how do i solve these?

Question

1/x = 4/3x + 1

1/R=1/R1+1/R2 solve for R1

a^2x+(a-1)=(a+1)x solve for x

 

How would I solve these? could you list the steps to solve these? Thanks!

Answer 1

1/x = 4/3x + 1

multiply by 3x to eliminate the fractions

3 = 4 + 3x

3x = 3-4 = -1

x = -1/3

UNLESS you really meant (4/3)x instead of 4/(3x)?

IF so, then 1/x = (4/3)x + 1

1/x = 4x/3 + 1

multiply by 3x to eliminate the fractions

3 = 4x^2 +3x

4x^2 +3x -3 = 0

x= -3/8 + or - (1/8)sqr(9+48) = (-3 + or - sqr57)/8 = about 0.569 or - 1.319

x= about 0.569 or - 1.319

1/R = 1/R1 + 1/R2

multiply by RR1R2 to eliminate the fractions

R!R2 = RR2 + R1R

R1R2 - R1R = RR2

R1(R2+R) = RR2

R1 = RR2/(R2 +R)

Answer 2

Mike:

 

(1) You have to understand something clearly. This is elementary stuff.  If you slur over it it will haunt you forever. Write your equations carefully, noting that 4/3x IS NOT THE SAME AS (4/3)x.  See below.

 

Treat 4/3x as 4/(3x), Solve

 

1/x = 4/3x + 1 = 4/(3x) +1

 

Sometimes it easier to multiply both sides by the x in the denominator. You get

 

1 = 4/3 + 1/x

 

1 - 4/3 = 1 / x

 

(3-4)/3 = 1/x

 

(-1)/3 = 1/x

 

x= - 3.

 

On the other  hand IF YOU TAKE 4/3x as (4/3)x, then

 

1/x = (4/3)x + 1

 

1 = (4/3) x² + x

 

3 = 4x² + 3x

 

4x² - 3x + 3 = 0

 

This is a quadratic equation in x.  It has two roots, given by x = {3 (+ or -) sqrt[9 - 4(4)(3)]}/8.   Perhaps at this time this is beyond what you are learning.  Nonetheless, it is important enough to realize the difference between 4/(3x) and (4/3)x.

 

(2) 1/R=1/R1+1/R2 solve for R1.

 

One way to get rid of those pesky reciprocals is to find the LCM (least common multiple) of them (here RxR1XR2) and multiply throughout:  You get

 

(R1)(R2) = (R)(R2) + (R)(R1)

 

Since you want to solve for R1, collect on one side all the terms in R!:

 

(R!) (R2 - R) = (R)(R2).

 

So

 

R1 = [(R)(R2)]/(R2 - R1).  Done.

 

(3) a^2x+(a-1)=(a+1)x solve for x

Once again, Mike, you have a BIG HEADACHE!  Look at a^2x.  Tell me whether you mean by it, (a^2)x  OR a^(2x).  Your answer is pretty well going to depend on what you mean.

 

(a) Take a^2x as (a^2) x.

 

a²x + (a-1) = (a+1) x

 

a² x+ a - 1 = ax + x

 

x(a² - a-1) = 1 - a

 

x = (1- a)/(a² - a - 1)  Done.

 

(b) Take a^2x as a^(2x)

 

a^2x  + (a-1) = (a+1)x

 

This equation is definitely beyond you scope as of now.  But you should STILL know how to write a^2x correctly.

 

Mike, please, learn carefully, what the symbols mean and how to write them correctly.  Best.

 

Dattaprabhakar (Dr. G.)

 

Answer 3

1/R = 1/R1 + 1/R2 (Rule for 2 parallel resistors)

Subtract 1/R2 from both sides of the equation.

1/R – 1/R2 = 1/R1

Combine fractions on the left using the common denominator method.

(R2 - R)/(R x R2)

Therefore

(R2 - R)/(R x R2) = 1/R1

Reciprocate both sides of the equation.

(R x R2)/(R2 – R) = R1

 

1/X = 4/3X + 1

1/X - 4/3X = 1

3/3X - 4/3X = 1

- 1/3X = 1

multiplay both sides by X

- 1/3 = X