Question

How do you simplify $ \dfrac{{2x - 1}}{{(x - 3)(x + 2)}} + \dfrac{{x - 4}}{{x - 3}}? $

Answer 1

Hint: As we know that this question consists of polynomials. We will first bring both the fractions with common denominators and then we factorise if needed and then solve it. Here in this question we have to find the product of two polynomials, we just multiply each term of the first polynomial by each term of the second polynomial and then simplify or if there is any algebraic identity possible we can apply that.

Complete step-by-step answer:
Here we have, $ \dfrac{{2x - 1}}{{(x - 3)(x + 2)}} + \dfrac{{x - 4}}{{x - 3}} $ .
Here we will first multiply the numerator and denominator of $ \dfrac{{x - 4}}{{x - 3}} $ with $ (x + 2) $ . So the new dfraction can be written as
 $ \dfrac{{2x - 1}}{{(x - 3)(x + 2)}} + \dfrac{{(x - 4)(x + 2)}}{{(x - 3)(x + 2)}} $ .
We will now solve the numerator by multiplying the factors of the second fraction: $ \dfrac{{2x - 1}}{{(x - 3)(x + 2)}} + \dfrac{{{x^2} - 2x - 8}}{{(x - 3)(x + 2)}} $ .
Since now both the denominators are same so we will add the numerators :
 $ \dfrac{{{x^2} + 2x - 2x - 8 - 1}}{{(x - 3)(x + 2)}} = \dfrac{{{x^2} - 9}}{{(x - 3)(x + 2)}} $ .
We can see the numerator can be written as the form by the difference of square formula $ {a^2} - {b^2} = (a + b)(a - b) $ , similarly we can write
 $ {x^2} - {3^2} = (x + 3)(x - 3) $ .
We will substitute the values now and we have:
 $ \dfrac{{(x - 3)(x + 3)}}{{(x - 3)(x + 2)}} $ .
By cancelling the common factors it gives us $ \dfrac{{x + 3}}{{x + 2}} $
Hence the required simplified value is $ \dfrac{{x + 3}}{{x + 2}} $ .
So, the correct answer is “ $ \dfrac{{x + 3}}{{x + 2}} $ ”.

Note: We should know that the algebraic identity used in the above solution is called the square of the difference of the terms formula, which is also a binomial expression or also called the special binomial product rule. It is expanded as the subtraction of two times the product of two terms from the sum of the squares of the given terms. Also we should be careful with the positive and negative sign as in the above solution which gives the positive value not the negative value