I am going to try again.

When you want to eliminate fractions from an

equation, you multiply both sides by the least common multiple of the denominators. You multiplied both sides by 2, WHICH IS NOT A

DENOMINATOR.

[tex]2 * \left (\dfrac{\dfrac{a - b}{b}}{0.5} + \dfrac{\dfrac{a - c}{c}}{0.5} + \dfrac{a - b}{b} + \dfrac{a - c}{c} \right ) = 2 * 2 \implies[/tex]

[tex]\dfrac{\dfrac{2(a - b)}{b}}{0.5} + \dfrac{\dfrac{2(a - c)}{c}}{0.5} + \dfrac{2(a - b)}{b} + \dfrac{2(a - c)}{c} = 4.[/tex]

So you have still not got rid of the denominators of 0.5.

One way to get rid of that decimal denominator is to multiply by 0.5, WHICH IS A DENOMINATOR.

[tex]0.5* \left (\dfrac{\dfrac{a - b}{b}}{0.5} + \dfrac{\dfrac{a - c}{c}}{0.5} + \dfrac{a - b}{b} + \dfrac{a - c}{c} \right ) = 0.5 * 2 \implies[/tex]

[tex]\dfrac{a - b}{b} + \dfrac{a - c}{c} + \dfrac{0.5(a - b)}{b} + \dfrac{0.5(a - c)}{c} = 1.[/tex]

Now you can get rid of the decimals by multiplying both sides of the equation by 2 to get:

[tex]\dfrac{2(a - b)}{b} + \dfrac{2(a - c)}{c} + \dfrac{(a - b)}{b} + \dfrac{(a - c)}{c} = 2 \implies[/tex]

[tex]\dfrac{3(a - b)}{b} + \dfrac{3(a - c)}{c} = 2.[/tex]

And you have yet to eliminate fractions. There is another way.

[tex]\dfrac{\dfrac{a - b}{b}}{0.5} + \dfrac{\dfrac{a - c}{c}}{0.5} + \dfrac{a - b}{b} + \dfrac{a - c}{c} = 2 \implies[/tex]

[tex]\dfrac{\dfrac{a - b}{b}}{\dfrac{1}{2}}+ \dfrac{\dfrac{a - c}{c}}{\dfrac{1}{2}} + \dfrac{a - b}{b} + \dfrac{a - c}{c} = 2 \implies[/tex]

[tex]\dfrac{a - b}{b} * \dfrac{2}{1} + \dfrac{a - c}{c} * \dfrac{2}{1} + \dfrac{a - b}{b} + \dfrac{a - c}{c} = 2 \implies[/tex]

[tex]\dfrac{2(a - b)}{b} + \dfrac{2(a - c)}{c} + \dfrac{(a - b)}{b} + \dfrac{(a - c)}{c} = 2 \implies[/tex]

[tex]\dfrac{3(a - b)}{b} + \dfrac{3(a - c)}{c} = 2.[/tex]

A different way to get to the same place: simplify fractions before eliminating them. Now eliminate fractions.

[tex]\dfrac{3(a - b)}{b} + \dfrac{3(a - c)}{c} = 2 \implies[/tex]

[tex]bc * \left ( \dfrac{3(a - b)}{b} + \dfrac{3(a - c)}{c} \right ) = bc * 2 \implies[/tex]

[tex]3c(a - b) + 3b(a - c) = 2bc \implies 3ac - 3bc + 3ab - 3bc = 2bc \implies[/tex]

[tex]a\{3(b + c)\} = 2bc + 3bc + 3bc = 8bc \implies a= \dfrac{8bc}{3(b + c)}.[/tex]

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