1. ## algebra

If

3b + 7s + c = 120
4b + 10s + c = 164.5
what is the value of b+s+c ?

My textbook gives 31 as the answer, I cannot understand.
thank you for any help.

2. Originally Posted by simone
If

3b + 7s + c = 120
4b + 10s + c = 164.5
what is the value of b+s+c ?

My textbook gives 31 as the answer, I cannot understand.
thank you for any help.
are you sure you shouldn't have 3 equations here?

3. thank you for your interest. that's what I'm banging my head on, there's no 3° equation. You think it's a typo?

4. Originally Posted by simone
thank you for your interest. that's what I'm banging my head on, there's no 3° equation. You think it's a typo?
ah! i got it.

subtract the first equation from the second, you get: b + 3s = 44.5 or b = 44.5 - 3s

then subtract 4 times the first from 3 times the second and you get 2s - c = 13.5 or c = 2s - 13.5

thus b + s + c = (44.5 - 3s) + s + (2s - 13.5) = 31

5. Hello, Simone!

$\begin{array}{cccc}3b + 7s + c & = & 120 & {\color{blue}[1]}\\
4b + 10s + c &= & 164.5& {\color{blue}[2]}\end{array}$

What is the value of $b+s+c$ ?

My textbook gives 31 as the answer.
There may be an elegant solution . . . but I can't find it.

Subtract [1] from [2]: . $b + 3s \:=\:44.5\quad\Rightarrow\quad s \:=\:\frac{44.5 - b}{3}\;\;{\color{blue}[3]}$

Substitute into [1]: . $3b + 7\left(\frac{44.4-b}{3}\right) + c\:=\:120\quad\Rightarrow\quad c \:=\:\frac{48.5-2b}{3}\;\;{\color{blue}[4]}$

Consider: . $b + s + c$

Substitute [3] and [4]: . $b + s + c \;=\;b + \frac{44.5-b}{3} + \frac{48.5-2b}{3} \;=\;\frac{3b + 48.5 - b + 44.5 - 2b}{3}$

. . . Therefore: . $b + s + c\;= \;\frac{93}{3}\;=\;31$

Too fast for me, Jhevon!

6. Originally Posted by Soroban
Too fast for me, Jhevon!
AND there were no rational functions in my solution

I'm just playing, we all know you're the better Mathematician here, i'm just a Nascent one