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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.

Advanced method

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

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

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

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

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

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

Add both algebraic equations

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.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$=\,\,\, 2$

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



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