Thread: Solving for A in terms of B & C

1. Solving for A in terms of B & C

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/t...rms-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.

(((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)

2. Originally Posted by questbest
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/t...rms-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.

(((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 YOUR MINOR ERROR LIES HERE
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 YOUR MAJOR ERROR LIES HERE
A(C+B)=3.33BC
A=(3.33BC)/(C+B)
$14 - 6 + 21 - 6 = 8 + 15 = 23.$

$14 - 6 + 21 - 6 \ne 14 + 21 = 35.$

Do you see your major error now? You were to subtract BC twice, which is the same as subtracting twice BC. Instead you cancelled what you needed to subtract.

$AC - BC + AB - BC = AC + AB - BC - BC = AC + AB - ( BC + BC) =$

$AC + AB - 2BC.$

Do you see why you were in error?

Your minor error is much harder to explain. Most people are taught in basic arithmetic to simplify fractions into their APPROXIMATE decimal equivalent right away. This is very bad training. Approximations always introduce errors. Avoid approximations if possible. If you must approximate, do so at the very end of your work because the need for an approximation may disappear by the end.

$\dfrac{4}{3} \ne 1.33 \text { because } 3 * 1.33 = 3.99 \ne 4.$

If you do not divide both sides of the equation by 3, you will end up with an answer that looks like

$A = \dfrac{4p}{3q}$ where p and q are algebraic expressions. When you substitute numbers into p, the result may be a multiple of 3, and the threes will cancel, giving you an exact answer.

3. Originally Posted by questbest
(((A-B)/B)/0.5)
In case it helps you:

Your bracketing is ok, but outer set not required:
((A-B)/B)/0.5

Also, above simplifies to 2(A - B)/B

4. Originally Posted by JeffM
$14 - 6 + 21 - 6 = 8 + 15 = 23.$

$14 - 6 + 21 - 6 \ne 14 + 21 = 35.$

Do you see your major error now? You were to subtract BC twice, which is the same as subtracting twice BC. Instead you cancelled what you needed to subtract.

$AC - BC + AB - BC = AC + AB - BC - BC = AC + AB - ( BC + BC) =$

$AC + AB - 2BC.$

Do you see why you were in error?

Your minor error is much harder to explain. Most people are taught in basic arithmetic to simplify fractions into their APPROXIMATE decimal equivalent right away. This is very bad training. Approximations always introduce errors. Avoid approximations if possible. If you must approximate, do so at the very end of your work because the need for an approximation may disappear by the end.

$\dfrac{4}{3} \ne 1.33 \text { because } 3 * 1.33 = 3.99 \ne 4.$

If you do not divide both sides of the equation by 3, you will end up with an answer that looks like

$A = \dfrac{4p}{3q}$ where p and q are algebraic expressions. When you substitute numbers into p, the result may be a multiple of 3, and the threes will cancel, giving you an exact answer.
I am confused about the major error line because after getting AC + AB - 2BC=(4/3)BC you then can add 2BC to both sides of the equation and then get AC+AB=(10/3)BC which is the same end result as before then solving for A gives the same answer

Are you talking about order of operations why 14 - 6 + 21 - 6 = 8 + 15 = 23 instead of what I am doing and getting 35? I am used to plugging numbers into a calculator and rolling from there so I guess I am not understanding... to get rid of the BC terms on the left side so that A is only left

5. Originally Posted by Denis
In case it helps you:

Your bracketing is ok, but outer set not required:
((A-B)/B)/0.5

Also, above simplifies to 2(A - B)/B
Thanks

6. Originally Posted by questbest
I am confused about the major error line because after getting AC + AB - 2BC=(4/3)BC you then can add 2BC to both sides of the equation and then get AC+AB=(10/3)BC which is the same end result as before then solving for A gives the same answer

Are you talking about order of operations why 14 - 6 + 21 - 6 = 8 + 15 = 23 instead of what I am doing and getting 35? I am used to plugging numbers into a calculator and rolling from there so I guess I am not understanding... to get rid of the BC terms on the left side so that A is only left

NO WONDER YOU ARE CONFUSED. I made an error. My original post made NO SENSE. My abject apologies. I do not have the time to correct myself right now. I shall be back in a little over an hour and try to straighten things out. I'm sorry for wasting your time.

