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Math Help - Questions on solving equations with multiple variables?

  1. #1
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    Questions on solving equations with multiple variables?

    Ok, I'm trying to prepare myself for the CLM test for college placement, and have a question in regards to a specific type of math problem. Let me show you an example of the type of problem below:

    f(x)=x^2
    g(x)=3x+2

    find (f(g(-2))

    Solution steps

    g(x)=3x+2
    find g(-2)
    replace x with -2
    g(x) becomes 3(-2)+2=-4
    g(-2)=-4

    find (f(g(-2))
    replace g(-2) with -4
    f(-4)
    replace x with g(-2)
    f(g(-2))=(g(-2))^2
    since g(-2)=-4, f(-4) is...
    (-4)^2=16

    Now, after seeing a couple examples of how to solve, I get how to do it, but not why it works the way it does.

    My problem is with step 1, if I replaced the x=-2 with any other number, the equation changes. For example, if I gave a similar example, and used two different numbers for the variable:

    8(x)=6(x)+10

    If x=5
    8(5)=6(5)+10
    40=40 is a true statement

    However, if x=8
    8(8)=6(8)+10
    64=58 is no longer a true statement

    So, I'm at a loss as to why you are allowed to replace x on both sides with -2 in the original problem, based on the fact that depending on what you plug in, g will have to be altered for it to still be true. Is that just how it works, that all numbers are indefinitive and defined by whatever variable you are able to replace with any numbers given? If someone can help clarify this for me and help make sense of why this works the way it does, I would really appreciate it.
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  2. #2
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    Re: Questions on solving equations with multiple variables?

    the "process" is called function composition ... one function "inside" another

    given your example for f(x) and g(x), the general composition is ...

    f[g(x)] = f(3x+2) = (3x+2)^2
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  3. #3
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    Re: Questions on solving equations with multiple variables?

    g(x)=3x+2 is not an equation, or at least it's a very different kind of equation from 8(x)=6(x)+10. The original expression is the definition of the function g. It defined the operation called g by specifying what it does with an arbitrary input number x. Thus, given x, the function g maps it to 3x + 2. E.g., given -2, g maps it to 3(-2) + 2 = -4.

    The important thing is that g(x) = 3x + 2 cannot be true or false. Consider this. When we have a statement all of whose terms are defined, then this statement is either true or false. E.g., "Indiana has on average 20 tornadoes a year" is either true or false. However, "By a tornado we will understand a violently rotating column of air" is neither true nor false; it's a definition of one of the terms that was previously undefined. (This phrase may only seem to be true or false if you already know some definition of "tornado.")

    For this reason, definitions are often written differently. E.g., g(x) := 3x + 2; g(x)\stackrel{\text{\tiny def}}{=}3x+2; g(x)\triangleq3x+2; g:x\mapsto3x+2 are all common notations for defining g.

    In contrast, 8(x) = 6(x) + 10 is an equation that is true for precisely one x (x = 5) and false for all other x. Note also that 8(x) is sometimes used as a notation for 8\cdot x, 8\times x or just 8x, i.e., 8 multiplied by x, whereas g(x) is not a multiplication; it's the result of applying the function g to x.
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  4. #4
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    Re: Questions on solving equations with multiple variables?

    Quote Originally Posted by Ascendant78 View Post
    My problem is with step 1, if I replaced the x=-2 with any other number, the equation changes. For example, if I gave a similar example, and used two different numbers for the variable:

    8(x)=6(x)+10

    If x=5
    8(5)=6(5)+10
    40=40 is a true statement

    However, if x=8
    8(8)=6(8)+10
    64=58 is no longer a true statement
    You understand, don't you, that this example has nothing to do with your question? Previously "g(x)" is NOT "g times x", it is g "of" x. The g is a function of x defined by "whatever x is multiply it by 3 and add 2. The equation 8x= 6x+10 is does NOT define a function- it is simply an equation- it says that x is a number such that multiplying it by 8 will give the same thing as multiplying it by 6 and then adding 10. You can solve that equation: from 8x= 6x+ 10, subtract 6x from both sides: 2x= 10 so x= 5 is the only value of x that makes that true.
    "g(x)= 3x+ 2" says that whatever x is, g(x) is defined as 3x+ 2.

    So, I'm at a loss as to why you are allowed to replace x on both sides with -2 in the original problem, based on the fact that depending on what you plug in, g will have to be altered for it to still be true.
    Yes, of course. That's what a "function" is. It may have a different value for every value of x. One definition of "function" is that it is a set of pairs of numbers, (x, f(x)). "x" can be any number in the "domain" of the function and y= f(x) depends upon x.

    Is that just how it works, that all numbers are indefinitive and defined by whatever variable you are able to replace with any numbers given? If someone can help clarify this for me and help make sense of why this works the way it does, I would really appreciate it.
    Well, I wouldn't say "that is just how it works". Rather, I would say that is the definition of the word "function" in mathematics.
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  5. #5
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    Re: Questions on solving equations with multiple variables?

    Thank you so much for clarifying this for me with all the responses. Now it makes perfect sense.

    My problem was that I was looking at it as an equation with variables to solve, rather than as a definition of a function. A part of the problem with trying to teach this stuff to yourself is that you can usually find this information online, but not always with a good explanation behind it (or for that matter, any explanation at all). Anyway, I found out this morning that a precalculus book I will be working on next has some of these types of problems in it, along with the explanations you have all helped me with. So, looks like I'll have more of them to work on. However, after I see what type of function it is, I see it is so much easier than what I originally thought it would be. Thanks again.
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