You are here:

Algebra/Reworking an equation

Advertisement


Question
Hi Scott

I'm trying to rework the following equation to get A=. can you help?

X = ((((A+1.99)/1.175)-(((A*15/100)+0.25))-(B+C))

Many thanks
Kevin

Answer
X = ((((A+1.99)/1.175)-(((A*15/100)+0.25))-(B+C))

This is the same as X = (A+1.99)/1.175 - A*0.15 + 0.25 - B - C.

Breaking down the 1st term gives X = A/1.75 + 1.99/1.175 - A*0.15 + 0.25 - B - C.

Note that 1.75 is 7/4, so 1/1.75 is 4/7.
On 1.99/1.175, 1.99 is the same as 199/100 and 1.175 is the same as 47/40,
so 1.99/1.175 comes out to (199/100)(40/47) = 398/235.
For the next term, note that -0.15 = -3/20.
For the last fraction, note that 0.25 = 1/4.

Rewriting it again using this gives X = 4A/7 + 398/235 - 3A/20 + 1/4 - B - C.
Now note that 4A/7 and -3A/20 need to be multiplied by 20/20 and 7/7, respectively.
This gives 80A/140 and -21A/140.
The combined A's are then 80A/140 - 21A/140 = 59A/140.
Taking the constants, 398/225 and 1/4, the 1st *4/4 is 1592/900 and
the 2nd time 225/225 is 225/900.  Adding these together gives 1817/900.
The other parts of the equation, X, -B, and -C, stay the same.

We now have X = 59A/140 + 1817/900 - B - C.
This can quickly be changed into 59A/140 = X - 1817/900 + B + C.
The only thing left to do is to mulitply each term by 140/59.
This gives A = 140(X+B+C)/59 - 12,719/2655.

I just did it quickly with no calculator, so check the multiplication and division.

Algebra

All Answers


Answers by Expert:


Ask Experts

Volunteer


Scott A Wilson

Expertise

Any algebraic question you've got, like linear, quadratic, exponential, etc.

Experience

solving story problems solving linear, parabolic, and 3rd order equations solving equations with multiple variables

Publications
documents at Boeing

Education/Credentials
MS at math OSU in mathematics at OSU BS at OSU in mathematical sciences (math, statistics, computer science)

Awards and Honors
both BS and MS degrees were given with honors

Past/Present Clients
students from all over since the 80's; over 1,000 in algebra

©2012 About.com, a part of The New York Times Company. All rights reserved.