7. Originally Posted by questbest
(((A-B)/B)/0.5)+(((A-C)/C)/0.5)+((A-B)/B)+((A-C)/C)=2
((A-B)/B)/0.5 + ((A-C)/C)/0.5 + (A-B)/B + (A-C)/C = 2

2(A-B)/B + 2(A-C)/C + (A-B)/B + (A-C)/C = 2

3(A-B)/B + 3(A-C)/C = 2

Continue...next is multiply by BC
you should end up with A = 8BC/(3(B+C))
....or A = 2.66667BC/(B+C) if you prefer!!

8. Originally Posted by questbest
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/t...rms-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.

(((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.

$2 * \left (\dfrac{\dfrac{a - b}{b}}{0.5} + \dfrac{\dfrac{a - c}{c}}{0.5} + \dfrac{a - b}{b} + \dfrac{a - c}{c} \right ) = 2 * 2 \implies$

$\dfrac{\dfrac{2(a - b)}{b}}{0.5} + \dfrac{\dfrac{2(a - c)}{c}}{0.5} + \dfrac{2(a - b)}{b} + \dfrac{2(a - c)}{c} = 4.$

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.

$0.5* \left (\dfrac{\dfrac{a - b}{b}}{0.5} + \dfrac{\dfrac{a - c}{c}}{0.5} + \dfrac{a - b}{b} + \dfrac{a - c}{c} \right ) = 0.5 * 2 \implies$

$\dfrac{a - b}{b} + \dfrac{a - c}{c} + \dfrac{0.5(a - b)}{b} + \dfrac{0.5(a - c)}{c} = 1.$

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

$\dfrac{2(a - b)}{b} + \dfrac{2(a - c)}{c} + \dfrac{(a - b)}{b} + \dfrac{(a - c)}{c} = 2 \implies$

$\dfrac{3(a - b)}{b} + \dfrac{3(a - c)}{c} = 2.$

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

$\dfrac{\dfrac{a - b}{b}}{0.5} + \dfrac{\dfrac{a - c}{c}}{0.5} + \dfrac{a - b}{b} + \dfrac{a - c}{c} = 2 \implies$

$\dfrac{\dfrac{a - b}{b}}{\dfrac{1}{2}}+ \dfrac{\dfrac{a - c}{c}}{\dfrac{1}{2}} + \dfrac{a - b}{b} + \dfrac{a - c}{c} = 2 \implies$

$\dfrac{a - b}{b} * \dfrac{2}{1} + \dfrac{a - c}{c} * \dfrac{2}{1} + \dfrac{a - b}{b} + \dfrac{a - c}{c} = 2 \implies$

$\dfrac{2(a - b)}{b} + \dfrac{2(a - c)}{c} + \dfrac{(a - b)}{b} + \dfrac{(a - c)}{c} = 2 \implies$

$\dfrac{3(a - b)}{b} + \dfrac{3(a - c)}{c} = 2.$

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

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

$bc * \left ( \dfrac{3(a - b)}{b} + \dfrac{3(a - c)}{c} \right ) = bc * 2 \implies$

$3c(a - b) + 3b(a - c) = 2bc \implies 3ac - 3bc + 3ab - 3bc = 2bc \implies$

$a\{3(b + c)\} = 2bc + 3bc + 3bc = 8bc \implies a= \dfrac{8bc}{3(b + c)}.$

9. Well, I'm one who reduces to the extreme!

u = (a-b)/b, v = (a-c)/c

Initial equation becomes:
u/.5 + v/.5 + u + v = 2
which is same as:
2u + 2v + u + v = 2
3u + 3v = 2
3(u + v) = 2
u + v = 2/3

Now substitute back in and you're almost finished....

10. Originally Posted by Denis
Well, I'm one who reduces to the extreme!

u = (a-b)/b, v = (a-c)/c

Initial equation becomes:
u/.5 + v/.5 + u + v = 2
which is same as:
2u + 2v + u + v = 2
3u + 3v = 2
3(u + v) = 2
u + v = 2/3

Now substitute back in and you're almost finished....
Lovely

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