Systems of equations problem in algebra A dog weighs 1/8 of a cow. Their combined total is 360kg, how much does the dog weigh and how much does the cow weigh? 
I got this question on my algebra test, my teacher said the answer was along the lines of 
$$x + 8x =9x$$
$$ \implies 360/9 = 40$$
$$360 - 40$$
$$=320$$
$$\text{The cow weighs 320 and the dog weighs 40}$$
My dad and I said that it would be $360/8 = 45$ so the dog is $45$ and the cow is $315 $
So who is correct? Do it for yourselves to make sure please. 
 A: Your dad's solution would be the correct one if the dog's weight were $1/8$ of the combined weight.  In that case, we would have the dog's weight as $360/8=45$, and then the cow's weight would be the difference, $360-45=315$.
But the problem doesn't say that the dog's weight is $1/8$ of the combined weight of $360$; it says that the dog's weight is $1/8$ of the cow's weight, which when combined with the dog's weight is $360$.  Not the same thing.  Notice that $45$ is $1/7$ of $315$, not $1/8$ of it.  In this case, the teacher's solution is correct.
A: Just in case this is a homework problem, I will solve a VERY similar problem. 
Suppose a dog weighs $1/3$ of a statue. The combined weight of the dog and the statue is $140$. Determine the weight of the cow and of the statue.
Let $x$ denote the weight of the dog. Let $y$ denote the weight of the statue. We know that the dog weighs $1/3$ of the weight of the statue, so $x=\frac{y}{3}$. Now the combined weight of the two is $140$, so $x+y=140$. Note that  $x=\frac{y}{3}$ implies $3x=y$. 
Substing $y=3x$ into $x+y=140$, we see that $4x=140$. Dividing by $4$ on both sides shows that $x=\frac{140}{4}=\frac{70}{2}=35$. Now putting $x=35$ into $x+y=140$, we see that $35+y=140$, and so $y=140-35=105$.
A: OK, I will answer a general question...hopefully to help you write down the necessary equations:
Thing A weighs $\frac{\diamondsuit}{\clubsuit}$ of thing B.  Thing A and thing B together weigh $\frac{\spadesuit}{\heartsuit}$.  Let's assume that thing A weighs $x$ and thing B weighs $y$.  The first statement tells me that:
$$
x = \frac{\diamondsuit}{\clubsuit}y
$$
The second statement tells me that, added together, they weigh $\frac{\spadesuit}{\heartsuit}$:
$$
x + y = \frac{\spadesuit}{\heartsuit}
$$
You can solve this through substitution: $x = \frac{\diamondsuit}{\clubsuit}y$ so plug that into the second equation to find $y$:
\begin{align}
\frac{\diamondsuit}{\clubsuit}y + y = \frac{\spadesuit}{\heartsuit} \\
y * \left(\frac{\diamondsuit}{\clubsuit} + 1\right) = \frac{\spadesuit}{\heartsuit} \\
y = \frac{\frac{\spadesuit}{\heartsuit}}{1 + \frac{\diamondsuit}{\clubsuit}} \\
y = \frac{\spadesuit \clubsuit}{\heartsuit\left(\clubsuit + \diamondsuit\right)}
\end{align}
Now that you have $y$ you can plug into the original equation to find that:
$$
x = \frac{\diamondsuit}{\clubsuit}y = \frac{\diamondsuit}{\clubsuit}*\frac{\spadesuit \clubsuit}{\heartsuit\left(\clubsuit + \diamondsuit\right)} = \frac{\diamondsuit \spadesuit}{\heartsuit\left(\clubsuit + \diamondsuit\right)}
$$
I used the symbolds $\diamondsuit$, $\clubsuit$, $\spadesuit$, and $\heartsuit$ to hopefully illustrate that variables are symbols--and you can use any symbols you like.
