If $x \,=\, \dfrac{6ab}{a+b}$, find the value of $\dfrac{x+3a}{x-3a}$ $+$ $\dfrac{x+3b}{x-3b}$

The value of $x$ is expressed as an algebraic expression in terms of $a$ and $b$.

$x \,=\, \dfrac{6ab}{a+b}$

On the basis of this algebraic equation, the following algebraic expression has to be evaluated.

$\dfrac{x+3a}{x-3a}$ $+$ $\dfrac{x+3b}{x-3b}$

The value of this algebraic expression can be evaluated in two mathematical approaches.

The value of the following algebraic expression can be evaluated by an advanced mathematical approach. It is a recommendable method for those who have good knowledge on mathematics.

$\dfrac{x+3a}{x-3a}$ $+$ $\dfrac{x+3b}{x-3b}$

In this method, the given algebraic equation is transformed as each term of the algebraic expression.

Find the value of first term

$\dfrac{x+3a}{x-3a}$ is a first term in the algebraic expression and its value can be evaluated by the given algebraic equation $x \,=\, \dfrac{6ab}{a+b}$

$x$ and $3a$ are two terms in both numerator and denominator of the expression. They both are connected by sum and subtraction. This term can be obtained possibly by making $3a$ as denominator to $x$. Then, use componendo and dividendo rule to complete the transformation of given algebraic equation as first algebraic term of the expression.

$\implies$ $x \,=\, \dfrac{3a \times 2b}{a+b}$

$\implies$ $\dfrac{x}{3a} \,=\, \dfrac{2b}{a+b}$

Use componendo and dividendo rule to get the value of first term of the expression.

$\implies$ $\dfrac{x+3a}{x-3a} \,=\, \dfrac{2b+a+b}{2b-(a+b)}$

$\implies$ $\dfrac{x+3a}{x-3a} \,=\, \dfrac{a+3b}{2b-a-b}$

$\,\,\, \therefore \,\,\,\,\,\,$ $\dfrac{x+3a}{x-3a} \,=\, \dfrac{a+3b}{b-a}$

Find the value of second term

In the same way, get the value of second algebraic term of the expression from the equation $x \,=\, \dfrac{6ab}{a+b}$

$\implies$ $x \,=\, \dfrac{2a \times 3b}{a+b}$

$\implies$ $\dfrac{x}{3b} \,=\, \dfrac{2a}{a+b}$

$\implies$ $\dfrac{x+3b}{x-3b} \,=\, \dfrac{2a+a+b}{2a-(a+b)}$

$\implies$ $\dfrac{x+3b}{x-3b} \,=\, \dfrac{3a+b}{2a-a-b}$

$\,\,\, \therefore \,\,\,\,\,\,$ $\dfrac{x+3b}{x-3b} \,=\, \dfrac{3a+b}{a-b}$

The values of two terms of the algebraic expression are evaluated in the above two steps.

$(1) \,\,\,\,\,\,$ $\dfrac{x+3a}{x-3a} \,=\, \dfrac{a+3b}{b-a}$

$(2) \,\,\,\,\,\,$ $\dfrac{x+3b}{x-3b} \,=\, \dfrac{3a+b}{a-b}$

Now, add both equations and then simplify it to obtain the value of the algebraic expression.

$\implies$ $\dfrac{x+3a}{x-3a}$ $+$ $\dfrac{x+3b}{x-3b}$ $\,=\,$ $\dfrac{a+3b}{b-a}$ $+$ $\dfrac{3a+b}{a-b}$

$\implies$ $\dfrac{x+3a}{x-3a}$ $+$ $\dfrac{x+3b}{x-3b}$ $\,=\,$ $\dfrac{a+3b}{b-a}$ $+$ $\dfrac{3a+b}{-(b-a)}$

$\implies$ $\dfrac{x+3a}{x-3a}$ $+$ $\dfrac{x+3b}{x-3b}$ $\,=\,$ $\dfrac{a+3b}{b-a}$ $-$ $\dfrac{3a+b}{b-a}$

$\implies$ $\dfrac{x+3a}{x-3a}$ $+$ $\dfrac{x+3b}{x-3b}$ $\,=\,$ $\dfrac{a+3b-(3a+b)}{b-a}$

$\implies$ $\dfrac{x+3a}{x-3a}$ $+$ $\dfrac{x+3b}{x-3b}$ $\,=\,$ $\dfrac{a+3b-3a-b}{b-a}$

$\implies$ $\dfrac{x+3a}{x-3a}$ $+$ $\dfrac{x+3b}{x-3b}$ $\,=\,$ $\dfrac{a-3a+3b-b}{b-a}$

$\implies$ $\dfrac{x+3a}{x-3a}$ $+$ $\dfrac{x+3b}{x-3b}$ $\,=\,$ $\dfrac{-2a+2b}{b-a}$

$\implies$ $\dfrac{x+3a}{x-3a}$ $+$ $\dfrac{x+3b}{x-3b}$ $\,=\,$ $\dfrac{2b-2a}{b-a}$

