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Rational expression is a fraction in which denominator and numerator are polynomials or we can say that the ratio of two polynomials. A term which is in the form $\frac{P(x)}{Q(x)}$, where, P and Q are the function of $x$, and $Q(x)$ is not equal to zero. “$P(x)$” is numerator, and “$Q(x)$” is denominator.

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## How to Add Rational Expressions?

With Like Denominator:

If denominators of the expressions are same, then directly add fractions numerators.

$\frac{a(x)}{b(x)}$ and $\frac{c(x)}{b(x)}$ (Expression with same denominator, $b(x))$

$\frac{a(x)}{b(x)}$ + $\frac{c(x)}{b(x)}$ = $\frac{a(x) + c(x)}{b(x)}$

Let us solve expression with same denominator:

$\frac{2x+7}{3x-4}$ + $\frac{4x-11}{3x-4}$

$\frac{2x+7}{3x-4}$ + $\frac{4x-11}{3x-4}$

= $\frac{2x+7 + 4x - 11}{3x-4}$

= $\frac{6x - 4}{3x-4}$

With Unlike Denominator:

Solve this expression using LCM method.

Add  $\frac{a(x)}{b(x)}$ and $\frac{c(x)}{d(x)}$ ($b(x)$ and $d(x)$ are different denominators)

$\frac{a(x)}{b(x)}$ + $\frac{c(x)}{d(x)}$

LCM of $b(x)$ and $d(x)$ = $b(x)d(x)$

$\frac{a(x)d(x) + c(x)b(x)}{d(x)b(x)}$

## Adding Rational Expressions with Different Denominators

Consider two rational numbers $\frac{a(x)}{b(x)}$ and $\frac{c(x)}{d(x)}$ .

Step 1: First, we have to take L.C.M(Lowest common multiplier) of denominators.

So, L.C.M. of “$b(x)$” and “$d(x)$” is $b(x)d(x)$.

Step 2: Now, L.C.M. of denominators will be divided by the denominator of first term.

$\frac{b(x)d(x)}{b(x)}$ = $d(x)$,

So, we multiply $d(x)$ with the numerator of first term “$a(x)$”.

So, $a(x)d(x)$.

Step 3: Now, L.C.M. of denominators will be divided by the denominator of second term.

$\frac{b(x)d(x)}{d(x)}$ = $b(x)$,

So, we multiply $b(x)$ with the numerator of second term “$c(x)$”.

$b(x)c(x)$.

Step 4: Now, we have to add result of Step 2 and Step 3: $a(x)d(x)\ +\ b(x)c(x)$

Step 5: So, $\frac{a(x)d(x) + b(x)c(x)}{b(x)d(x)}$. (This is the final answer.)

Given below are some solved examples:

### Solved Examples

Question 1: Solve $\frac{2x + 5}{3x - 4} + \frac{6x - 11}{3x - 4}$
Solution:
L.C.M of (3x - 4) and (3x - 4) is (3x - 4).

This is the expression with same denominators. So, simply add the numerators.

So, $\frac{3x - 4}{3x - 4} = 1$ and (2x + 5)(1) = (2x + 5).

So, $\frac{3x - 4}{3x - 4} = 1$ and (6x - 11)(1) = (6x - 11).

Now, $\frac{(2x + 5) + (6x - 11)}{3x - 4}$ = $\frac{8x - 6}{3x - 4}$

Question 2: $\frac{3x + 2}{5x - 2} + \frac{8x - 3}{5x + 2}$
Solution:
L.C.M of (5x - 2) and (5x + 2) is (5x)^2 – 2^2.

So, $\frac{5x^2 - 2^2}{5x - 2}$ = (5x + 2) and (3x + 2)(5x + 2)

= 15x2 + 16x + 4.

So, $\frac{5x^2 - 2^2}{5x - 2}$ = (5x - 2) and (8x - 3)(5x - 2)

= 40x2 - 31x + 6 .

Now, $\frac{(15x^2 + 16x + 4) + (40x^2 - 31x + 6)}{(5x)^2 - 2^2}$

= $\frac{55x^2 - 15x + 10}{5x^2 - 2^2}$

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