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 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.

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

Steps for simplifying the rational expressions:

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

Let p(x) = x^{2} - 5x - 6 = (x - 6)(x + 1)

Let q(x) = x^{2} + 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))

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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.

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 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.

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

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

To solve complex expressions follow the steps given below:

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

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

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

Canceling common terms, we get

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

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

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)

= $\frac{4x(x + 1)}{(x^2 - 1)(x - 1)}$ = $\frac{4x(x + 1)}{(x - 1)(x + 1)(x - 1)}$ [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)

$\frac{A}{B}$ = $\frac{\frac{x}{x + 1}}{\frac{x^2}{x^2 - 1}}$

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

= $\frac{x - 1}{x}$ (Cancel x(x + 1) common factor from both numerator and denominator)

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

= $\frac{x - 1}{x}$ (Cancel x(x + 1) common factor from both numerator and denominator)

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 $\frac{18}{33}$.

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

2(x + 2) = y + 1

2x + 4 = y + 1

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

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

5x = 3y + 9

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

Solving (i) and (ii), we have

x = -18 and y = -33

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

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 $\frac{3}{2}$.

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 $\frac{3}{2}$.

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