Question

If it is given that \[x > 3\], then which of the following is equivalent to \[\dfrac{1}{{\dfrac{1}{{x + 2}} + \dfrac{1}{{x + 3}}}}\]?
A. \[\dfrac{{2x + 5}}{{{x^2} + 5x + 6}}\]
B. \[\dfrac{{{x^2} + 5x + 6}}{{2x + 5}}\]
C. \[2x + 5\]
D. \[{x^2} + 5x + 6\]

Answer 1

Hint: We use the concepts of variables and expressions related to algebra of mathematics. We will clearly know about variables and also about like terms and unlike terms. And using those concepts and definitions, we will solve the given question.

Complete step by step answer:
A variable is an algebraic substituent, whose value will be changing according to situations and conditions. We represent variables by any lower-case alphabet or some Greek or Latin letters like \[x,y,z,a,b,c,\alpha ,\beta \] and etcetera. Terms containing the same variables with the same powers are said to be like terms. For example, take the terms \[4{x^2}\] and \[\dfrac{{{x^2}}}{2}\]. Here, both of them have the same variable with the same power which is \[{x^2}\]. So, these two terms are like terms.

Now, take other terms like, \[5xy\] and \[ - 3{x^2}y\]. These two terms have the same variables but the powers are not the same. So, these are not like terms. So, we can say that these are unlike terms. In variable algebra, like terms can be added or subtracted but unlike terms cannot be. Now the given question is \[\dfrac{1}{{\dfrac{1}{{x + 2}} + \dfrac{1}{{x + 3}}}}\]. In denominator, there are two fractions. So, up on taking LCM and simplifying, we get,
\[ \Rightarrow \dfrac{1}{{\dfrac{{(x + 3) + (x + 2)}}{{(x + 2)(x + 3)}}}}\]
Using the rule that like terms can be added, we get,
\[ \Rightarrow \dfrac{1}{{\dfrac{{2x + 5}}{{(x + 2)(x + 3)}}}}\]

Now we use a distributing rule which says that, \[(a + b)(c + d) = a(c + d) + b(c + d) = ac + ad + bc + bd\]
\[ \Rightarrow \dfrac{1}{{\dfrac{{2x + 5}}{{x(x + 3) + 2(x + 3)}}}} = \dfrac{1}{{\dfrac{{2x + 5}}{{{x^2} + 3x + 2x + 6}}}}\]
On simplification, we get,
\[ \Rightarrow \dfrac{1}{{\dfrac{{2x + 5}}{{{x^2} + 5x + 6}}}}\]
We know that \[\dfrac{1}{{\dfrac{a}{b}}} = \dfrac{b}{a}\]
\[ \therefore \dfrac{{{x^2} + 5x + 6}}{{2x + 5}}\]
So, this the value which is equivalent to \[\dfrac{1}{{\dfrac{1}{{x + 2}} + \dfrac{1}{{x + 3}}}}\].

Therefore, option B is the correct answer.

Note: Unlike terms cannot be added or subtracted, but both like terms and unlike terms can be multiplied or divided easily. We can use the formulas like \[{x^m} \times {x^n} = {x^{m + n}}\] and \[\dfrac{{{x^m}}}{{{x^n}}} = {x^{m - n}}\] also. Expressions with an addition or subtraction symbol between different unlike terms is called a polynomial.