# Is this a typo in my math book?

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:

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The homework tag should never be used alone. Please add at least another one explaining the topic(s) covered by this homework – M Turgeon Sep 5 '12 at 2:46

5
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. – Dilip Sarwate Sep 5 '12 at 2:47

## 4 Answers

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: $$b = 7/3$$

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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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@Joe, appreciate the Tex work. – Emmad Kareem Sep 5 '12 at 4:09
No problem - was just minor while I was reading this problem. :) – Joe Sep 5 '12 at 4:13
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