Could I please get some help with solving this one? I think that it is similiar to the last post I had here:
https://www.freemathhelp.com/forum/...A-B)-B)-((A-C)-C)-1-for-A-in-terms-of-B-amp-C
But I am not sure about what I need to change. This is for personal use and not for school. I really appreciate the help I got the last time I posted.
Please see my work below:
(((A-B)/B)/0.5)+(((A-C)/C)/0.5)+((A-B)/B)+((A-C)/C)=2
Multiply everything by 2 to get rid of the decimals in the denominator:
((A-B)/B)+((A-C)/C)+2((A-B)/B)+2((A-C)/C)=4
Simplifying:
3((A-B)/B)+3((A-C)/C)=4
Simplifying:
((A-B)/B)+((A-C)/C)=1.33
Similar to last thread to solve from here - Multiply all by BC to remove denominators:
BC*((A/B)-1)+((A/C)-1)=1.33BC
AC-BC+AB-BC=1.33BC
AC+AB=3.33BC
A(C+B)=3.33BC
A=(3.33BC)/(C+B)
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.
\(\displaystyle 2 * \left (\dfrac{\dfrac{a - b}{b}}{0.5} + \dfrac{\dfrac{a - c}{c}}{0.5} + \dfrac{a - b}{b} + \dfrac{a - c}{c} \right ) = 2 * 2 \implies\)
\(\displaystyle \dfrac{\dfrac{2(a - b)}{b}}{0.5} + \dfrac{\dfrac{2(a - c)}{c}}{0.5} + \dfrac{2(a - b)}{b} + \dfrac{2(a - c)}{c} = 4.\)
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.
\(\displaystyle 0.5* \left (\dfrac{\dfrac{a - b}{b}}{0.5} + \dfrac{\dfrac{a - c}{c}}{0.5} + \dfrac{a - b}{b} + \dfrac{a - c}{c} \right ) = 0.5 * 2 \implies\)
\(\displaystyle \dfrac{a - b}{b} + \dfrac{a - c}{c} + \dfrac{0.5(a - b)}{b} + \dfrac{0.5(a - c)}{c} = 1.\)
Now you can get rid of the decimals by multiplying both sides of the equation by 2 to get:
\(\displaystyle \dfrac{2(a - b)}{b} + \dfrac{2(a - c)}{c} + \dfrac{(a - b)}{b} + \dfrac{(a - c)}{c} = 2 \implies\)
\(\displaystyle \dfrac{3(a - b)}{b} + \dfrac{3(a - c)}{c} = 2.\)
And you have yet to eliminate fractions. There is another way.
\(\displaystyle \dfrac{\dfrac{a - b}{b}}{0.5} + \dfrac{\dfrac{a - c}{c}}{0.5} + \dfrac{a - b}{b} + \dfrac{a - c}{c} = 2 \implies\)
\(\displaystyle \dfrac{\dfrac{a - b}{b}}{\dfrac{1}{2}}+ \dfrac{\dfrac{a - c}{c}}{\dfrac{1}{2}} + \dfrac{a - b}{b} + \dfrac{a - c}{c} = 2 \implies\)
\(\displaystyle \dfrac{a - b}{b} * \dfrac{2}{1} + \dfrac{a - c}{c} * \dfrac{2}{1} + \dfrac{a - b}{b} + \dfrac{a - c}{c} = 2 \implies\)
\(\displaystyle \dfrac{2(a - b)}{b} + \dfrac{2(a - c)}{c} + \dfrac{(a - b)}{b} + \dfrac{(a - c)}{c} = 2 \implies\)
\(\displaystyle \dfrac{3(a - b)}{b} + \dfrac{3(a - c)}{c} = 2.\)
A different way to get to the same place: simplify fractions before eliminating them. Now eliminate fractions.
\(\displaystyle \dfrac{3(a - b)}{b} + \dfrac{3(a - c)}{c} = 2 \implies\)
\(\displaystyle bc * \left ( \dfrac{3(a - b)}{b} + \dfrac{3(a - c)}{c} \right ) = bc * 2 \implies\)
\(\displaystyle 3c(a - b) + 3b(a - c) = 2bc \implies 3ac - 3bc + 3ab - 3bc = 2bc \implies\)
\(\displaystyle a\{3(b + c)\} = 2bc + 3bc + 3bc = 8bc \implies a= \dfrac{8bc}{3(b + c)}.\)