Question

If \[a = \dfrac{{4xy}}{{x + y}}\], the value of \[\dfrac{{a + 2x}}{{a - 2x}} + \dfrac{{a + 2y}}{{a - 2y}}\] in most simplified form is
A. 0
B. 1
C. \[ - 1\]
D. 2

Answer 1

Hint:
Here we will first put the value of \[a\] in the equation. Then we will take the LCM of the equation and simplify it. Then we will solve the equation to get the value of the expression in the most simplified form.

Complete Step by step Solution:
Given equation is \[a = \dfrac{{4xy}}{{x + y}}\].
We will find the value of the expression \[\dfrac{{a + 2x}}{{a - 2x}} + \dfrac{{a + 2y}}{{a - 2y}}\] by putting the value of \[a\] given in the expression. Therefore, we get
\[ \Rightarrow \dfrac{{a + 2x}}{{a - 2x}} + \dfrac{{a + 2y}}{{a - 2y}} = \dfrac{{\dfrac{{4xy}}{{x + y}} + 2x}}{{\dfrac{{4xy}}{{x + y}} - 2x}} + \dfrac{{\dfrac{{4xy}}{{x + y}} + 2y}}{{\dfrac{{4xy}}{{x + y}} - 2y}}\]
Now we will take the LCM of \[x + y\] in the numerator as well as in the denominator of the above equation. Therefore, we get
\[ \Rightarrow \dfrac{{a + 2x}}{{a - 2x}} + \dfrac{{a + 2y}}{{a - 2y}} = \dfrac{{\dfrac{{4xy + 2x\left( {x + y} \right)}}{{x + y}}}}{{\dfrac{{4xy - 2x\left( {x + y} \right)}}{{x + y}}}} + \dfrac{{\dfrac{{4xy + 2y\left( {x + y} \right)}}{{x + y}}}}{{\dfrac{{4xy - 2y\left( {x + y} \right)}}{{x + y}}}}\]
We will cancel out the term \[x + y\] which is present in the numerator as well as in the denominator. Therefore, we get
\[ \Rightarrow \dfrac{{a + 2x}}{{a - 2x}} + \dfrac{{a + 2y}}{{a - 2y}} = \dfrac{{4xy + 2x\left( {x + y} \right)}}{{4xy - 2x\left( {x + y} \right)}} + \dfrac{{4xy + 2y\left( {x + y} \right)}}{{4xy - 2y\left( {x + y} \right)}}\]
Now we will expand the equation by opening the brackets of the equation, we get
\[ \Rightarrow \dfrac{{a + 2x}}{{a - 2x}} + \dfrac{{a + 2y}}{{a - 2y}} = \dfrac{{4xy + 2{x^2} + 2xy}}{{4xy - 2{x^2} - 2xy}} + \dfrac{{4xy + 2xy + 2{y^2}}}{{4xy - 2xy - 2{y^2}}}\]
Simplifying the equation further, we get
\[ \Rightarrow \dfrac{{a + 2x}}{{a - 2x}} + \dfrac{{a + 2y}}{{a - 2y}} = \dfrac{{6xy + 2{x^2}}}{{2xy - 2{x^2}}} + \dfrac{{6xy + 2{y^2}}}{{2xy - 2{y^2}}}\]
Now we will take the common terms in the numerator and denominator as well. Therefore, we get
\[ \Rightarrow \dfrac{{a + 2x}}{{a - 2x}} + \dfrac{{a + 2y}}{{a - 2y}} = \dfrac{{2x\left( {3y + x} \right)}}{{2x\left( {y - x} \right)}} + \dfrac{{2y\left( {3x + y} \right)}}{{2y\left( {x - y} \right)}}\]
Cancelling out the similar terms in the numerator as well as in the denominator of the equation, we get
\[ \Rightarrow \dfrac{{a + 2x}}{{a - 2x}} + \dfrac{{a + 2y}}{{a - 2y}} = \dfrac{{3y + x}}{{y - x}} + \dfrac{{3x + y}}{{x - y}}\]
We can write the above equation as
\[ \Rightarrow \dfrac{{a + 2x}}{{a - 2x}} + \dfrac{{a + 2y}}{{a - 2y}} = \dfrac{{3y + x}}{{y - x}} - \dfrac{{3x + y}}{{y - x}}\]
Now we will take the LCM \[y - x\] for both the terms. Therefore, we get
\[ \Rightarrow \dfrac{{a + 2x}}{{a - 2x}} + \dfrac{{a + 2y}}{{a - 2y}} = \dfrac{{3y + x - \left( {3x + y} \right)}}{{y - x}}\]
\[ \Rightarrow \dfrac{{a + 2x}}{{a - 2x}} + \dfrac{{a + 2y}}{{a - 2y}} = \dfrac{{3y + x - 3x - y}}{{y - x}}\]
Adding and subtracting the like terms, we get
\[ \Rightarrow \dfrac{{a + 2x}}{{a - 2x}} + \dfrac{{a + 2y}}{{a - 2y}} = \dfrac{{2y - 2x}}{{y - x}}\]
Taking 2 common in the numerator of the equation, we get
\[ \Rightarrow \dfrac{{a + 2x}}{{a - 2x}} + \dfrac{{a + 2y}}{{a - 2y}} = \dfrac{{2\left( {y - x} \right)}}{{y - x}}\]
Cancelling the like terms, we get
\[ \Rightarrow \dfrac{{a + 2x}}{{a - 2x}} + \dfrac{{a + 2y}}{{a - 2y}} = 2\]
Hence, the value of \[\dfrac{{a + 2x}}{{a - 2x}} + \dfrac{{a + 2y}}{{a - 2y}}\] in most simplified form is equal to 2.

So, option D is the correct option.

Note:
Here, we need to know how to take the LCM of the numbers in the form of the ratio. We have to simplify the equation according to the given options and solve accordingly. We have to perform the basic arithmetic operation to simplify and solve the solve equation. To get the common terms we have to take the maximum factor of the terms. Factors are the smallest numbers with which the given number is divisible and their multiplication will give the original number. We can cancel out the terms which are present in the numerator and present in the denominator of the equation.