Question

If x=6ab/a+b, find the value of x+3a/x-3a + x+3b/x-3b

Answer 1

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Answer 2

Therefore the value of  $\frac{x+3a}{x-3a} +\frac{x+3b}{x-3b}$ is $2$

Step-by-step explanation:

Given;

If $x=\frac{6ab}{a+b}$ then find $\frac{x+3a}{x-3a} +\frac{x+3b}{x-3b}=?$

∵$x=\frac{6ab}{a+b}$

⇒$x=\frac{3a\times2b}{a+b}$

⇒$\frac{x}{3a} =\frac{2b}{a+b}$

⇒$\frac{x+3a}{x-3a} =\frac{2b+a+b}{2b-a-b}$

∴$\frac{x+3a}{x-3a} =\frac{a+3b}{b-a}$___equation-1

Again;

$x=\frac{2a\times3b}{a+b}$

⇒$\frac{x}{3b} =\frac{2a}{a+b}$

⇒$\frac{x+3b}{x-3b} =\frac{2a+a+b}{2a-a-b}$

⇒$\frac{x+3b}{x-3b} =\frac{3a+b}{a-b}$___equation-2

By adding equation-1 and equation-2;

  $\frac{x+3a}{x-3a}+\frac{x+3b}{x-3b}=\frac{a+3b}{b-a}+\frac{3a+b}{a-b}$

⇒$\frac{x+3a}{x-3a}+\frac{x+3b}{x-3b}=-\frac{a+3b}{a-b}+\frac{3a+b}{a-b}$

⇒$\frac{x+3a}{x-3a}+\frac{x+3b}{x-3b}=\frac{-a-3b+3a+b}{a-b}$

⇒$\frac{x+3a}{x-3a}+\frac{x+3b}{x-3b}=\frac{2a-2b}{a-b}$

⇒$\frac{x+3a}{x-3a}+\frac{x+3b}{x-3b}=\frac{2(a-b)}{a-b}$

∴$\frac{x+3a}{x-3a}+\frac{x+3b}{x-3b}=2$

Thus, the value of  $\frac{x+3a}{x-3a} +\frac{x+3b}{x-3b}$ is $2$