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I am doing homework which is submitted online. I came across a question asking if two functions are equal. $f(x)=3x+4$ and $g(x)=14+(8/x)+b(x-4)$. I set the two equations together and got 7 for an answer but the book says the answer is $\frac {7}{3}$.

Here is an image of the solution in the book:

  • 6
    Perhaps you should edit your query here to state the **question** that was asked clearly. The two functions ae _not_ equal as functions. The solution seems to suggest that what was asked is something like "Find all values of $b$ suc that $f(-3) = g(-3)$ and $f(4) = g(4)$", that is the functions have equal value for two specific values of $x$. This is quite different from saying that the functions are equal.2012-09-05

4 Answers 4

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If you set $f(-3) = g(-3)$, you end up with:

$-5 = 14 -(8/3) -7b$

If you multiply by 3 to remove fractions, you are left with

$-15 = 42 - 8 -21b$

Collecting like terms, we are left with

$21b = 49$

reducing yields: $b = 7/3$

Can you find your error?

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The book is correct. Setting $f(-3)=g(-3)$ gives us

$-5=14-\frac{8}{3}-7b \Rightarrow -19=-\frac{8}{3}-7b \Rightarrow 49=21b \Rightarrow b=\frac{7}{3}$.

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Well, at least to your specific question "Is this a typo?", the answer is "No." You are making an arithmetic error somewhere, because: $\frac{1}{7}\bigg(\frac{34}{3} + \frac{15}{3}\bigg) = \frac{1}{7}\frac{49}{3} = \frac{7}{3}.$

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$ -5 = \left(\frac{34}{3}\right) - 7B$

Multiply both sides by 3:

$-15= 3\cdot\left(\frac{34}{3}\right) - 3\cdot(7B)$

that is:

$-15 = 34 - 21B$

so

$-15-34 = -21B$

$B=\frac{49}{21}=\frac{7}{3}$

The book is correct on this one.

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    No problem - was just mi$n$or while I was reading this proble$m$. :)2012-09-05