^2$ - 13$x$ + 6 = 0

6$x^2$ - 9$x$ - 4$x$ + 6 = 0

3$x$(2$x$ - 3) - 2(2$x$ - 3) = 0

(3$x$ - 2)(2$x$ - 3) = 0

3$x$ - 2 = 0   or  2$x$ - 3 = 0

$x$ =  $\frac{2}{3}$   or  x = $\frac{3}{2}$

The number may be $\frac{2}{3}$   or  $\frac{3}{2}$.

More topics in Rational Expressions
Adding and Subtracting Rational Expressions Multiplying Rational Expressions
Dividing Rational Expressions Simplifying Rational Expressions
Rational Expressions Word Problems
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