$\implies$ $\dfrac{x+3a}{x-3a}$ $+$ $\dfrac{x+3b}{x-3b}$ $\,=\,$ $\dfrac{2(b-a)}{b-a}$

$\implies$ $\dfrac{x+3a}{x-3a}$ $+$ $\dfrac{x+3b}{x-3b}$ $\,=\,$ $2 \times \dfrac{b-a}{b-a}$

$\implies$ $\dfrac{x+3a}{x-3a}$ $+$ $\dfrac{x+3b}{x-3b}$ $\,=\,$ $2 \times \require{cancel} \dfrac{\cancel{b-a}}{\cancel{b-a}}$

$\implies$ $\dfrac{x+3a}{x-3a}$ $+$ $\dfrac{x+3b}{x-3b}$ $\,=\,$ $2 \times 1$

$\,\,\, \therefore \,\,\,\,\,\,$ $\dfrac{x+3a}{x-3a}$ $+$ $\dfrac{x+3b}{x-3b}$ $\,=\,$ $2$

Direct method

The value of algebraic expression $\dfrac{x+3a}{x-3a}$ $+$ $\dfrac{x+3b}{x-3b}$ can be evaluated directly by replacing $x$ by $\dfrac{6ab}{a+b}$

$\dfrac{x+3a}{x-3a}$ $+$ $\dfrac{x+3b}{x-3b}$ $\,=\,$ $\dfrac{\dfrac{6ab}{a+b}+3a}{\dfrac{6ab}{a+b}-3a}$ $+$ $\dfrac{\dfrac{6ab}{a+b}+3b}{\dfrac{6ab}{a+b}-3b}$

Simplify the right hand side algebraic expression.

$=\,\,\,$ $\dfrac{\dfrac{6ab+3a(a+b)}{a+b}}{\dfrac{6ab-3a(a+b)}{a+b}}$ $+$ $\dfrac{\dfrac{6ab+3b(a+b)}{a+b}}{\dfrac{6ab-3b(a+b)}{a+b}}$

$=\,\,\,$ $\dfrac{6ab+3a(a+b)}{a+b}$ $\times$ $\dfrac{a+b}{6ab-3a(a+b)}$ $+$ $\dfrac{6ab+3b(a+b)}{a+b}$ $\times$ $\dfrac{a+b}{6ab-3b(a+b)}$

$=\,\,\,$ $\require{cancel} \dfrac{6ab+3a(a+b)}{\cancel{a+b}}$ $\times$ $\require{cancel} \dfrac{\cancel{a+b}}{6ab-3a(a+b)}$ $+$ $\require{cancel} \dfrac{6ab+3b(a+b)}{\cancel{a+b}}$ $\times$ $\require{cancel} \dfrac{\cancel{a+b}}{6ab-3b(a+b)}$

$=\,\,\,$ $\dfrac{6ab+3a(a+b)}{6ab-3a(a+b)}$ $+$ $\dfrac{6ab+3b(a+b)}{6ab-3b(a+b)}$

$=\,\,\,$ $\dfrac{6ab+3a^2+3ab}{6ab-3a^2-3ab}$ $+$ $\dfrac{6ab+3ba+3b^2}{6ab-3ba-3b^2}$

$=\,\,\,$ $\dfrac{6ab+3ab+3a^2}{6ab-3ab-3a^2}$ $+$ $\dfrac{6ab+3ab+3b^2}{6ab-3ab-3b^2}$

$=\,\,\,$ $\dfrac{9ab+3a^2}{3ab-3a^2}$ $+$ $\dfrac{9ab+3b^2}{3ab-3b^2}$

$=\,\,\,$ $\dfrac{3a(3b+a)}{3a(b-a)}$ $+$ $\dfrac{3b(3a+b)}{3b(a-b)}$

$=\,\,\,$ $\require{cancel} \dfrac{\cancel{3a}(3b+a)}{\cancel{3a}(b-a)}$ $+$ $\require{cancel} \dfrac{\cancel{3b}(3a+b)}{\cancel{3b}(a-b)}$

$=\,\,\,$ $\dfrac{3b+a}{b-a}$ $+$ $\dfrac{3a+b}{a-b}$

$=\,\,\,$ $\dfrac{3b+a}{b-a}$ $+$ $\dfrac{3a+b}{-(b-a)}$

$=\,\,\,$ $\dfrac{3b+a}{b-a}$ $-$ $\dfrac{3a+b}{b-a}$

$=\,\,\,$ $\dfrac{3b+a-(3a+b)}{b-a}$

$=\,\,\,$ $\dfrac{3b+a-3a-b}{b-a}$

$=\,\,\,$ $\dfrac{3b-b+a-3a}{b-a}$

$=\,\,\,$ $\dfrac{2b-2a}{b-a}$

$=\,\,\,$ $\dfrac{2(b-a)}{b-a}$

$=\,\,\,$ $\require{cancel} \dfrac{2\cancel{(b-a)}}{\cancel{b-a}}$

$=\,\,\, 2$

$\,\,\, \therefore \,\,\,\,\,\,$ $\dfrac{x+3a}{x-3a}$ $+$ $\dfrac{x+3b}{x-3b}$ $\,=\,$ $2$