How to Solve for r and c in Two Equations with Known Quantity w

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In summary, the conversation discusses two equations involving r, c, and a known quantity w. The first equation can be solved easily by noticing the left hand side is just the left hand side of the second equation times r. The second equation can be rewritten using a new variable, y, and solved using the quadratic formula. The solutions for r and c can then be found using the value of x, which is equal to wc.
  • #1
zekester
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i have two equations r^2/wc
divided by r^2 + (1/(wc)^2)
equals 13026

and r/wc
divided by r^2 + (1/(wc)^2)
equals 951

where w is a known quantity. how do I solve for r and c.
 
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  • #2
You want to solve
[tex]\frac{\frac{r^2}{wc}}{r^2+ \frac{1}{w^2c^2}}= 13026[/tex]
and
[tex]\frac{\frac{r}{wc}}{r^2+ \frac{1}{w^2c^2}}= 13026[/tex]
for r and c given that w is a known constant.

The first thing I would do is notice that the left hand side of the first equation is just the left hand side of the second equation times r:
[tex]\frac{\frac{r^2}{wc}}{r^2+ \frac{1}{w^2c^2}}= r\frac{\frac{r}{wc}}{r^2+ \frac{1}{w^2c^2}}= r(951)= 13026[/tex]
so r= 13026/951= 13.7 (approximately). That was easy!

If we let x= wc then the second equation can be written
[tex]\frac{\frac{r}{x}}{r^2+\frac{1}{x^2}}= 951.[/tex]
Multiply both numerator and denominator of the fraction by x2 to get
[tex]\frac{rx}{r^2x^2+ 1}= 951[/tex]
Now, I would be inclined to let y= rx so that
[tex]\frac{y}{y^2+ 1}= 951[/tex]
Multiplying on both sides by y2+1 gives the quadratic equation
951y2- y- 951= 0. That can be solved by the quadratic formula:
the two solutions are approximately 0.0329 and -0.0319. Since I don't know where this problem came from I don't know if both of these are valid or not.
Now we have rx= 0.0329 and rx= -0.0319 and we know that r= 13.7 so
x= 0.0329/13.7= 0.0024 and x= -0.0319/13.7= -0.0023.
If x= wc= 0.0024, c= 0.0024/w.
If x= wc= -0.0023, c= -0.0023/w.
 
  • #3
thanks for the help it is appreciated
 

Related to How to Solve for r and c in Two Equations with Known Quantity w

1. What is the purpose of solving for a variable?

Solving for a variable allows us to find the unknown value in an equation or problem. This is useful in many fields of science, such as physics, chemistry, and engineering, where we need to determine the value of a certain variable in order to make predictions or solve problems.

2. How do I solve for a variable?

To solve for a variable, we use algebraic techniques such as combining like terms, isolating the variable on one side of the equation, and using inverse operations to cancel out any constants or coefficients. The goal is to get the variable by itself on one side of the equation in order to determine its value.

3. Can you give an example of solving for a variable?

Sure! Let's say we have the equation 3x + 5 = 20. To solve for x, we first subtract 5 from both sides to get 3x = 15. Then, we divide both sides by 3 to get x = 5. So the value of the variable x in this equation is 5.

4. What if there are multiple variables in an equation?

If there are multiple variables in an equation, we can still solve for one variable by treating the other variables as constants. This means we treat them as known values and use the same algebraic techniques to isolate the desired variable.

5. Why is it important to check our solution when solving for a variable?

It is important to check our solution because there may be cases where our initial solution is incorrect due to an error in our calculations. By plugging our solution back into the original equation and checking that it satisfies the equation, we can ensure that our solution is accurate.

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