What is an Expression for b in Terms of a?

  • Thread starter LLS
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In summary, what you need to do is take the derivative of both sides with respect to f, and then use the chain rule.
  • #1
LLS
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Did I do this one correctly?

Homework Statement



If a = f - 2 and b = 4f + 5, what is an expression for b in terms of a?

It's multiple choice:

4a
4a - 4
4a + 2
4a + 7
4a + 13

The Attempt at a Solution



a = f - 2 and b = 4f + 5

add 2 to each side ----> a = f and b = 4f + 7

my answer is 4a + 7
 
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  • #2
You can add 2 to both sides of one equation, but you cannot just add two to one side of two equations. For example, what would happen if a = f, and b = 3f? When f = 1, we have a = 1, b = 3. Add two to both sides. a = f + 2 and b = 3f + 2. Now, to get the same answer of a = 1, we plug in -1 for f. This time, b = -1 instead of 3.

In this problem, what is the only thing that connects the two equations, i.e. what is the only thing to appear in both equations? How could you use that fact?
 
  • #3
I am having trouble posting a response-getting an error message. I can't quote the above post.

"F" is common to both sides. I'm not sure what to do if that is the correct path.
 
  • #4
Yes, you're right. What is f equal to? How can you get rid of the f in the second equation?
 
  • #5
LLS said:
If a = f - 2 and b = 4f + 5, what is an expression for b in terms of a?

Hi LLS! :smile:

It's a sort of chain-rule:

To write b in terms of a, you must write b in terms of f, and write f in terms of a.
So what is f in terms of a? :smile:
 
  • #6
I need to divide by f on both sides.

a = -2 and b = 9

Would that make f = 0? And then b = 5?

Oy I'm confused.

I'll be out for the night at a family dinner in a bit. Please don't be upset if I don't get back to to the problem until tomorrow. I do appreciate the help.
 
  • #7
No problem. Whenever you do anything to one side of an equation, you need to do it to the other side. In your case, dividing by f, what you end up with is

[tex]\frac{a}{f} = -2\hspace{0.35in}\text{ and }\hspace{0.35in}\frac{b}{f} = 4 + \frac{5}{f}\hspace{0.35in}\text{ if } f \neq 0[/tex]​

You cannot just divide one side of an equation by something. Always do it to both sides of the equal sign.

Read Tim's comment again. Why would finding f in terms of a help if you know b in terms of f?
 
  • #8
? Why in the world would you "need to divide by f"?

You know that a = f - 2 . Okay, so solve for f "in terms of a": f= ?

Now just replace f by that in b= 4f + 5.
 
  • #9
b = 4(f - 2) + 5

b = 4f - 8 + 5

b = 4f - 13

Thank you
 
  • #10
LLS said:
b = 4(f - 2) + 5

Nooo … b = 4f + 5, and you must write f in terms of a.

So f = … ? :smile:
 
  • #11
LLS said:
b = 4(f - 2) + 5

b = 4f - 8 + 5

b = 4f - 13

Thank you

Since that is not any of the given answers, I'm going to hope that was a typo!
 

What does it mean to "solve for B in terms of A"?

Solving for B in terms of A means finding an expression for the variable B in a given equation or problem, using the other variable A as a reference point.

Why is it important to solve for B in terms of A?

Solving for B in terms of A allows us to understand the relationship between the two variables and how they affect each other. It also allows us to manipulate the equation to solve for a specific value or to graph the relationship between the variables.

What are the steps to solve for B in terms of A?

The steps to solve for B in terms of A may vary depending on the specific equation or problem, but in general, you would isolate the variable B on one side of the equation by using inverse operations to get rid of any constants or coefficients. Then, you would substitute any known values for A and solve for B.

What are some common techniques used to solve for B in terms of A?

Some common techniques used to solve for B in terms of A include using the distributive property, combining like terms, and using properties of exponents. Other techniques may include factoring, substitution, or setting up and solving a system of equations.

Can you give an example of solving for B in terms of A?

Sure, let's say we have the equation 2B + 5 = 3A. To solve for B in terms of A, we would first subtract 5 from both sides to isolate the variable B. This gives us 2B = 3A - 5. Then, we would divide both sides by 2 to get B = (3A - 5)/2. Therefore, we have solved for B in terms of A, expressing B as a function of A.

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